sample problem solving in speed

Physics Problems with Solutions

Velocity and speed: solutions to problems.

Solutions to the problems on velocity and speed of moving objects. More tutorials can be found in this website.

A man walks 7 km in 2 hours and 2 km in 1 hour in the same direction. a) What is the man's average speed for the whole journey? b) What is the man's average velocity for the whole journey? Solution to Problem 1: a)

You start walking from a point on a circular field of radius 0.5 km and 1 hour later you are at the same point. a) What is your average speed for the whole journey? b) What is your average velocity for the whole journey? Solution to Problem 3: a) If you walk around a circular field and come back to the same point, you have covered a distance equal to the circumference of the circle.

John drove South 120 km at 60 km/h and then East 150 km at 50 km/h. Determine a) the average speed for the whole journey? b) the magnitude of the average velocity for the whole journey? Solution to Problem 4: a)

If I can walk at an average speed of 5 km/h, how many miles I can walk in two hours? Solution to Problem 5: distance = (average speed) * (time) = 5 km/h * 2 hours = 10 km using the rate of conversion 0.62 miles per km, the distance in miles is given by distance = 10 km * 0.62 miles/km = 6.2 miles

A train travels along a straight line at a constant speed of 60 mi/h for a distance d and then another distance equal to 2d in the same direction at a constant speed of 80 mi/h. a)What is the average speed of the train for the whole journey?

A car travels 22 km south, 12 km west, and 14 km north in half an hour. a) What is the average speed of the car? b) What is the final displacement of the car? c) What is the average velocity of the car?

Solution to Problem 7: a)

Solution to Problem 8: a)

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Problems on Calculating Speed | Speed Questions and Answers

Solve different types of problems on calculating speed and get acquainted with various models of questions asked in your exams. Be aware of the Formula to Calculate and Relationship between Speed Time and Distance. Practice Speed Problems on a regular basis so that you can be confident while attempting the exams. We even provided solutions for all the Questions provided and explained everything in detail for better understanding. Try to solve the Speed Questions on your own and then cross-check where you are lagging.

We know the Speed of the Object is nothing but the distance traveled by the object in unit time.

Formula to find out Speed is given by Speed = Distance/Time

Word Problems on Calculating Speed

1. A man walks 25 km in 6 hours. Find the speed of the man?

Solution: Distance traveled = 25 km Time taken to travel = 6 hours Speed of Man = Distance traveled/Time taken = 25km/6hr = 4.16 km/hr Therefore, a man travels at a speed of 4.16 km/hr

2. A car covers a distance of 420 m in 1 minute whereas a train covers 70 km in 30 minutes. Find the ratio of their speeds?

Solution: Speed of the Car = Distance Traveled/Time Taken = 420m/60 sec = 7 m/sec

Speed of the Train = Distance Traveled/Time Taken = 70 km/1/2 hr = 140 km/hr

To convert it into m/sec multiply with 5/18 = 140*5/18 = 38.8 m/sec = 39 m/sec (Approx) Ratio of Speeds = 7:39

3. A car moves from A to B at a speed of 70 km/hr and comes back from B to A at a speed of 40 km/hr. Find its average speed during the journey?

Solution: Since the distance traveled is the same the Average Speed= (x+y)/2 where x, y are two different speeds Substitute the Speeds in the given formula Average Speed = (70+40)/2 = 110/2 = 55 km/hr The Average Speed of the Car is 55 km/hr

4. A bus covers a certain distance in 45 minutes if it runs at a speed of 50 km/hr. What must be the speed of the bus in order to reduce the time of journey by 20 minutes?

Solution: Speed = Distance/Time 50 = x/3/4 50 = 4x/3 4x = 150 x = 150/4 = 37.5 km

Now by applying the same formula we can find the speed

Now, time = 40 mins or 0.66 hr since the journey is reduced by 20 mins

S = Distance/Time = 37.5/0.66 = 56.81 km/hr

5. Ram traveled 200 km in 3 hours by train and then traveled 140 km in 3 hours by car and 5 km in 1/2 hour by cycle. What is the average speed during the whole journey?

Solution: Distance traveled by Train is 200 km in 3 hours Distance Traveled by Car is 140 km in 3 hours Distance Traveled by Cycle is 5 km in 1/2 hour Average Speed = Total Distance/Total Time = (200+140+5)/(3+3+1/2) = 345/6 1/2 = 345/(13/2)” = 345*2/13 = 53.07 km/hr

6. A train covers 150 km in 3 hours. Find its speed?

Solution: Speed = Distance/Time = 150 km/3 hr = 50 km/hr Therefore, Speed of the Train is 50 km/hr.

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Chapter 8: Rational Expressions

8.8 Rate Word Problems: Speed, Distance and Time

Distance, rate and time problems are a standard application of linear equations. When solving these problems, use the relationship rate (speed or velocity) times time equals distance .

[latex]r\cdot t=d[/latex]

For example, suppose a person were to travel 30 km/h for 4 h. To find the total distance, multiply rate times time or (30km/h)(4h) = 120 km.

The problems to be solved here will have a few more steps than described above. So to keep the information in the problem organized, use a table. An example of the basic structure of the table is below:

The third column, distance, will always be filled in by multiplying the rate and time columns together. If given a total distance of both persons or trips, put this information in the distance column. Now use this table to set up and solve the following examples.

Example 8.8.1

Joey and Natasha start from the same point and walk in opposite directions. Joey walks 2 km/h faster than Natasha. After 3 hours, they are 30 kilometres apart. How fast did each walk?

The distance travelled by both is 30 km. Therefore, the equation to be solved is:

[latex]\begin{array}{rrrrrrl} 3r&+&3(r&+&2)&=&30 \\ 3r&+&3r&+&6&=&30 \\ &&&-&6&&-6 \\ \hline &&&&\dfrac{6r}{6}&=&\dfrac{24}{6} \\ \\ &&&&r&=&4 \text{ km/h} \end{array}[/latex]

This means that Natasha walks at 4 km/h and Joey walks at 6 km/h.

Example 8.8.2

Nick and Chloe left their campsite by canoe and paddled downstream at an average speed of 12 km/h. They turned around and paddled back upstream at an average rate of 4 km/h. The total trip took 1 hour. After how much time did the campers turn around downstream?

The distance travelled downstream is the same distance that they travelled upstream. Therefore, the equation to be solved is:

[latex]\begin{array}{rrlll} 12(t)&=&4(1&-&t) \\ 12t&=&4&-&4t \\ +4t&&&+&4t \\ \hline \dfrac{16t}{16}&=&\dfrac{4}{16}&& \\ \\ t&=&0.25&& \end{array}[/latex]

This means the campers paddled downstream for 0.25 h and spent 0.75 h paddling back.

Example 8.8.3

Terry leaves his house riding a bike at 20 km/h. Sally leaves 6 h later on a scooter to catch up with him travelling at 80 km/h. How long will it take her to catch up with him?

The distance travelled by both is the same. Therefore, the equation to be solved is:

[latex]\begin{array}{rrrrr} 20(t)&=&80(t&-&6) \\ 20t&=&80t&-&480 \\ -80t&&-80t&& \\ \hline \dfrac{-60t}{-60}&=&\dfrac{-480}{-60}&& \\ \\ t&=&8&& \end{array}[/latex]

This means that Terry travels for 8 h and Sally only needs 2 h to catch up to him.

Example 8.8.4

On a 130-kilometre trip, a car travelled at an average speed of 55 km/h and then reduced its speed to 40 km/h for the remainder of the trip. The trip took 2.5 h. For how long did the car travel 40 km/h?

[latex]\begin{array}{rrrrrrr} 55(t)&+&40(2.5&-&t)&=&130 \\ 55t&+&100&-&40t&=&130 \\ &-&100&&&&-100 \\ \hline &&&&\dfrac{15t}{15}&=&\dfrac{30}{15} \\ \\ &&&&t&=&2 \end{array}[/latex]

This means that the time spent travelling at 40 km/h was 0.5 h.

Distance, time and rate problems have a few variations that mix the unknowns between distance, rate and time. They generally involve solving a problem that uses the combined distance travelled to equal some distance or a problem in which the distances travelled by both parties is the same. These distance, rate and time problems will be revisited later on in this textbook where quadratic solutions are required to solve them.

For Questions 1 to 8, find the equations needed to solve the problems. Do not solve.

A is 60 kilometres from B. An automobile at A starts for B at the rate of 20 km/h at the same time that an automobile at B starts for A at the rate of 25 km/h. How long will it be before the automobiles meet?
Two automobiles are 276 kilometres apart and start to travel toward each other at the same time. They travel at rates differing by 5 km/h. If they meet after 6 h, find the rate of each.
Two trains starting at the same station head in opposite directions. They travel at the rates of 25 and 40 km/h, respectively. If they start at the same time, how soon will they be 195 kilometres apart?
Two bike messengers, Jerry and Susan, ride in opposite directions. If Jerry rides at the rate of 20 km/h, at what rate must Susan ride if they are 150 kilometres apart in 5 hours?
A passenger and a freight train start toward each other at the same time from two points 300 kilometres apart. If the rate of the passenger train exceeds the rate of the freight train by 15 km/h, and they meet after 4 hours, what must the rate of each be?
Two automobiles started travelling in opposite directions at the same time from the same point. Their rates were 25 and 35 km/h, respectively. After how many hours were they 180 kilometres apart?
A man having ten hours at his disposal made an excursion by bike, riding out at the rate of 10 km/h and returning on foot at the rate of 3 km/h. Find the distance he rode.
A man walks at the rate of 4 km/h. How far can he walk into the country and ride back on a trolley that travels at the rate of 20 km/h, if he must be back home 3 hours from the time he started?

Solve Questions 9 to 22.

A boy rides away from home in an automobile at the rate of 28 km/h and walks back at the rate of 4 km/h. The round trip requires 2 hours. How far does he ride?
A motorboat leaves a harbour and travels at an average speed of 15 km/h toward an island. The average speed on the return trip was 10 km/h. How far was the island from the harbour if the trip took a total of 5 hours?
A family drove to a resort at an average speed of 30 km/h and later returned over the same road at an average speed of 50 km/h. Find the distance to the resort if the total driving time was 8 hours.
As part of his flight training, a student pilot was required to fly to an airport and then return. The average speed to the airport was 90 km/h, and the average speed returning was 120 km/h. Find the distance between the two airports if the total flying time was 7 hours.
Sam starts travelling at 4 km/h from a campsite 2 hours ahead of Sue, who travels 6 km/h in the same direction. How many hours will it take for Sue to catch up to Sam?
A man travels 5 km/h. After travelling for 6 hours, another man starts at the same place as the first man did, following at the rate of 8 km/h. When will the second man overtake the first?
A motorboat leaves a harbour and travels at an average speed of 8 km/h toward a small island. Two hours later, a cabin cruiser leaves the same harbour and travels at an average speed of 16 km/h toward the same island. How many hours after the cabin cruiser leaves will it be alongside the motorboat?
A long distance runner started on a course, running at an average speed of 6 km/h. One hour later, a second runner began the same course at an average speed of 8 km/h. How long after the second runner started will they overtake the first runner?
Two men are travelling in opposite directions at the rate of 20 and 30 km/h at the same time and from the same place. In how many hours will they be 300 kilometres apart?
Two trains start at the same time from the same place and travel in opposite directions. If the rate of one is 6 km/h more than the rate of the other and they are 168 kilometres apart at the end of 4 hours, what is the rate of each?
Two cyclists start from the same point and ride in opposite directions. One cyclist rides twice as fast as the other. In three hours, they are 72 kilometres apart. Find the rate of each cyclist.
Two small planes start from the same point and fly in opposite directions. The first plane is flying 25 km/h slower than the second plane. In two hours, the planes are 430 kilometres apart. Find the rate of each plane.
On a 130-kilometre trip, a car travelled at an average speed of 55 km/h and then reduced its speed to 40 km/h for the remainder of the trip. The trip took a total of 2.5 hours. For how long did the car travel at 40 km/h?
Running at an average rate of 8 m/s, a sprinter ran to the end of a track and then jogged back to the starting point at an average of 3 m/s. The sprinter took 55 s to run to the end of the track and jog back. Find the length of the track.

Answer Key 8.8

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Speed, Velocity, and Acceleration Problems

by Abnurlion
December 26, 2022 November 2, 2023

What is Speed?

Definition of Speed : Speed is the rate of change of distance with time . Speed is different from velocity because it’s not in a specified direction. In this article, you will learn how to solve speed, velocity, and acceleration problems.

Additionally, you need to know that speed is a scalar quantity and we can write its symbol as S. The formula for calculating the speed of an object is:

Speed, S = Distance (d) / Time (t)

Thus, s = d/t

Note: In most cases, we also use S as a symbol for distance.

The S.I unit for speed is meter per second (m/s) or ms -1

Non-Uniform or Average Speed : This is a non-steady distance covered by an object at a particular period of time. We can also define non-uniform speed as the type of distance that an object covered at an equal interval of time.

The formula for calculating non-uniform speed is

Average speed = Total distance covered by the object / Total time taken

Actual speed: This is also known as the instantaneous speed of an object which is the distance covered by an object over a short interval of time.

What is Velocity?

Definition of Velocity: Velocity is the rate of displacement with time. Velocity is the speed of an object in a specified direction. The unit of velocity is the same as that of speed which is meter per second (ms -1 ). We use V as a symbol for velocity.

Note: We often use U to indicate initial speed, and V to indicate final speed.

The formula for calculating Velocity (V) = displacement (S) / time (t)

The difference between velocity and speed is the presence of displacement and distance respectively. Because displacement is a measure of separation in a specified direction, while distance is not in a specified direction. Velocity is a vector quantity.

Uniform Velocity

Definition of uniform velocity: The rate of change of displacement is constant no matter how small the time interval may be. Also, uniform velocity is the distance covered by an object in a specified direction in an equal time interval.

The formula for uniform velocity = Total displacement / Total time taken

What is Acceleration?

Definition of Acceleration: Acceleration is the rate of change of velocity with time. Acceleration is measured in meters per second square (ms -2 ). The symbol for acceleration is a. Acceleration is also a vector quantity.

The formula for acceleration , a = change in velocity (v)/time taken (t)

Thus, a = v/t

We can also write acceleration as

a = change in velocity/time = ΔV/Δt = (v – u)/t

[where v = final velocity, u = initial velocity, and t = time taken]

Uniform Acceleration

In the case of uniform acceleration , the rate of change of velocity with time is constant.

Average velocity of the object = (Initial velocity + final velocity)/2

Average velocity = (v + u)/2

What is Retardation?

Retardation is the decreasing rate of change in velocity moved, covered, or traveled by an object.

The formula for calculating retardation is

Retardation = Change in a decrease in velocity/time taken

Equations of Motion

You can also apply the following equations of motion to calculate the speed, velocity, and acceleration of the body:

s = [ (v + u)/2 ]t
v 2 = u 2 + 2as

s = ut + (1/2)at 2

Where v = final velocity, u = initial velocity, t = time, a = acceleration, and s = distance

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Solved Problems of Speed, Velocity, and Acceleration

Here are solved problems to help you understand how to calculate speed, velocity, and acceleration:

A train moves at a speed of 54km/h for a one-quarter minute. Find the distance travelled by train.

Speed, Velocity, and Acceleration Problems

From the question above

Speed = 54 km/h = (54 x 1000)/(60 x 60) = 54,000/3,600 = 15 m/s

Time = one quarter minute = 1/4 minute = (1/4) x 60 = 15 seconds

Since we have

speed = distance/time

After cross-multiplication, we will now have

Distance = speed x time

We can now insert our data into the above expression

Distance = 15 m/s x 15 s = 225 m

Therefore, the distance travelled by train is 225 meters.

A car travelled a distance of 5km in 50 seconds. Find the speed in meters per second.

Distance = 5km = 5 x 1000m = 5,000m

Time = 50 seconds

and the formula for speed

speed = distance/time = 5000/50 = 100m/s

A motorcycle starting from rest moves with a uniform acceleration until it attains a speed of 108 kilometres per hour after 15 seconds. Find its acceleration.

Initial velocity, u = 0 (because the motorcycle starts from rest)

Final velocity, v = speed = 108 km/h = (108 x 1000m) / (60 x 60s) = 108,000/3,600 = 30m/s

Time taken, t = 15 seconds

Therefore, we can now apply the formula that says

Acceleration = change in velocity/ time = (v-u)/t = (30-0)/15 =30/15 = 2ms -2

Therefore, the acceleration of the motorcycle is 2ms -2

A bus covers 50 kilometres in 1 hour. What is it is the average speed?

Total distance covered = 50 km = 50 x 1000m = 50,000m

Time taken = 1 hour = 1 x 60 x 60s =3,600s

Therefore, we can now calculate the average speed of the bus by substituting our data into the above formula

Average speed = 50,000/3,600 = 13.9 m/s

Therefore, the average speed of the bus is 13.9m/s or approximately 14 meters per second.

A car travels 80 km in 1 hour and then another 20 km in the next hour. Find the average velocity of the car.

Initial displacement of the car = 80km

Final displacement of the car = 20km

The total displacement of the car = initial displacement of the car + final displacement of the car

The total displacement of the car = 80km + 20km = 100km

The time for 80km is 1hr

And the time for 20km is 1hr

Total time taken = The time for 80km (1hr) + The time for 20km (1hr)

Total time taken = 1hr + 1hr = 2hrs

Now, to calculate the average velocity of the car, we apply the formula that says

Average velocity = total displacement/total time taken = 100km/2hrs = 50km/h

We can further convert the above answer into meters per second

Average velocity = 50km/h = (50 x 1000m)/(1 x 60 x 60s) = 50,000/3,600 = 13.9m/s or 14ms -1

Therefore, the average velocity of the car is 14ms -1

How to Conduct Physics Practical

and How to Calculate Velocity Ratio of an Inclined Plane

A body falls from the top of a tower 100 meters high and hits the ground in 5 seconds. Find its acceleration.

Displacement = 100m

Time = 5 seconds

and we can apply the formula for acceleration that says

acceleration, a = Displacement/time 2 = 100/5 2 = 100/25 = 4ms -2

Therefore the acceleration due to the gravity of the body is 4ms -2

Note: The acceleration due to the gravity of a body on the surface of the earth is constant (9.8ms -2 ), even though there may be a slight difference due to the mass and altitude of the body.

An object is thrown vertically upward at an initial velocity of 10ms -1 and reaches a maximum height of 50 meters. Find its initial upward acceleration.

Initial velocity, u = 10ms -1

Final velocity, v = 0

maximum height = displacement = 50m

Initial upward acceleration, a =?

When we apply the formula that says a = (v 2 – u 2 )/2d we will have

a = (0 – 10 2 )/(2 x 50) = -100/100 = -1 ms -2

Hence, since our acceleration is negative, we can now say that we are dealing with retardation or deceleration.

Therefore, the retardation is -1ms -2

Note: Retardation is the negative of acceleration, thus it is written in negative form.

A car is traveling at a velocity of 8ms -1 and experiences an acceleration of 5ms -2 . How far does it travel in 4 seconds?

Initial velocity, u = 8ms -1

acceleration, a = 5ms -2

Distance, s =?

time, t = 4s

We can apply the formula that says

s = 8 x 4 + (1/2) x 5 x 4 2

s = 32 + (1/2) x 80 = 32 + 40 = 72m

Therefore, the distance covered by the car in 4s is 72 meters.

A body is traveling at a velocity of 10m/s and experiences a deceleration of 5ms -2 . How long does it take the body to come to a complete stop?

Initial velocity, u = 10m/s

acceleration , a = retardation = -5ms -2

time, t = ?

We already know that acceleration, a = change in velocity/time

This implies that

Time, t = change in velocity/acceleration

t = (v – u)/t = (0-10)/-5 = -10/-5 = 2s

Therefore, the time it takes the car to stop is 2 seconds .

A body is traveling at a velocity of 20 m/s and has a mass of 10 kg. How much force is required to change its velocity by 10 m/s in 5 seconds?

Change in velocity, v =10 m/s

mass of the object, m = 10kg

time, t = 5s

We can apply newton’s second law of motion which says f = ma

and since a = change in velocity/time

we will have an acceleration equal to

a = 10/5 = 2ms -2

Therefore, to find the force, we can now say

f = ma = 10 x 2 = 20N

Therefore, the force that can help us to change the velocity by 10 m/s in 5 seconds is 20-Newton.

Drop a comment if there is anything you don’t understand about speed, velocity, and acceleration Problems.

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Speed and Velocity

Speed is how fast something moves.

Velocity is speed with a direction .

Saying Ariel the Dog runs at 9 km/h (kilometers per hour) is a speed.

But saying he runs 9 km/h Westwards is a velocity.

Imagine something moving back and forth very fast: it has a high speed, but a low (or zero) velocity.

Speed is measured as distance moved over time.

Speed = Distance Time

Example: A car travels 50 km in one hour.

Its average speed is 50 km per hour (50 km/h)

Speed = Distance Time = 50 km 1 hour

We can also use these symbols:

Speed = Δs Δt

Where Δ (" Delta ") means "change in", and

s means distance ("s" for "space")
t means time

Example: You run 360 m in 60 seconds.

So your speed is 6 meters per second (6 m/s).

Speed is commonly measured in:

meters per second (m/s or m s -1 ), or
kilometers per hour (km/h or km h -1 )

A km is 1000 m, and there are 3600 seconds in an hour, so we can convert like this (see Unit Conversion Method to learn more):

So 1 m/s is equal to 3.6 km/h

Example: What is 20 m/s in km/h ?

20 m/s × 3.6 km/h 1 m/s = 72 km/h

Example: What is 120 km/h in m/s ?

120 km/h × 1 m/s 3.6 km/h = 33.333... m/s

Average vs Instantaneous Speed

The examples so far calculate average speed : how far something travels over a period of time.

But speed can change as time goes by. A car can go faster and slower, maybe even stop at lights.

So there is also instantaneous speed : the speed at an instant in time. We can try to measure it by using a very short span of time (the shorter the better).

Example: Sam uses a stopwatch and measures 1.6 seconds as the car travels between two posts 20 m apart. What is the instantaneous speed ?

Well, we don't know exactly, as the car may have been speeding up or slowing down during that time, but we can estimate:

20 m 1.6 s = 12.5 m/s = 45 km/h

It is really still an average, but is close to an instantaneous speed.

Constant Speed

When the speed does not change it is constant .

For constant speed, the average and instantaneous speeds are the same.

Because the direction is important velocity uses displacement instead of distance:

Velocity = Displacement Time in a direction.

Example: You walk from home to the shop in 100 seconds, what is your speed and what is your velocity?

Speed = 220 m 100 s = 2.2 m/s

Velocity = 130 m 100 s East = 1.3 m/s East

You forgot your money so you turn around and go back home in 120 more seconds: what is your round-trip speed and velocity?

The total time is 100 s + 120 s = 220 s:

Speed = 440 m 220 s = 2.0 m/s

Velocity = 0 m 220 s = 0 m/s

Yes, the velocity is zero as you ended up where you started.

Learn more at Vectors .

Motion is relative. When we say something is "at rest" or "moving at 4 m/s" we forget to say "in relation to me" or "in relation to the ground", etc.

Think about this: are you really standing still? You are on planet Earth which is spinning at 40,075 km per day (about 1675 km/h or 465 m/s), and moving around the Sun at about 100,000 km/h, which is itself moving through the Galaxy.

Next time you are out walking, imagine you are still and it is the world that moves under your feet. Feels great.

It is all relative!

Speed and Velocity

Practice problem 1.

Average speed

This problem is deceptively easy. Averaging is taught in elementary school, which makes this an elementary problem. Right?

The add-and-divide method of averaging only works when averaging items of equal weight. The average age of the students in a classroom is the sum of their ages divided by the number of students only because each student is considered to have the same significance (a student, is a student, is a student, … ). In this problem, however, the two segments of the walk are significantly different. The second "half" was actually the majority of the walk. It carries more weight than the shorter first "half". Thus, the add-and-divide method won't work.

Let's return to our definition. Since speed is the rate of change of distance with time, we'll need both the distance traveled and the time it took to complete the walk. After we determine both of these numbers, the rest is easy.

Look closely at the calculations on the right side. Notice that the formula contains ∆ (delta) symbols and yet I added the distances in the numerator and the times in the denominator. That's because ∆ doesn't mean difference, it means change. During the walk my position didn't change from 6.0 km to 10 km, it changed first by 6.0 km and then by 10 km for a total change of 16 km.

Average velocity

Velocity is the rate of change of displacement with respect to time. Velocity is a vector, which means the problem should be solved graphically. Draw an arrow pointing toward the top of the page (north). Label it 6 km. Draw another arrow to the left (west) starting from the previous one (arranged head to tail). Make it slightly longer and label it 10 km. Draw a third arrow starting on the tail of the first and ending on the head of the second. Since north and west are at right angles to one another, the resultant displacement is the hypotenuse of a right triangle. Use Pythagorean theorem to find its magnitude and tangent to find its direction.

Divide displacement by time to get velocity.

practice problem 2

Notice that no numbers are stated in this problem. When a numerical value is needed to solve a problem and that number is not given, it could mean one of several things.

Look it up! It may appear somewhere in the textbook you are using — on the inside covers, in an appendix , or in the text of the chapter you are currently working on. It may be found in the reference table that some teachers distribute. Standardized exams usually also have their own reference table .
Know it! Some numbers are numbers that you should just know. In this problem, there is one relevant number that nearly everyone knows. You may also be expected to memorize certain numbers by an instructor or professor.
Calculate it! Maybe there's a way to find the number you need to know using other numbers given in the problem.
Forget about it! Maybe you don't really need the number you think you do. Maybe you are on the wrong track. Especially under test conditions, it is highly unlikely that you could be asked a question that requires a numerical value that you can not find, do not know, or can not calculate. Perhaps there is another method to solve this problem.

In order to calculate speed, you will need distance and time. What distance does a point on the equator move in a convenient period of time? Well, I hope you know that the Earth rotates once on its axis every day. You should also know how to calculate the length of a day in seconds. (A day is the period of the Earth's rotation, for which an upper case T is the symbol.) During a day, a point on the Earth's equator would have traveled a distance equal to the circumference of the Earth. The radius of the Earth is a number commonly found in textbooks and on reference tables. The problem can now be solved.

That's about 75% faster than the speed of sound in air at room temperature (343 m/s). An interesting problem to be dealt with later is that if the Earth is spinning so rapidly, how come things on the equator don't fly off into space?

practice problem 3

Astronomical distances are so large that using meters is cumbersome. For really large distances the light year is the best unit. A light year is the distance that light would travel in one year in a vacuum. Since the speed of light is fast, and a year is long, the light year is a pretty good unit for astronomy. One light year is about ten petameters (ten quadrillion meters) as the following calculation shows.

Start with the definition of speed and solve it for distance. The traditional symbol for the speed of light is c from the Latin word for swiftness — celeritas .

Numbers in, answer out.

Since both the speed of light and the year have exactly defined values in the International System of Units, the light year can be stated with an unnecessarily large number of significant digits.

Some distances in light years are provided below.

The distance to Proxima Centauri (the star nearest the Sun) is 4.3 light years .
The diameter of the Milky Way (the collection of stars that includes the Sun and all the stars visible to the naked eye) is about 100,000 light years .
The distance to Andromeda (the nearest galaxy outside the Milky Way) is about 2 million light years.
The distance to the edge of the universe (the observable part of it) is 13.77 billion light years .

practice problem 4

Everyone should know (or at least realize after a bit of thought) that there are…

∆ t = 60 × 60 = 3,600 s

in an hour. Many Americans who are fans of track and field know that four laps around a 400 m outdoor track is almost one mile.

∆ s = 1 mile ≈ 4 × 400 m = 1,600 m = 1.6 km

More precisely… actually, most precisely… actually, exactly by definition…

∆ s = 1 mile = 1,609.344 m = 1.609344 km

The first answer…

For comparison, the speed limit on many Canadian highways is 100 km/h.

The second answer…

You should note that the number with International units is a little bit less than half the value of the number with British-American units. I've gotten used to mph, but I have to be conversant in m/s for my job. A good rule of thumb for comparing speeds is to…

divide by 2 and subtract a little when converting from mph to m/s

multiply by 2 and add a little when converting from m/s to mph.

2.2 Speed and Velocity

Section learning objectives.

By the end of this section, you will be able to do the following:

Calculate the average speed of an object
Relate displacement and average velocity

Teacher Support

The learning objectives in this section will help your students master the following standards:

(B) describe and analyze motion in one dimension using equations with the concepts of distance, displacement, speed, average velocity, instantaneous velocity, and acceleration.

In addition, the High School Physics Laboratory Manual addresses content in this section in the lab titled: Position and Speed of an Object, as well as the following standards:

Section Key Terms

In this section, students will apply what they have learned about distance and displacement to the concepts of speed and velocity.

[BL] [OL] Before students read the section, ask them to give examples of ways they have heard the word speed used. Then ask them if they have heard the word velocity used. Explain that these words are often used interchangeably in everyday life, but their scientific definitions are different. Tell students that they will learn about these differences as they read the section.

[AL] Explain to students that velocity, like displacement, is a vector quantity. Ask them to speculate about ways that speed is different from velocity. After they share their ideas, follow up with questions that deepen their thought process, such as: Why do you think that? What is an example? How might apply these terms to motion that you see every day?

There is more to motion than distance and displacement. Questions such as, “How long does a foot race take?” and “What was the runner’s speed?” cannot be answered without an understanding of other concepts. In this section we will look at time , speed, and velocity to expand our understanding of motion.

A description of how fast or slow an object moves is its speed. Speed is the rate at which an object changes its location. Like distance, speed is a scalar because it has a magnitude but not a direction. Because speed is a rate, it depends on the time interval of motion. You can calculate the elapsed time or the change in time, Δ t Δ t , of motion as the difference between the ending time and the beginning time

The SI unit of time is the second (s), and the SI unit of speed is meters per second (m/s), but sometimes kilometers per hour (km/h), miles per hour (mph) or other units of speed are used.

When you describe an object's speed, you often describe the average over a time period. Average speed , v avg , is the distance traveled divided by the time during which the motion occurs.

You can, of course, rearrange the equation to solve for either distance or time

Suppose, for example, a car travels 150 kilometers in 3.2 hours. Its average speed for the trip is

A car's speed would likely increase and decrease many times over a 3.2 hour trip. Its speed at a specific instant in time, however, is its instantaneous speed . A car's speedometer describes its instantaneous speed.

[OL] [AL] Caution students that average speed is not always the average of an object's initial and final speeds. For example, suppose a car travels a distance of 100 km. The first 50 km it travels 30 km/h and the second 50 km it travels at 60 km/h. Its average speed would be distance /(time interval) = (100 km)/[(50 km)/(30 km/h) + (50 km)/(60 km/h)] = 40 km/h. If the car had spent equal times at 30 km and 60 km rather than equal distances at these speeds, its average speed would have been 45 km/h.

[BL] [OL] Caution students that the terms speed, average speed, and instantaneous speed are all often referred to simply as speed in everyday language. Emphasize the importance in science to use correct terminology to avoid confusion and to properly communicate ideas.

Worked Example

Calculating average speed.

A marble rolls 5.2 m in 1.8 s. What was the marble's average speed?

We know the distance the marble travels, 5.2 m, and the time interval, 1.8 s. We can use these values in the average speed equation.

Average speed is a scalar, so we do not include direction in the answer. We can check the reasonableness of the answer by estimating: 5 meters divided by 2 seconds is 2.5 m/s. Since 2.5 m/s is close to 2.9 m/s, the answer is reasonable. This is about the speed of a brisk walk, so it also makes sense.

Practice Problems

A pitcher throws a baseball from the pitcher’s mound to home plate in 0.46 s. The distance is 18.4 m. What was the average speed of the baseball?

The vector version of speed is velocity. Velocity describes the speed and direction of an object. As with speed, it is useful to describe either the average velocity over a time period or the velocity at a specific moment. Average velocity is displacement divided by the time over which the displacement occurs.

Velocity, like speed, has SI units of meters per second (m/s), but because it is a vector, you must also include a direction. Furthermore, the variable v for velocity is bold because it is a vector, which is in contrast to the variable v for speed which is italicized because it is a scalar quantity.

Tips For Success

It is important to keep in mind that the average speed is not the same thing as the average velocity without its direction. Like we saw with displacement and distance in the last section, changes in direction over a time interval have a bigger effect on speed and velocity.

Suppose a passenger moved toward the back of a plane with an average velocity of –4 m/s. We cannot tell from the average velocity whether the passenger stopped momentarily or backed up before he got to the back of the plane. To get more details, we must consider smaller segments of the trip over smaller time intervals such as those shown in Figure 2.9 . If you consider infinitesimally small intervals, you can define instantaneous velocity , which is the velocity at a specific instant in time. Instantaneous velocity and average velocity are the same if the velocity is constant.

Earlier, you have read that distance traveled can be different than the magnitude of displacement. In the same way, speed can be different than the magnitude of velocity. For example, you drive to a store and return home in half an hour. If your car’s odometer shows the total distance traveled was 6 km, then your average speed was 12 km/h. Your average velocity, however, was zero because your displacement for the round trip is zero.

Watch Physics

Calculating average velocity or speed.

This video reviews vectors and scalars and describes how to calculate average velocity and average speed when you know displacement and change in time. The video also reviews how to convert km/h to m/s.

A scalar quantity is fully described by its magnitude, while a vector needs both magnitude and direction to fully describe it. Displacement is an example of a scalar quantity and time is an example of a vector quantity.
A scalar quantity is fully described by its magnitude, while a vector needs both magnitude and direction to fully describe it. Time is an example of a scalar quantity and displacement is an example of a vector quantity.
A scalar quantity is fully described by its magnitude and direction, while a vector needs only magnitude to fully describe it. Displacement is an example of a scalar quantity and time is an example of a vector quantity.
A scalar quantity is fully described by its magnitude and direction, while a vector needs only magnitude to fully describe it. Time is an example of a scalar quantity and displacement is an example of a vector quantity.

This video does a good job of reinforcing the difference between vectors and scalars. The student is introduced to the idea of using ‘s’ to denote displacement, which you may or may not wish to encourage. Before students watch the video, point out that the instructor uses s → s → for displacement instead of d, as used in this text. Explain the use of small arrows over variables is a common way to denote vectors in higher-level physics courses. Caution students that the customary abbreviations for hour and seconds are not used in this video. Remind students that in their own work they should use the abbreviations h for hour and s for seconds.

Calculating Average Velocity

A student has a displacement of 304 m north in 180 s. What was the student's average velocity?

We know that the displacement is 304 m north and the time is 180 s. We can use the formula for average velocity to solve the problem.

Since average velocity is a vector quantity, you must include direction as well as magnitude in the answer. Notice, however, that the direction can be omitted until the end to avoid cluttering the problem. Pay attention to the significant figures in the problem. The distance 304 m has three significant figures, but the time interval 180 s has only two, so the quotient should have only two significant figures.

Note the way scalars and vectors are represented. In this book d represents distance and displacement. Similarly, v represents speed, and v represents velocity. A variable that is not bold indicates a scalar quantity, and a bold variable indicates a vector quantity. Vectors are sometimes represented by small arrows above the variable.

Use this problem to emphasize the importance of using the correct number of significant figures in calculations. Some students have a tendency to include many digits in their final calculations. They incorrectly believe they are improving the accuracy of their answer by writing many of the digits shown on the calculator. Point out that doing this introduces errors into the calculations. In more complicated calculations, these errors can propagate and cause the final answer to be wrong. Instead, remind students to always carry one or two extra digits in intermediate calculations and to round the final answer to the correct number of significant figures.

Solving for Displacement when Average Velocity and Time are Known

Layla jogs with an average velocity of 2.4 m/s east. What is her displacement after 46 seconds?

We know that Layla's average velocity is 2.4 m/s east, and the time interval is 46 seconds. We can rearrange the average velocity formula to solve for the displacement.

The answer is about 110 m east, which is a reasonable displacement for slightly less than a minute of jogging. A calculator shows the answer as 110.4 m. We chose to write the answer using scientific notation because we wanted to make it clear that we only used two significant figures.

Dimensional analysis is a good way to determine whether you solved a problem correctly. Write the calculation using only units to be sure they match on opposite sides of the equal mark. In the worked example, you have m = (m/s)(s). Since seconds is in the denominator for the average velocity and in the numerator for the time, the unit cancels out leaving only m and, of course, m = m.

Solving for Time when Displacement and Average Velocity are Known

Phillip walks along a straight path from his house to his school. How long will it take him to get to school if he walks 428 m west with an average velocity of 1.7 m/s west?

We know that Phillip's displacement is 428 m west, and his average velocity is 1.7 m/s west. We can calculate the time required for the trip by rearranging the average velocity equation.

Here again we had to use scientific notation because the answer could only have two significant figures. Since time is a scalar, the answer includes only a magnitude and not a direction.

4 km/h north
4 km/h south
64 km/h north
64 km/h south

A bird flies with an average velocity of 7.5 m/s east from one branch to another in 2.4 s. It then pauses before flying with an average velocity of 6.8 m/s east for 3.5 s to another branch. What is the bird’s total displacement from its starting point?

Virtual Physics

The walking man.

In this simulation you will put your cursor on the man and move him first in one direction and then in the opposite direction. Keep the Introduction tab active. You can use the Charts tab after you learn about graphing motion later in this chapter. Carefully watch the sign of the numbers in the position and velocity boxes. Ignore the acceleration box for now. See if you can make the man’s position positive while the velocity is negative. Then see if you can do the opposite.

Grasp Check

Which situation correctly describes when the moving man’s position was negative but his velocity was positive?

Man moving toward 0 from left of 0
Man moving toward 0 from right of 0
Man moving away from 0 from left of 0
Man moving away from 0 from right of 0

This is a powerful interactive animation, and it can be used for many lessons. At this point it can be used to show that displacement can be either positive or negative. It can also show that when displacement is negative, velocity can be either positive or negative. Later it can be used to show that velocity and acceleration can have different signs. It is strongly suggested that you keep students on the Introduction tab. The Charts tab can be used after students learn about graphing motion later in this chapter.

Check Your Understanding

Yes, because average velocity depends on the net or total displacement.
Yes, because average velocity depends on the total distance traveled.
No, because the velocities of both runners must remain exactly the same throughout the journey.
No, because the instantaneous velocities of the runners must remain the same at the midpoint but can vary at other points.

If you divide the total distance traveled on a car trip (as determined by the odometer) by the time for the trip, are you calculating the average speed or the magnitude of the average velocity, and under what circumstances are these two quantities the same?

Average speed. Both are the same when the car is traveling at a constant speed and changing direction.
Average speed. Both are the same when the speed is constant and the car does not change its direction.
Magnitude of average velocity. Both are same when the car is traveling at a constant speed.
Magnitude of average velocity. Both are same when the car does not change its direction.
Yes, if net displacement is negative.
Yes, if the object’s direction changes during motion.
No, because average velocity describes only the magnitude and not the direction of motion.
No, because average velocity only describes the magnitude in the positive direction of motion.

Use the Check Your Understanding questions to assess students’ achievement of the sections learning objectives. If students are struggling with a specific objective, the Check Your Understanding will help identify which and direct students to the relevant content. Assessment items in TUTOR will allow you to reassess.

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Straight Line Motion

Solved Speed, Velocity, and Acceleration Problems

Simple problems on speed, velocity, and acceleration with descriptive answers are presented for the AP Physics 1 exam and college students. In each solution, you can find a brief tutorial.

Speed and velocity Problems:

Problem (1): What is the speed of a rocket that travels $8000\,{\rm m}$ in $13\,{\rm s}$?

Solution : Speed is defined in physics as the total distance divided by the elapsed time, so the rocket's speed is \[\text{speed}=\frac{8000}{13}=615.38\,{\rm m/s}\]

Problem (2): How long will it take if you travel $400\,{\rm km}$ with an average speed of $100\,{\rm m/s}$?

Solution : Average speed is the ratio of the total distance to the total time. Thus, the elapsed time is \begin{align*} t&=\frac{\text{total distance}}{\text{average speed}}\\ \\ &=\frac{400\times 10^{3}\,{\rm m}}{100\,{\rm m/s}}\\ \\ &=4000\,{\rm s}\end{align*} To convert it to hours, it must be divided by $3600\,{\rm s}$ which gives $t=1.11\,{\rm h}$.

Problem (3): A person walks $100\,{\rm m}$ in $5$ minutes, then $200\,{\rm m}$ in $7$ minutes, and finally $50\,{\rm m}$ in $4$ minutes. Find its average speed.

Solution : First find its total distance traveled ($D$) by summing all distances in each section, which gets $D=100+200+50=350\,{\rm m}$. Now, by definition of average speed, divide it by the total time elapsed $T=5+7+4=16$ minutes.

But keep in mind that since the distance is in SI units, so the time traveled must also be in SI units, which is $\rm s$. Therefore, we have\begin{align*}\text{average speed}&=\frac{\text{total distance} }{\text{total time} }\\ \\ &=\frac{350\,{\rm m}}{16\times 60\,{\rm s}}\\ \\&=0.36\,{\rm m/s}\end{align*}

Problem (4): A person walks $750\,{\rm m}$ due north, then $250\,{\rm m}$ due east. If the entire walk takes $12$ minutes, find the person's average velocity.

Solution : Average velocity , $\bar{v}=\frac{\Delta x}{\Delta t}$, is displacement divided by the elapsed time. Displacement is also a vector that obeys the addition vector rules. Thus, in this velocity problem, add each displacement to get the total displacement .

In the first part, displacement is $\Delta x_1=750\,\hat{j}$ (due north) and in the second part $\Delta x_2=250\,\hat{i}$ (due east). The total displacement vector is $\Delta x=\Delta x_1+\Delta x_2=750\,\hat{i}+250\,\hat{j}$ with magnitude of \begin{align*}|\Delta x|&=\sqrt{(750)^{2}+(250)^{2}}\\ \\&=790.5\,{\rm m}\end{align*} In addition, the total elapsed time is $t=12\times 60$ seconds. Therefore, the magnitude of the average velocity is \[\bar{v}=\frac{790.5}{12\times 60}=1.09\,{\rm m/s}\]

Problem (5): An object moves along a straight line. First, it travels at a velocity of $12\,{\rm m/s}$ for $5\,{\rm s}$ and then continues in the same direction with $20\,{\rm m/s}$ for $3\,{\rm s}$. What is its average speed?

Solution: Average velocity is displacement divided by elapsed time, i.e., $\bar{v}\equiv \frac{\Delta x_{tot}}{\Delta t_{tot}}$.

Here, the object goes through two stages with two different displacements, so add them to find the total displacement. Thus,\[\bar{v}=\frac{x_1 + x_2}{t_1 +t_2}\] Again, to find the displacement, we use the same equation as the average velocity formula, i.e., $x=vt$. Thus, displacements are obtained as $x_1=v_1\,t_1=12\times 5=60\,{\rm m}$ and $x_2=v_2\,t_2=20\times 3=60\,{\rm m}$. Therefore, we have \begin{align*} \bar{v}&=\frac{x_1+x_2}{t_1+t_2}\\ \\&=\frac{60+60}{5+3}\\ \\&=\boxed{15\,{\rm m/s}}\end{align*}

Problem (6): A plane flies the distance between two cities in $1$ hour and $30$ minutes with a velocity of $900\,{\rm km/h}$. Another plane covers that distance at $600\,{\rm km/h}$. What is the flight time of the second plane?

Solution: first find the distance between two cities using the average velocity formula $\bar{v}=\frac{\Delta x}{\Delta t}$ as below \begin{align*} x&=vt\\&=900\times 1.5\\&=1350\,{\rm km}\end{align*} where we wrote one hour and a half minutes as $1.5\,\rm h$. Now use again the same kinematic equation above to find the time required for another plane \begin{align*} t&=\frac xv\\ \\ &=\frac{1350\,\rm km}{600\,\rm km/h}\\ \\&=2.25\,{\rm h}\end{align*} Thus, the time for the second plane is $2$ hours and $0.25$ of an hour, which converts to minutes as $2$ hours and ($0.25\times 60=15$) minutes.

Problem (7): To reach a park located south of his jogging path, Henry runs along a 15-kilometer route. If he completes the journey in 1.5 hours, determine his speed and velocity.

Solution: Henry travels his route to the park without changing direction along a straight line. Therefore, the total distance traveled in one direction equals the displacement, i.e, \[\text{distance traveled}=\Delta x=15\,\rm km\]Velocity is displacement divided by the time of travel \begin{align*} \text{velocity}&=\frac{\text{displacement}}{\text{time of travel}} \\\\ &=\frac{15\,\rm km}{1.5\,\rm h} \\\\ &=\boxed{10\,\rm km/h}\end{align*} and by definition, its average speed is \begin{align*} \text{speed}&=\frac{\text{distance covered}}{\text{time interval}}\\\\&=\frac{15\,\rm km}{1.5\,\rm h}\\\\&=\boxed{10\,\rm km/h}\end{align*} Thus, Henry's velocity is $10\,\rm km/h$ to the south, and its speed is $10\,\rm km/h$. As you can see, speed is simply a positive number, with units but velocity specifies the direction in which the object is moving.

Problem (8): In 15 seconds, a football player covers the distance from his team's goal line to the opposing team's goal line and back to the midway point of the field having 100-yard-length. Find, (a) his average speed, and (b) the magnitude of the average velocity.

Solution: The total length of the football field is $100$ yards or in meters, $L=91.44\,\rm m$. Going from one goal's line to the other and back to the midpoint of the field takes $15\,\rm s$ and covers a distance of $D=100+50=150\,\rm yd$.

average speed and velocity at football field

Distance divided by the time of travel gets the average speed, \[\text{speed}=\frac{150\times 0.91}{15}=9.1\,\rm m/s\] To find the average velocity, we must find the displacement of the player between the initial and final points.

The initial point is her own goal line and her final position is the midpoint of the field, so she has displaced a distance of $\Delta x=50\,\rm yd$ or $\Delta x=50\times 0.91=45.5\,\rm m$. Therefore, her velocity is calculated as follows \begin{align*} \text{velocity}&=\frac{\text{displacement}}{\text{time elapsed}} \\\\ &=\frac{45.5\,\rm m}{15\,\rm s} \\\\&=\boxed{3.03\quad \rm m/s}\end{align*} Contrary to the previous problem, here the motion is not in one direction, hence, the displacement is not equal to the distance traveled. Accordingly, the average speed is not equal to the magnitude of the average velocity.

Problem (9): You begin at a pillar and run towards the east (the positive $x$ direction) for $250\,\rm m$ at an average speed of $5\,\rm m/s$. After that, you run towards the west for $300\,\rm m$ at an average speed of $4\,\rm m/s$ until you reach a post. Calculate (a) your average speed from pillar to post, and (b) your average velocity from pillar to post.

Solution : First, you traveled a distance of $L_1=250\,\rm m$ toward east (or $+x$ direction) at $5\,\rm m/s$. Time of travel in this route is obtained as follows \begin{align*} t_1&=\frac{L_1}{v_1}\\\\ &=\frac{250}{5}\\\\&=50\,\rm s\end{align*} Likewise, traveling a distance of $L_2=300\,\rm m$ at $v_2=4\,\rm m/s$ takes \[t_2=\frac{300}{4}=75\,\rm s\] (a) Average speed is defined as the distance traveled (or path length) divided by the total time of travel \begin{align*} v&=\frac{\text{path length}}{\text{time of travel}} \\\\ &=\frac{L_1+L_2}{t_1+t_2}\\\\&=\frac{250+300}{50+75} \\\\&=4.4\,\rm m/s\end{align*} Therefore, you travel between these two pillars in $125\,\rm s$ and with an average speed of $4.4\,\rm m/s$.

(b) Average velocity requires finding the displacement between those two points. In the first case, you move $250\,\rm m$ toward $+x$ direction, i.e., $L_1=+250\,\rm m$. Similarly, on the way back, you move $300\,\rm m$ toward the west ($-x$ direction) or $L_2=-300\,\rm m$. Adding these two gives us the total displacement between the initial point and the final point, \begin{align*} L&=L_1+L_2 \\\\&=(+250)+(-300) \\\\ &=-50\,\rm m\end{align*} The minus sign indicates that you are generally displaced toward the west.

Finally, the average velocity is obtained as follows: \begin{align*} \text{average velocity}&=\frac{\text{displacement}}{\text{time of travel}} \\\\ &=\frac{-50}{125} \\\\&=-0.4\,\rm m/s\end{align*} A negative average velocity indicating motion to the left along the $x$-axis.

This speed problem better makes it clear to us the difference between average speed and average speed. Unlike average speed, which is always a positive number, the average velocity in a straight line can be either positive or negative.

Problem (10): What is the average speed for the round trip of a car moving uphill at 40 km/h and then back downhill at 60 km/h?

Solution : Assuming the length of the hill to be $L$, the total distance traveled during this round trip is $2L$ since $L_{up}=L_{down}=L$. However, the time taken for going uphill and downhill was not provided. We can write them in terms of the hill's length $L$ as $t=\frac L v$.

Applying the definition of average speed gives us \begin{align*} v&=\frac{\text{distance traveled}}{\text{total time}} \\\\ &=\frac{L_{up}+L_{down}}{t_{up}+t_{down}} \\\\ &=\cfrac{2L}{\cfrac{L}{v_{up}}+\cfrac{L}{v_{down}}} \end{align*} By reorganizing this expression, we obtain a formula that is useful for solving similar problems in the AP Physics 1 exams. \[\text{average speed}=\frac{2v_{up} \times v_{down}}{v_{up}+v_{down}}\] Substituting the numerical values into this, yields \begin{align*} v&=\frac{2(40\times 60)}{40+60} \\\\ &=\boxed{48\,\rm m/s}\end{align*} What if we were asked for the average velocity instead? During this round trip, the car returns to its original position, and thus its displacement, which defines the average velocity, is zero. Therefore, \[\text{average velocity}=0\,\rm m/s\]

Acceleration Problems

Problem (9): A car moves from rest to a speed of $45\,\rm m/s$ in a time interval of $15\,\rm s$. At what rate does the car accelerate?

Solution : The car is initially at rest, $v_1=0$, and finally reaches $v_2=45\,\rm m/s$ in a time interval $\Delta t=15\,\rm s$. Average acceleration is the change in velocity, $\Delta v=v_2-v_1$, divided by the elapsed time $\Delta t$, so \[\bar{a}=\frac{45-0}{15}=\boxed{3\,\rm m/s^2} \]

Problem (10): A car moving at a velocity of $15\,{\rm m/s}$, uniformly slows down. It comes to a complete stop in $10\,{\rm s}$. What is its acceleration?

Solution: Let the car's uniform velocity be $v_1$ and its final velocity $v_2=0$. Average acceleration is the difference in velocities divided by the time taken, so we have: \begin{align*}\bar{a}&=\frac{\Delta v}{\Delta t}\\\\&=\frac{v_2-v_1}{\Delta t}\\\\&=\frac{0-15}{10}\\\\ &=\boxed{-1.5\,{\rm m/s^2}}\end{align*}The minus sign indicates the direction of the acceleration vector, which is toward the $-x$ direction.

Problem (11): A car moves from rest to a speed of $72\,{\rm km/h}$ in $4\,{\rm s}$. Find the acceleration of the car.

Solution: Known: $v_1=0$, $v_2=72\,{\rm km/h}$, $\Delta t=4\,{\rm s}$. Average acceleration is defined as the difference in velocities divided by the time interval between those points \begin{align*}\bar{a}&=\frac{v_2-v_1}{t_2-t_1}\\\\&=\frac{20-0}{4}\\\\&=5\,{\rm m/s^2}\end{align*} In above, we converted $\rm km/h$ to the SI unit of velocity ($\rm m/s$) as \[1\,\frac{km}{h}=\frac {1000\,m}{3600\,s}=\frac{10}{36}\, \rm m/s\] so we get \[72\,\rm km/h=72\times \frac{10}{36}=20\,\rm m/s\]

Problem (12): A race car accelerates from an initial velocity of $v_i=10\,{\rm m/s}$ to a final velocity of $v_f = 30\,{\rm m/s}$ in a time interval of $2\,{\rm s}$. Determine its average acceleration.

Solution: A change in the velocity of an object $\Delta v$ over a time interval $\Delta t$ is defined as an average acceleration. Known: $v_i=10\,{\rm m/s}$, $v_f = 30\,{\rm m/s}$, $\Delta t=2\,{\rm s}$. Applying definition of average acceleration, we get \begin{align*}\bar{a}&=\frac{v_f-v_i}{\Delta t}\\&=\frac{30-10}{2}\\&=10\,{\rm m/s^2}\end{align*}

Problem (13): A motorcycle starts its trip along a straight line with a velocity of $10\,{\rm m/s}$ and ends with $20\,{\rm m/s}$ in the opposite direction in a time interval of $2\,{\rm s}$. What is the average acceleration of the car?

Solution: Known: $v_i=10\,{\rm m/s}$, $v_f=-20\,{\rm m/s}$, $\Delta t=2\,{\rm s}$, $\bar{a}=?$. Using average acceleration definition we have \begin{align*}\bar{a}&=\frac{v_f-v_i}{\Delta t}\\\\&=\frac{(-20)-10}{2}\\\\ &=\boxed{-15\,{\rm m/s^2}}\end{align*}Recall that in the definition above, velocities are vector quantities. The final velocity is in the opposite direction from the initial velocity so a negative must be included.

Problem (14): A ball is thrown vertically up into the air by a boy. After $4$ seconds, it reaches the highest point of its path. How fast does the ball leave the boy's hand?

Solution : At the highest point, the ball has zero speed, $v_2=0$. It takes the ball $4\,\rm s$ to reach that point. In this problem, our unknown is the initial speed of the ball, $v_1=?$. Here, the ball accelerates at a constant rate of $g=-9.8\,\rm m/s^2$ in the presence of gravity.

When the ball is tossed upward, the only external force that acts on it is the gravity force.

Using the average acceleration formula $\bar{a}=\frac{\Delta v}{\Delta t}$ and substituting the numerical values into this, we will have \begin{gather*} \bar{a}=\frac{\Delta v}{\Delta t} \\\\ -9.8=\frac{0-v_1}{4} \\\\ \Rightarrow \boxed{v_1=39.2\,\rm m/s} \end{gather*} Note that $\Delta v=v_2-v_1$.

Problem (15): A child drops crumpled paper from a window. The paper hit the ground in $3\,\rm s$. What is the velocity of the crumpled paper just before it strikes the ground?

Solution : The crumpled paper is initially in the child's hand, so $v_1=0$. Let its speed just before striking be $v_2$. In this case, we have an object accelerating down in the presence of gravitational force at a constant rate of $g=-9.8\,\rm m/s^2$. Using the definition of average acceleration, we can find $v_2$ as below \begin{gather*} \bar{a}=\frac{\Delta v}{\Delta t} \\\\ -9.8=\frac{v_2-0}{3} \\\\ \Rightarrow v_2=3\times (-9.8)=\boxed{-29.4\,\rm m/s} \end{gather*} The negative shows us that the velocity must be downward, as expected!

Problem (16): A car travels along the $x$-axis for $4\,{\rm s}$ at an average velocity of $10\,{\rm m/s}$ and $2\,{\rm s}$ with an average velocity of $30\,{\rm m/s}$ and finally $4\,{\rm s}$ with an average velocity $25\,{\rm m/s}$. What is its average velocity across the whole path?

Solution: There are three different parts with different average velocities. Assume each trip is done in one dimension without changing direction. Thus, displacements associated with each segment are the same as the distance traveled in that direction and is calculated as below: \begin{align*}\Delta x_1&=v_1\,\Delta t_1\\&=10\times 4=40\,{\rm m}\\ \\ \Delta x_2&=v_2\,\Delta t_2\\&=30\times 2=60\,{\rm m}\\ \\ \Delta x_3&=v_3\,\Delta t_3\\&=25\times 4=100\,{\rm m}\end{align*}Now use the definition of average velocity, $\bar{v}=\frac{\Delta x_{tot}}{\Delta t_{tot}}$, to find it over the whole path\begin{align*}\bar{v}&=\frac{\Delta x_{tot}}{\Delta t_{tot}}\\ \\&=\frac{\Delta x_1+\Delta x_2+\Delta x_3}{\Delta t_1+\Delta t_2+\Delta t_3}\\ \\&=\frac{40+60+100}{4+2+4}\\ \\ &=\boxed{20\,{\rm m/s}}\end{align*}

Problem (17): An object moving along a straight-line path. It travels with an average velocity $2\,{\rm m/s}$ for $20\,{\rm s}$ and $12\,{\rm m/s}$ for $t$ seconds. If the total average velocity across the whole path is $10\,{\rm m/s}$, then find the unknown time $t$.

Solution: In this velocity problem, the whole path $\Delta x$ is divided into two parts $\Delta x_1$ and $\Delta x_2$ with different average velocities and times elapsed, so the total average velocity across the whole path is obtained as \begin{align*}\bar{v}&=\frac{\Delta x}{\Delta t}\\\\&=\frac{\Delta x_1+\Delta x_2}{\Delta t_1+\Delta t_2}\\\\&=\frac{\bar{v}_1\,t_1+\bar{v}_2\,t_2}{t_1+t_2}\\\\10&=\frac{2\times 20+12\times t}{20+t}\\\Rightarrow t&=80\,{\rm s}\end{align*}

Note : whenever a moving object, covers distances $x_1,x_2,x_3,\cdots$ in $t_1,t_2,t_3,\cdots$ with constant or average velocities $v_1,v_2,v_3,\cdots$ along a straight-line without changing its direction, then its total average velocity across the whole path is obtained by one of the following formulas

Distances and times are known:\[\bar{v}=\frac{x_1+x_2+x_3+\cdots}{t_1+t_2+t_3+\cdots}\]
Velocities and times are known: \[\bar{v}=\frac{v_1\,t_1+v_2\,t_2+v_3\,t_3+\cdots}{t_1+t_2+t_3+\cdots}\]
Distances and velocities are known:\[\bar{v}=\frac{x_1+x_2+x_3+\cdots}{\frac{x_1}{v_1}+\frac{x_2}{v_2}+\frac{x_3}{v_3}+\cdots}\]

Problem (18): A car travels one-fourth of its path with a constant velocity of $10\,{\rm m/s}$, and the remaining with a constant velocity of $v_2$. If the total average velocity across the whole path is $16\,{\rm m/s}$, then find the $v_2$?

Solution: This is the third case of the preceding note. Let the length of the path be $L$ so \begin{align*}\bar{v}&=\frac{x_1+x_2}{\frac{x_1}{v_1}+\frac{x_2}{v_2}}\\\\16&=\frac{\frac 14\,L+\frac 34\,L}{\frac{\frac 14\,L}{10}+\frac{\frac 34\,L}{v_2}}\\\\\Rightarrow v_2&=20\,{\rm m/s}\end{align*}

Problem (19): An object moves along a straight-line path. It travels for $t_1$ seconds with an average velocity $50\,{\rm m/s}$ and $t_2$ seconds with a constant velocity of $25\,{\rm m/s}$. If the total average velocity across the whole path is $30\,{\rm m/s}$, then find the ratio $\frac{t_2}{t_1}$?

Solution: the velocities and times are known, so we have \begin{align*}\bar{v}&=\frac{v_1\,t_1+v_2\,t_2}{t_1+t_2}\\\\30&=\frac{50\,t_1+25\,t_2}{t_1+t_2}\\\\ \Rightarrow \frac{t_2}{t_1}&=4\end{align*}

TIME SPEED AND DISTANCE PROBLEMS

Formulas .

To know the shortcuts required to solve problems on time, speed and distance,

please click here

Problem 1 :

If a person drives his car in the speed 50 miles per hour, how far can he cover in 2.5 hours?

Given : Speed is 50 miles per hour.

So, the distance covered in 1 hour is

Then, the distance covered in 2.5 hours is

= 2.5 ⋅ 50 miles

= 125 miles

So, the person can cover 125 miles of distance in 2.5 hours.

Problem 2 :

If a person travels at a speed of 40 miles per hour. At the same rate, how long will he take to cover 160 miles distance?

Given : Speed is 40 miles per hour.

The formula to find the time when distance and speed are given is

Time taken to cover the distance of 160 miles is

So, the person will take 4 hours to cover 160 miles distance at the rate of 40 miles per hour.

Problem 3 :

A person travels at a speed of 60 miles per hour. How far will he travel in 4.5 hours?

Given : Speed is 60 miles per hour.

The distance covered in 1 hour is

Then, the distance covered in 4.5 hours is

= 4.5 ⋅ 60 miles

= 270 miles

So, the person will travel 270 miles distance in 4.5 hours.

Problem 4 :

A person travels at a speed of 60 kms per hour. Then how many meters can he travel in 5 minutes?

Given : Speed is 60 kms per hour.

The distance covered in 1 hour or 60 minutes is

= 60 ⋅ 1000 meters

= 60000 meters

Then the distance covered in 1 minute is

The distance covered in 5 minutes is

= 5 ⋅ 1000

= 5000 meters

So, the person can cover 5000 meters distance in 5 minutes.

Problem 5 :

A person covers 108 kms in 3 hours. What is his speed in meter per second?

Given : Distance is 108 kms and time is 3 hours.

The given distance in meters :

= 108 ⋅ 1000

= 108,000 meters

The given time in seconds :

= 3 ⋅ 60 minutes

= 180 minutes

= 180 ⋅ 60 seconds

= 10,800 seconds

The formula to find the speed is

Speed in meter per second is

= ¹⁰⁸⁰⁰⁰⁄₁₀₈₀₀

So, his speed in meter per second is 10.

Problem 6 :

A person covers 90 kms in 2 hours 30 minutes. Find the speed in meter per second.

Given : Distance is 90 kms and time is 2 hrs 30 min.

= 90 ⋅ 1000

= 90,000 meters

= 2 hrs 30 min

= (120 + 30) min

= 150 minutes

= 150 ⋅ 60 seconds

= 9,000 seconds

= ⁹⁰⁰⁰⁰⁄₉₀₀₀

Problem 7 :

A person travels at the rate of 60 miles per hour and covers 300 miles in 5 hours. If he reduces his speed by 10 miles per hour, how long will he take to cover the same distance?

Original speed is 60 miles per hour.

If the speed is reduced by 10 miles per hour, then the new speed is

= 50 miles per hour

Distance to be covered is 300 miles.

The formula to find time is

Time taken to cover 300 miles distance at the speed of 50 miles per hour is

So, if the person reduces his speed by 10 miles per hour, he will take 6 hours to cover 300 miles distance.

Problem 8 :

A person travels 50 kms per hour. If he increases his speed by 10 kms per hour, how many minutes will he take to cover 8000 meters?

Original speed is 50 kms per hour.

If the speed is increased by 10 kms per hour, then the new speed is

= 60 kms per hour

Because, we have to find the time in minutes for the distance given in meters, let us change the speed from kms per hour into meters per minute.

1 hour ----> 60 kms

1 ⋅ 60 minutes ----> 60 ⋅ 1000 meters

60 minutes ----> 60000 meters

1 minute ----> ⁶⁰⁰⁰⁰⁄₆₀ meters

1 minute ----> 1000 meters

So, the speed is 1000 meters/minute.

Time = Distance/Speed

Time taken to cover 8000 meters distance at the speed of 1000 meters per minute is

= ⁸⁰⁰⁰⁄₁₀₀₀

= 8 minutes

So, if the person increases his speed by 10 kms per hour, he will take 8 minutes to cover 8000 meters distance.

Problem 9 :

A person can travel at the speed of 40 miles per hour. If the speed is increased by 50%, how long will it take to cover 330 miles?

Original speed is 40 miles per hour.

If the speed is increased by 50%, then the new speed is

= 150% of 40

= 1.5 ⋅ 40

= 60 miles per hour

Distance to be covered is 330 miles.

Time taken to cover 330 miles distance at the speed of 60 miles per hour is

= 5.5 hours

= 5 hrs 30 minutes

So, if the person is increased by 50%, it will take 5 hrs 30 minutes to cover 330 miles distance.

Problem 10 :

A person speed at a rate of 40 kms per hour. If he increases his speed by 20%, what is his new speed in meter per minute?

Original speed is 40 kms per hour

If the speed is increased by 20%, then the new speed is

= 120 % of 40

= 1.2 ⋅ 40

= 48 kms per hour

Now, let us change the speed from kms per hour into meters per minute.

1 hour ----> 48 kms

1 ⋅ 60 minutes ----> 48 ⋅ 1000 meters

60 minutes ----> 48,000 meters

1 minute ----> ⁴⁸⁰⁰⁰⁄₆₀ meters

1 minute ----> 800 meters

So, the speed is 800 meters/minute.

If the person increases his speed by 20%, his new speed will be 800 meter per minute.

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Speed Formula

The speed formula can be defined as the rate at which an object covers some distance. Speed can be measured as the distance traveled by a body in a given period of time. The SI unit of speed is m/s. In this section, we will learn more about the speed formula and its applications.

What is the Speed Formula?

Let's move further and explore more about the speed formula in this section. Different units can be used to express speed, like m/s, km/hr, miles/hr, etc. The dimensional formula of speed is [LT -1 ]. Speed is a measure of how fast a body is moving. The formula for the speed of a given body can be expressed as,

Formula for Speed

Speed = Distance ÷ Time

How to Use Speed Formula?

Speed formula can be used to find the speed of objects, given the distance and time taken to cover that distance. We can also use the speed formula to calculate the distance or time by substituting the known values in the formula for speed and further evaluating, Distance = Speed × Time or, Time = Distance/Speed Let's take a quick look at an example showing how to use the formula for speed.

Example: What is your speed if you travel 3600 m in 30 minutes?

Solution: Using Formula for Speed,

Speed = Distance ÷ Time Speed = 3600 ÷ (30 × 60) = 2

Answer: Your speed if you travel 3600m in 30 minutes is 2 m/s.

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Examples on Speed Formula

Let us solve some interesting problems using the speed formula.

Example 1: A train covered a distance of 120 km in an hour. Determine the speed of the train in m/s using the speed formula.

To find: The speed of the train. The distance covered by a train in meters = 120×1000m = 120000m Time taken by train in seconds = 60×60 = 3600 second Using Formula for Speed, Speed = Distance/Time = 120000/3600 = 33.3m/sec

Answer: The speed of the train is 33.3 m/s.

Example 2: A cyclist covers 20 km in 50 minutes. Use the speed formula to calculate the speed of the cyclist in m/s.

Solution: To find: The speed of a cyclist. The distance covered by a cyclist in meters = 20×1000 =20000m Time taken by cyclist in seconds = 50×60= 3000 second Using formula for speed, Speed = Distance/Time = 20000/3000 = 6.67m/sec

Answer: The speed of the cyclist is 6.67 m/s.

Example 3: Using the speed formula calculate the speed of a person in kilometers per hour if the distance he travels is 40 kilometers in 2 hours?

The formula for speed is [Speed = Distance ÷ Time] Distance = 40 kilometers Time = 2 hours Speed = (40 ÷ 2) km/hr = 20 kilometers per hour

Answer: The speed of the person is 20 kilometers per hour.

FAQs on Speed Formula

How to calculate distance using speed formula .

The formula for Speed is given as [Speed = Distance ÷ Time]. To calculate the distance, the speed formula can be molded as [Distance = Speed × Time].

How to Calculate Time Using Speed Formula?

The formula for speed is given as [Speed = Distance ÷ Time].To calculate the time, the speed formula will be molded as [Time = Distance Travelled ÷ Speed].

How to Use the Formula for Speed?

Speed formula can be used in our day-to-day lives to find the speed of objects. To understand how to use the formula for speed let us consider an example. Example: What is your speed if you travel 4000m in 40 minutes? Solution: Using Formula for Speed, Speed = Distance ÷ Time Speed = 4000 ÷ (40 × 60) = 1.67m/s.

Answer: Your speed if you travel 4000m in 40 minutes is 1.67 m/s.

What will be the General Speed Formula for an Object?

The general speed formula for an object is given as [Speed = Distance ÷ Time]. SI units of speed is m/s.

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Real World Algebra by Edward Zaccaro

Algebra is often taught abstractly with little or no emphasis on what algebra is or how it can be used to solve real problems. Just as English can be translated into other languages, word problems can be "translated" into the math language of algebra and easily solved. Real World Algebra explains this process in an easy to understand format using cartoons and drawings. This makes self-learning easy for both the student and any teacher who never did quite understand algebra. Includes chapters on algebra and money, algebra and geometry, algebra and physics, algebra and levers and many more. Designed for children in grades 4-9 with higher math ability and interest but could be used by older students and adults as well. Contains 22 chapters with instruction and problems at three levels of difficulty.

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Mechanics (Essentials) - Class 11th

Course: mechanics (essentials) - class 11th > unit 4.

Intro to vectors and scalars
Average speed & velocity (with examples)

Average velocity and speed in one direction: word problems

Average velocity and speed with direction changes: word problems
Average speed for entire journey - solved numerical
Average velocity and speed review
Your answer should be
an exact decimal, like 0.75 ‍

Average Speed Problems

Related Pages Rate, Time, Distance Solving Speed, Time, Distance Problems Using Algebra More Algebra Lessons

In these lessons, we will learn how to solve word problems involving average speed.

There are three main types of average problems commonly encountered in school algebra: Average (Arithmetic Mean) Weighted Average and Average Speed.

How to calculate Average Speed?

The following diagram shows the formula for average speed. Scroll down the page for more examples and solutions on calculating the average speed.

Examples Of Average Speed Problems

Example: John drove for 3 hours at a rate of 50 miles per hour and for 2 hours at 60 miles per hour. What was his average speed for the whole journey?

Solution: Step 1: The formula for distance is

Distance = Rate × Time Total distance = 50 × 3 + 60 × 2 = 270

Step 2: Total time = 3 + 2 = 5

Step 3: Using the formula:

Answer: The average speed is 54 miles per hour.

Be careful! You will get the wrong answer if you add the two speeds and divide the answer by two.

How To Solve The Average Speed Problem?

How to calculate the average speed?

Example: The speed paradox: If I drive from Oxford to Cambridge at 40 miles per hour and then from Cambridge to Oxford at 60 miles per hour, what is my average speed for the whole journey?

How To Find The Average Speed For A Round Trip?

Example: On Alberto’s drive to his aunt’s house, the traffic was light, and he drove the 45-mile trip in one hour. However, the return trip took his two hours. What was his average trip for the round trip?

How To Find The Average Speed Of An Airplane With Good And Bad Weather?

Example: Mae took a non-stop flight to visit her grandmother. The 750-mile trip took three hours and 45 minutes. Because of the bad weather, the return trip took four hours and 45 minutes. What was her average speed for the round trip?

How To Relate Speed To Distance And Time?

If you are traveling in a car that travels 80km along a road in one hour, we say that you are traveling at an average of 80kn/h.

Average speed is the total distance divided by the total time for the trip. Therefore, speed is distance divided by time.

Instantaneous speed is the speed at which an object is traveling at any particular instant.

If the instantaneous speed of a car remains the same over a period of time, then we say that the car is traveling with constant speed.

The average speed of an object is the same as its instantaneous speed if that object is traveling at a constant speed.

How To Calculate Average Speed In Word Problems?

Example: Keri rollerblades to school, a total distance of 4.5km. She has to slow down twice to cross busy streets, but overall the journey takes her 0.65h. What is Keri’s average speed during the trip?

How To Use Average Speed To Calculate The Distance Traveled?

Example: Elle drives 169 miles from Sheffield to London. Her average speed is 65 mph. She leaves Sheffield at 6:30 a.m. Does she arrive in London by 9:00 a.m.?

How To Use Average Speed To Calculate The Time Taken?

Example: Marie Ann is trying to predict the time required to ride her bike to the nearby beach. She knows that the distance is 45 km and, from other trips, that she can usually average about 20 km/h. Predict how long the trip will take.

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JEE-IIT-NCERT Physics & Math

Widget atas posting, average velocity and average speed problems and solutions.

Problem #1

Sample Problems

Last updated
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Page ID 63035

Tom Weideman
University of California, Davis

All of the problems below have had their basic features discussed in an "Analyze This" box in this chapter. This means that the solutions provided here are incomplete, as they will refer back to the analysis performed for information (i.e. the full solution is essentially split between the analysis earlier and details here). If you have not yet spent time working on (not simply reading!) the analysis of these situations, these sample problems will be of little benefit to your studies.

Problem 1.1

The acceleration of a particle moving along the $x$-axis is given by the equation:

\[a \left( t \right) = \left(0.300 \frac{m}{s^3}\right)t + \left(2.40 \frac{m}{s^2}\right) \nonumber\]

The particle is at position $x = +4.60m$ and is moving in the $-x$ direction at a speed of $12.0 \dfrac{m}{s}$ at time $t = 0s$.

Find the time at which the particle (briefly) comes to rest.
Find the position where the particle (briefly) comes to rest.

a. From the analysis , we have an equation for the velocity of the particle at all times. Here we are given all the constants we need, namely:

\[\lambda = 0.300 \frac{m}{s^3},\;\;\; a_o = 2.40 \frac{m}{s^2},\;\;\; v_o=-12.0 \frac{m}{s}\nonumber\]

So all we need to do is plug these into the velocity equation, set the velocity equal to zero, and solve for the time in the quadratic equation:

\[v\left(t\right) = 0 = \frac{1}{2}\lambda t^2 + a_o t + v_o \;\;\;\Rightarrow\;\;\; t = \frac{ - a_o \pm \sqrt { a_o^2 - 4 \left( \frac{1}{2} \lambda \right) v_o}}{2 \left( \frac{1}{2} \lambda \right)} = 4.00s\nonumber\]

b. We just computed the time at which it comes to rest, and we already derived the equation for position in the analysis, so we can just plug the values in, noting that the position at time $t=0$ is given to be $x_o = +4.60m$:

\[x \left( t=4 \right) = -21.0m \nonumber \]

Problem 1.2

A ball is thrown vertically upward at the same instant that a second ball is dropped from rest directly above it. The two balls are $12.0m$ apart when they start their motion. Find the maximum speed at which the first ball can be thrown such that it doesn't collide with the second ball before it returns to its starting height. Treat the balls as being very small (i.e. ignore their diameters).

The balls will collide at the point in time derived in the analysis , with the starting difference in height being given as $y_o=12.0m$. The problem states that this time must be at least as long as it takes the lower ball to return to its starting point. In such a flight, the lower ball makes a total displacement of zero, so since we know its acceleration, we can solve for the time of travel in terms of the initial speed:

\[y\left(t\right) = 0 = -\frac{1}{2}gt^2+v_o t\;\;\;\Rightarrow\;\;\; t=\frac{2v_o}{g}\nonumber\]

If we plug this into the equation found in the analysis that relates the starting speed to the time of collision, we will find the starting velocity for which the balls will collide exactly at the lower ball's starting height.:

\[v_o t_{collision} = v_o\left(\frac{2v_o}{g}\right) = y_o\;\;\;\Rightarrow\;\;\; v_o = \sqrt {\frac{g y_o}{2}}=7.67\frac{m}{s}\nonumber\]

Clearly is the lower ball starts at any speed greater than this, then the balls will collide sooner, and they will have not yet fallen to the starting position of the lower ball.

Problem 1.3

A particle moves through space with a velocity vector that varies with time according to:

\[ \overrightarrow v \left( t \right) = \alpha \; \widehat i - \beta t \; \widehat j \;, \nonumber \]

where $\alpha$ and $\beta$ are positive constants. Find the rate at which the speed of this particle is changing at time $t=0$. Does this rate remain the same for all later times?

We already did all the math we need in the analysis . The rate of speed change is just $\dfrac{dv}{dt}$, which is computed in the analysis. Plugging-in $t=0$ gives a rate of speed change that equals zero! We see this does not remain true for all values of $t$, because it only vanishes at $t=0$.

The reason for this is that the acceleration vector is a constant, and is initially perpendicular to the velocity vector:

\[\overrightarrow v\left(0\right) = \alpha \hat i\;\;\;\;\; \overrightarrow a\left(0\right) = -\beta \hat j\nonumber\]

So at that moment, the acceleration only changes the direction of motion (does not speed it up). But after $t=0$, the constant acceleration has not changed, and the particle is moving in a new direction, so the acceleration then does change the speed.

Problem 1.4

A bead is threaded onto a circular hoop of wire which lies in a vertical plane. The bead starts at the bottom of the hoop from rest, and is pushed around the hoop such that it speeds up at a steady rate. Find the angle that the bead's acceleration vector makes with the horizontal when it gets back to the bottom of the hoop.

As stated in the analysis , we can treat the motion tangent to the circle like any other 1-dimensional accelerated motion. In this case, the distance the bead travels is given, so the "no time" kinematics equation ( Equation 1.4.3 ) is most applicable. Let's call the radius of the circle $ R$ and the final velocity $v$. The tangential acceleration is constant, the bead starts from rest, and the bead travels one circumference, so we get:

\[2a \Delta x = {v_f}^2 - {v_o}^2 \;\;\; \Rightarrow \;\;\; a_\parallel=\dfrac{\left(v^2-0^2\right)}{2\left(2\pi R \right)}=\dfrac{v^2}{4\pi R} \nonumber\]

The centripetal acceleration is toward the center of the circle, so it points upward and its magnitude is simply:

\[a_\bot=\dfrac{v^2}{R} \nonumber\]

The tangent of the angle that the full acceleration vector makes with the horizontal is the vertical component divided by the horizontal component, so:

\[\theta = \tan^{-1} \left(\dfrac{\dfrac{v^2}{R}}{ \dfrac{v^2}{4\pi R}}\right)=\tan^{-1} \left(4\pi \right) = 85^o \nonumber\]

Problem 1.5

A cannonball is fired at an angle $\theta$ up from the horizontal at a speed of $v_o$ along level ground. A second cannonball is fired at the same speed, but at a different angle. Both cannonballs travel the same horizontal distance before landing, but one of the cannonballs takes twice as long to make the journey as the other. Find the two angles at which the cannonballs are launched.

In the analysis , we found that except for a $45^o$ launch angle, there are two values that correspond to the same range for a given launch velocity, and that these angles are complementary. From the vertical equation in the analysis, we have the following flight time for a given angle and launch speed:

\[t=\dfrac{2v_o\sin\theta}{g}\nonumber\]

This applies to both cannonballs, so accumulating everything together we get:

\[\left. \begin{array}{l} t_1=\dfrac{2v_o\sin\theta_1}{g} \\ t_2=\dfrac{2v_o\sin\theta_2}{g} \\ \theta_1=90^o-\theta_2\ \\ t_1=2t_2 \end{array}\right\}\;\;\;\Rightarrow\;\;\; \sin\left(90^o-\theta_2\right) = 2\sin\theta_2\;\;\;\Rightarrow\;\;\; \cos\theta_2=2\sin\theta_2\;\;\;\Rightarrow\;\;\; \theta_2=\tan^{-1}0.5=26.6^o,\;\;\theta_1=63.4^o\nonumber\]

Problem 1.6

Two warlords aim identical catapults (i.e. they both release rocks at the same speed) at each other, with both of them being at the same altitude. The warlords have made the necessary computations to crush the other, and fire their catapults simultaneously. Amazingly, the two stones do not collide with each other in mid-air, but instead the stone Alexander fired passes well below the stone that Genghis shot. Genghis is annihilated 8.0s after the catapults are fired, and Alexander only got to celebrate his victory for 4.0s before he too was destroyed.

Find the maximum height reached by each of the rocks.
Find the amount of time that elapses from the launch to the moment that the rocks pass each other in the air.
Find the angles at which each warlord fires his rock.

Conceptual analysis of this problem is found here .

a. The time it takes a rock to travel to its peak height and back down again is equal to twice the time it takes to travel down from its peak height. Traveling down from its peak height, it starts with zero initial velocity, so we can calculate the height immediately for each rock:

\[ h_{A} = \frac{1}{2}g\left(\frac{t_A}{2}\right)^2 = \frac{1}{2} \left( 9.8 \dfrac{m}{s^2} \right) \left( \frac{8.0s}{2} \right)^2 = 78.4m \\ h_{G} = \frac{1}{2}g\left(\frac{t_G}{2}\right)^2 = \frac{1}{2} \left( 9.8 \dfrac{m}{s^2} \right) \left( \frac{8.0s+4.0s}{2} \right)^2 = 176.4m \nonumber \]

b. The x–components of the velocities of the rocks never change, and since it takes 12s for Genghis’s rock to travel the same horizontal distance as Alexander’s rock traveled in 8s, Alexander’s rock is traveling in the x-direction at a rate 1.5 times as great as Genghis’s rock is traveling in the x–direction. When they are at the same x–position (passing each other), the distance each has traveled is each one’s velocity times the time we are looking for, and we can express both of these distances in terms of the x–component of Genghis’s rock using the ratio described above:

\[ \begin{array}{l} x_A=v_{Ax}t, \;\;\;\;\; v_{Ax}=1.5v_{Gx} \;\;\; \Rightarrow \;\;\; x_A=1.5v_{Gx}t \\ x_G=v_{Gx}t \nonumber \end{array} \]

Since the rocks travel from both ends and are now at the same horizontal position, the sum of the distances they travel equals the total separation of the two warlords. This allows us to calculate the time:

\[ x_A + x_G = x_{tot} = v_G t_{tot} \;\;\; \Rightarrow \;\;\; \left( 1.5v_G \right) t + v_G t = v_G \left( 12.0s \right) \;\;\; \Rightarrow \;\;\; t=\dfrac{12.0s}{2.5} = 4.8s \nonumber \]

c. Clearly there are two different angles that will result in the rock traveling the same distance. One can see this from the range equation, but from a physical standpoint, this happens because one rock spends less time in the air but has a greater x–velocity, while the other spends more time in the air with a smaller x–velocity. To spend 1.5 times as long in the air, Genghis’s rock needs to start with 1.5 times as much vertical component of velocity as Alexander’s rock. This means that the ratios of the x and y components of the two rock velocities are inverses of one another, which means that the two angles are complimentary (i.e. $\theta_A = 90^o – \theta_G$). But the total speeds of the rocks are the same, so:

\[\left. \begin{array}{l} v_{Ax} = v_o \cos \theta _A = v_o \cos \left( 90^o - \theta _G \right) = v_o\sin \theta _G \\ v_{Gx} = v_o\cos \theta _G \end{array} \right\}\;\;\; \Rightarrow \;\;\;\dfrac{v_{Ax}}{v_{Gx}} = 1.5 = \dfrac{\sin \theta _G}{\cos \theta _G}\;\;\; \Rightarrow \;\;\;\left\{ \begin{array}{l} \theta _G = \tan ^{ - 1}1.5 = 56.3^o\\ \theta _A = 90^o - \theta _G = 33.7^o \end{array} \right. \nonumber\]

Problem 1.7

You stand on the bank of a river, contemplating swimming across, but the place where you hope to cross is just upstream of a dangerous waterfall. When you look at the speed of the river, you estimate that it is about the same speed as you are able to swim. You realize that you can only swim so far in the cold water at this speed before your muscles shut down, and in still water you estimate that this distance is about $100m$. The width of the river is about $80m$.

Find the minimum distance that you must start upstream of the waterfall in order to not be swept over it.
If the river flows west-to-east and you start on its south shore, compute the direction in which you must swim in order to get safely across if you leave from the starting point computed in part (a).

The analysis discusses the relevant reference frames in this problem: the river, the swimmer, and the Earth.

a. Clearly to minimize the distance upstream that you need to start, you must swim with a component of your velocity relative to the river being upstream. The more you are able to turn yourself upstream, the less you will float downstream, and the closer you can start to the waterfall. But there is a limit to how far you can swim relative to the water, so your angle with the river must be such that when you reach your limit relative to the river, you reach the other side. The velocities are all constant and the time spans are all equal, so they are proportional to the displacements, which we can draw:

We are given that the speed of the river relative to the earth is the same as the speed of the swimmer relative to the water, so we’ll call that quantity $v$, and the width of the river (which we know), we’ll call w. From the Pythagorean theorem we can get the distance swum upstream against the current:

\[distance \; swum \; against \; water = \sqrt{\left(vt \right)^2 - \left(w \right)^2} \nonumber \]

The distance the water moves downstream relative to the earth is clearly vt, so the total distance the swimmer moves downstream is:

\[x = vt - \sqrt{\left(vt \right)^2 - \left(w \right)^2} \nonumber \]

But we actually know the value of $vt$, because it is the maximum distance that the swimmer can go in the water. Plugging in all the values therefore gives our answer:

\[x = \left(100m \right) - \sqrt{\left(100m \right)^2 - \left(80m \right)^2} = 40m \nonumber \]

b. The angle is easy to determine, since we know the length of the displacement vector of the swimmer relative to the water and the width of the river:

\[\cos\theta = \dfrac{w}{vt} = \dfrac{80m}{100m} \;\;\; \Rightarrow \;\;\; \theta = 37^o west \; of \; north \nonumber \]

Distance, Time and Speed Word Problems | GMAT GRE Maths

Before you get into distance, time and speed word problems, check out how you can add a little power to your resume by getting our mini-MBA certificate.

Problems involving Time, Distance and Speed are solved based on one simple formula.

Distance = Speed * Time

Which implies →

Speed = Distance / Time and

Time = Distance / Speed

Let us take a look at some simple examples of distance, time and speed problems. Example 1. A boy walks at a speed of 4 kmph. How much time does he take to walk a distance of 20 km?

Time = Distance / speed = 20/4 = 5 hours. Example 2. A cyclist covers a distance of 15 miles in 2 hours. Calculate his speed.

Speed = Distance/time = 15/2 = 7.5 miles per hour. Example 3. A car takes 4 hours to cover a distance, if it travels at a speed of 40 mph. What should be its speed to cover the same distance in 1.5 hours?

Distance covered = 4*40 = 160 miles

Speed required to cover the same distance in 1.5 hours = 160/1.5 = 106.66 mph Now, take a look at the following example:

Example 4. If a person walks at 4 mph, he covers a certain distance. If he walks at 9 mph, he covers 7.5 miles more. How much distance did he actually cover?

Now we can see that the direct application of our usual formula Distance = Speed * Time or its variations cannot be done in this case and we need to put in extra effort to calculate the given parameters.

Let us see how this question can be solved.

For these kinds of questions, a table like this might make it easier to solve.

Let the distance covered by that person be ‘d’.

Walking at 4 mph and covering a distance ‘d’ is done in a time of ‘d/4’

IF he walks at 9 mph, he covers 7.5 miles more than the actual distance d, which is ‘d+7.5’.

He does this in a time of (d+7.5)/9.

Since the time is same in both the cases →

d/4 = (d+7.5)/9 → 9d = 4(d+7.5) → 9d=4d+30 → d = 6.

So, he covered a distance of 6 miles in 1.5 hours. Example 5. A train is going at 1/3 of its usual speed and it takes an extra 30 minutes to reach its destination. Find its usual time to cover the same distance.

Here, we see that the distance is same.

Let us assume that its usual speed is ‘s’ and time is ‘t’, then

s*t = (1/3)s*(t+30) → t = t/3 + 10 → t = 15.

So the actual time taken to cover the distance is 15 minutes.

Note: Note the time is expressed in terms of ‘minutes’. When we express distance in terms of miles or kilometers, time is expressed in terms of hours and has to be converted into appropriate units of measurement.

Solved Questions on Trains

Example 1. X and Y are two stations which are 320 miles apart. A train starts at a certain time from X and travels towards Y at 70 mph. After 2 hours, another train starts from Y and travels towards X at 20 mph. At what time do they meet?

Let the time after which they meet be ‘t’ hours.

Then the time travelled by second train becomes ‘t-2’.

Distance covered by first train+Distance covered by second train = 320 miles

70t+20(t-2) = 320

Solving this gives t = 4.

So the two trains meet after 4 hours. Example 2. A train leaves from a station and moves at a certain speed. After 2 hours, another train leaves from the same station and moves in the same direction at a speed of 60 mph. If it catches up with the first train in 4 hours, what is the speed of the first train?

Let the speed of the first train be ‘s’.

Distance covered by the first train in (2+4) hours = Distance covered by second train in 4 hours

Therefore, 6s = 60*4

Solving which gives s=40.

So the slower train is moving at the rate of 40 mph.

Questions on Boats/Airplanes

For problems with boats and streams,

Speed of the boat upstream (against the current) = Speed of the boat in still water – speed of the stream

[As the stream obstructs the speed of the boat in still water, its speed has to be subtracted from the usual speed of the boat]

Speed of the boat downstream (along with the current) = Speed of the boat in still water + speed of the stream

[As the stream pushes the boat and makes it easier for the boat to reach the destination faster, speed of the stream has to be added]

Similarly, for airplanes travelling with/against the wind,

Speed of the plane with the wind = speed of the plane + speed of the wind

Speed of the plane against the wind = speed of the plane – speed of the wind

Let us look at some examples.

Example 1. A man travels at 3 mph in still water. If the current’s velocity is 1 mph, it takes 3 hours to row to a place and come back. How far is the place?

Let the distance be ‘d’ miles.

Time taken to cover the distance upstream + Time taken to cover the distance downstream = 3

Speed upstream = 3-1 = 2 mph

Speed downstream = 3+1 = 4 mph

So, our equation would be d/2 + d/4 = 3 → solving which, we get d = 4 miles. Example 2. With the wind, an airplane covers a distance of 2400 kms in 4 hours and against the wind in 6 hours. What is the speed of the plane and that of the wind?

Let the speed of the plane be ‘a’ and that of the wind be ‘w’.

Our table looks like this:

4(a+w) = 2400 and 6(a-w) = 2400

Expressing one unknown variable in terms of the other makes it easier to solve, which means

a+w = 600 → w=600-a

Substituting the value of w in the second equation,

a-(600-a) = 400 → a = 500

The speed of the plane is 500 kmph and that of the wind is 100 kmph.

More solved examples on Speed, Distance and Time

Example 1. A boy travelled by train which moved at the speed of 30 mph. He then boarded a bus which moved at the speed of 40 mph and reached his destination. The entire distance covered was 100 miles and the entire duration of the journey was 3 hours. Find the distance he travelled by bus.

Let the time taken by the train be ‘t’. Then that of bus is ‘3-t’.

The entire distance covered was 100 miles

So, 30t + 40(3-t) = 100

Solving which gives t=2.

Substituting the value of t in 40(3-t), we get the distance travelled by bus is 40 miles.

Alternatively, we can add the time and equate it to 3 hours, which directly gives the distance.

d/30 + (100-d)/40 = 3

Solving which gives d = 60, which is the distance travelled by train. 100-60 = 40 miles is the distance travelled by bus. Example 2. A plane covered a distance of 630 miles in 6 hours. For the first part of the trip, the average speed was 100 mph and for the second part of the trip, the average speed was 110 mph. what is the time it flew at each speed?

Our table looks like this.

Assuming the distance covered in the 1 st part of journey to be ‘d’, the distance covered in the second half becomes ‘630-d’.

Assuming the time taken for the first part of the journey to be ‘t’, the time taken for the second half becomes ‘6-t’.

From the first equation, d=100t

The second equation is 630-d = 110(6-t).

Substituting the value of d from the first equation, we get

630-100t = 110(6-t)

Solving this gives t=3.

So the plane flew the first part of the journey in 3 hours and the second part in 3 hours. Example 2. Two persons are walking towards each other on a walking path that is 20 miles long. One is walking at the rate of 3 mph and the other at 4 mph. After how much time will they meet each other?

Assuming the distance travelled by the first person to be ‘d’, the distance travelled by the second person is ’20-d’.

The time is ‘t’ for both of them because when they meet, they would have walked for the same time.

Since time is same, we can equate as

d/3 = (20-d)/4

Solving this gives d=60/7 miles (8.5 miles approximately)

Then t = 20/7 hours

So the two persons meet after 2 6/7 hours.

Practice Questions for you to solve

Problem 1: Click here

A boat covers a certain distance in 2 hours, while it comes back in 3 hours. If the speed of the stream is 4 kmph, what is the boat’s speed in still water?

A) 30 kmph B) 20 kmph C) 15 kmph D) 40 kmph

Answer 1: Click here

Explanation

Let the speed of the boat be ‘s’ kmph.

Then, 2(s+4) = 3(s-4) → s = 20

Problem 2: Click here

A cyclist travels for 3 hours, travelling for the first half of the journey at 12 mph and the second half at 15 mph. Find the total distance he covered.

A) 30 miles B) 35 miles C) 40 miles D) 180 miles

Answer 2: Click here

Since it is mentioned, that the first ‘half’ of the journey is covered in 12 mph and the second in 15, the equation looks like

(d/2)/12 + (d/2)/15 = 3

Solving this gives d = 40 miles

17 thoughts on “Distance, Time and Speed Word Problems | GMAT GRE Maths”

Meera walked to school at a speed of 3 miles per hour. Once she reached the school, she realized that she forgot to bring her books, so rushed back home at a speed of 6 miles per hour. She then walked back to school at a speed of 4 miles per hour. All the times, she walked in the same route. please explain above problem

When she walks faster the time she takes to reach her home and school is lower. There is nothing wrong with the statement. They never mentioned how long she took every time.

a man covers a distance on a toy train.if the train moved 4km/hr faster,it would take 30 min. less. if it moved 2km/hr slower, it would have taken 20 min. more .find the distance.

Let the speed be x. and time be y. A.T.Q, (x+4)(y-1/2)=d and (x-2)(y+1/3)=d. Equate these two and get the answer

Could you explain how ? you have two equations and there are 3 variables.

The 3rd equation is d=xy. Now, you have 3 equations with 3 unknowns. The variables x and y represent the usual speed and usual time to travel distance d.

Speed comes out to be 20 km/hr and the time taken is 3 hrs. The distance traveled is 60 km.

(s + 4) (t – 1/2)= st 1…new equotion = -1/2s + 4t = 2

(s – 2) (t + 1/3)= st 2…new equotion = 1/3s – 2t = 2/3

Multiply all by 6 1… -3s + 24t = 12 2… 2s – 12t = 4 Next, use elimination t= 3 Find s: -3s + 24t = 12 -3s + 24(3) = 12 -3s = -60 s= 20

st or distance = 3 x 20 = 60 km/h

It’s probably the average speed that we are looking for here. Ave. Speed= total distance/ total time. Since it’s harder to look for one variable since both are absent, you can use, 3d/ d( V2V3 + V1V3 + V1V2/ V1V2V3)

2 girls meenu and priya start at the same time to ride from madurai to manamadurai, 60 km away.meenu travels 4kmph slower than priya. priya reaches manamadurai and at turns back meeting meenu 12km from manamaduai. find meenu’s speed?

Hi, when the two girls meet, they have taken equal time to travel their respective distance. So, we just need to equate their time equations

Distance travelled by Meenu = 60 -12 = 48 Distance travelled by Priya = 60 + 12 = 72 Let ‘s’ be the speed of Meenu

Time taken by Meenu => t1 = 48/s Time taken by Priya => t2 = 72/(s+4)

t1=t2 Thus, 48/s = 72/(s+4) => 24s = 192 => s = 8Km/hr

A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 KMS away from A at the same time. On the way, however the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:

Let speed of the CAR BE x kmph.

Then, speed of the train = 3/2(x) .’. 75/x – 75/(3/2)x= 125/(10*60) — subtracting the times travelled by two them hence trains wastage time

therefore x= 120 kmph

A cyclist completes a distance of 60 km at the same speed throughout. She travels 10 km in one hour. She stops every 20 km for one hour to have a break. What are the two variables involved in this situation?

For the answer, not variables: 60km divided by 10km/h=6 hours 60 divided by 20= 3 hours 3 hours+6 hours= 9 hours Answer: 9 hours

Let the length of the train to prod past a point be the intrinsic distance (D) of the train and its speed be S. Its speed, S in passing the electric pole of negligible length is = D/12. The length of the platform added to the intrinsic length of the train. So, the total distance = D + 200. The time = 20 secs. The Speed, S = (D + 200)/20 At constant speed, D/12 = (D + 200)/20 Cross-multiplying, 20D = 12D + 200*12 20D – 12D = 200*12; 8D = 200*12 D = 200*12/8 = 300m. 4th Aug, 2018

Can anyone solve this? Nathan and Philip agree to meet up at the park at 5:00 pm. Nathan lives 300 m due north of the park, and Philip lives 500 m due west of the park. Philip leaves his house at 4:54 pm and walks towards the park at a pace of 1.5 m/s, but Nathan loses track of time and doesn’t leave until 4:59 pm. Trying to avoid being too late, he jogs towards the park at 2.5 m/s. At what rate is the distance between the two friends changing 30 seconds after Nathan has departed?

Check Your Understanding

Answer: d = 1720 m

Answer: a = 8.10 m/s/s

Answers: d = 33.1 m and v f = 25.5 m/s

Answers: a = 11.2 m/s/s and d = 79.8 m

Answer: t = 1.29 s

Answers: a = 243 m/s/s

Answer: a = 0.712 m/s/s

Answer: d = 704 m

Answer: d = 28.6 m

Answer: v i = 7.17 m/s

Answer: v i = 5.03 m/s and hang time = 1.03 s (except for in sports commericals)

Answer: a = 1.62*10 5 m/s/s

Answer: d = 48.0 m

Answer: t = 8.69 s

Answer: a = -1.08*10^6 m/s/s

Answer: d = -57.0 m (57.0 meters deep)

Answer: v i = 47.6 m/s

Answer: a = 2.86 m/s/s and t = 30. 8 s

Answer: a = 15.8 m/s/s

Answer: v i = 94.4 mi/hr

Solutions to Above Problems

d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s 2 )*(32.8 s) 2

Return to Problem 1

110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s) 2

110 m = (13.57 s 2 )*a

a = (110 m)/(13.57 s 2 )

a = 8.10 m/ s 2

Return to Problem 2

d = (0 m/s)*(2.60 s)+ 0.5*(-9.8 m/s 2 )*(2.60 s) 2

d = -33.1 m (- indicates direction)

v f = v i + a*t

v f = 0 + (-9.8 m/s 2 )*(2.60 s)

v f = -25.5 m/s (- indicates direction)

Return to Problem 3

a = (46.1 m/s - 18.5 m/s)/(2.47 s)

a = 11.2 m/s 2

d = v i *t + 0.5*a*t 2

d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s 2 )*(2.47 s) 2

d = 45.7 m + 34.1 m

(Note: the d can also be calculated using the equation v f 2 = v i 2 + 2*a*d)

Return to Problem 4

-1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s 2 )*(t) 2

-1.40 m = 0+ (-0.835 m/s 2 )*(t) 2

(-1.40 m)/(-0.835 m/s 2 ) = t 2

1.68 s 2 = t 2

Return to Problem 5

a = (444 m/s - 0 m/s)/(1.83 s)

a = 243 m/s 2

d = (0 m/s)*(1.83 s)+ 0.5*(243 m/s 2 )*(1.83 s) 2

d = 0 m + 406 m

Return to Problem 6

(7.10 m/s) 2 = (0 m/s) 2 + 2*(a)*(35.4 m)

50.4 m 2 /s 2 = (0 m/s) 2 + (70.8 m)*a

(50.4 m 2 /s 2 )/(70.8 m) = a

a = 0.712 m/s 2

Return to Problem 7

(65 m/s) 2 = (0 m/s) 2 + 2*(3 m/s 2 )*d

4225 m 2 /s 2 = (0 m/s) 2 + (6 m/s 2 )*d

(4225 m 2 /s 2 )/(6 m/s 2 ) = d

Return to Problem 8

d = (22.4 m/s + 0 m/s)/2 *2.55 s

d = (11.2 m/s)*2.55 s

Return to Problem 9

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(2.62 m)

0 m 2 /s 2 = v i 2 - 51.35 m 2 /s 2

51.35 m 2 /s 2 = v i 2

v i = 7.17 m/s

Return to Problem 10

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(1.29 m)

0 m 2 /s 2 = v i 2 - 25.28 m 2 /s 2

25.28 m 2 /s 2 = v i 2

v i = 5.03 m/s

To find hang time, find the time to the peak and then double it.

0 m/s = 5.03 m/s + (-9.8 m/s 2 )*t up

-5.03 m/s = (-9.8 m/s 2 )*t up

(-5.03 m/s)/(-9.8 m/s 2 ) = t up

t up = 0.513 s

hang time = 1.03 s

Return to Problem 11

(521 m/s) 2 = (0 m/s) 2 + 2*(a)*(0.840 m)

271441 m 2 /s 2 = (0 m/s) 2 + (1.68 m)*a

(271441 m 2 /s 2 )/(1.68 m) = a

a = 1.62*10 5 m /s 2

Return to Problem 12

(NOTE: the time required to move to the peak of the trajectory is one-half the total hang time - 3.125 s.)

First use: v f = v i + a*t

0 m/s = v i + (-9.8 m/s 2 )*(3.13 s)

0 m/s = v i - 30.7 m/s

v i = 30.7 m/s (30.674 m/s)

Now use: v f 2 = v i 2 + 2*a*d

(0 m/s) 2 = (30.7 m/s) 2 + 2*(-9.8 m/s 2 )*(d)

0 m 2 /s 2 = (940 m 2 /s 2 ) + (-19.6 m/s 2 )*d

-940 m 2 /s 2 = (-19.6 m/s 2 )*d

(-940 m 2 /s 2 )/(-19.6 m/s 2 ) = d

Return to Problem 13

-370 m = (0 m/s)*(t)+ 0.5*(-9.8 m/s 2 )*(t) 2

-370 m = 0+ (-4.9 m/s 2 )*(t) 2

(-370 m)/(-4.9 m/s 2 ) = t 2

75.5 s 2 = t 2

Return to Problem 14

(0 m/s) 2 = (367 m/s) 2 + 2*(a)*(0.0621 m)

0 m 2 /s 2 = (134689 m 2 /s 2 ) + (0.1242 m)*a

-134689 m 2 /s 2 = (0.1242 m)*a

(-134689 m 2 /s 2 )/(0.1242 m) = a

a = -1.08*10 6 m /s 2

(The - sign indicates that the bullet slowed down.)

Return to Problem 15

d = (0 m/s)*(3.41 s)+ 0.5*(-9.8 m/s 2 )*(3.41 s) 2

d = 0 m+ 0.5*(-9.8 m/s 2 )*(11.63 s 2 )

d = -57.0 m

(NOTE: the - sign indicates direction)

Return to Problem 16

(0 m/s) 2 = v i 2 + 2*(- 3.90 m/s 2 )*(290 m)

0 m 2 /s 2 = v i 2 - 2262 m 2 /s 2

2262 m 2 /s 2 = v i 2

v i = 47.6 m /s

Return to Problem 17

( 88.3 m/s) 2 = (0 m/s) 2 + 2*(a)*(1365 m)

7797 m 2 /s 2 = (0 m 2 /s 2 ) + (2730 m)*a

7797 m 2 /s 2 = (2730 m)*a

(7797 m 2 /s 2 )/(2730 m) = a

a = 2.86 m/s 2

88.3 m/s = 0 m/s + (2.86 m/s 2 )*t

(88.3 m/s)/(2.86 m/s 2 ) = t

t = 30. 8 s

Return to Problem 18

( 112 m/s) 2 = (0 m/s) 2 + 2*(a)*(398 m)

12544 m 2 /s 2 = 0 m 2 /s 2 + (796 m)*a

12544 m 2 /s 2 = (796 m)*a

(12544 m 2 /s 2 )/(796 m) = a

a = 15.8 m/s 2

Return to Problem 19

v f 2 = v i 2 + 2*a*d

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(91.5 m)

0 m 2 /s 2 = v i 2 - 1793 m 2 /s 2

1793 m 2 /s 2 = v i 2

v i = 42.3 m/s

Now convert from m/s to mi/hr:

v i = 42.3 m/s * (2.23 mi/hr)/(1 m/s)

v i = 94.4 mi/hr

Return to Problem 20

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Published: 10 April 2024

A hybrid particle swarm optimization algorithm for solving engineering problem

Jinwei Qiao 1 , 2 ,
Guangyuan Wang 1 , 2 ,
Zhi Yang 1 , 2 ,
Xiaochuan Luo 3 ,
Jun Chen 1 , 2 ,
Kan Li 4 &
Pengbo Liu 1 , 2

Scientific Reports volume 14 , Article number: 8357 ( 2024 ) Cite this article

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Computational science
Mechanical engineering

To overcome the disadvantages of premature convergence and easy trapping into local optimum solutions, this paper proposes an improved particle swarm optimization algorithm (named NDWPSO algorithm) based on multiple hybrid strategies. Firstly, the elite opposition-based learning method is utilized to initialize the particle position matrix. Secondly, the dynamic inertial weight parameters are given to improve the global search speed in the early iterative phase. Thirdly, a new local optimal jump-out strategy is proposed to overcome the "premature" problem. Finally, the algorithm applies the spiral shrinkage search strategy from the whale optimization algorithm (WOA) and the Differential Evolution (DE) mutation strategy in the later iteration to accelerate the convergence speed. The NDWPSO is further compared with other 8 well-known nature-inspired algorithms (3 PSO variants and 5 other intelligent algorithms) on 23 benchmark test functions and three practical engineering problems. Simulation results prove that the NDWPSO algorithm obtains better results for all 49 sets of data than the other 3 PSO variants. Compared with 5 other intelligent algorithms, the NDWPSO obtains 69.2%, 84.6%, and 84.6% of the best results for the benchmark function ( ${f}_{1}-{f}_{13}$ ) with 3 kinds of dimensional spaces (Dim = 30,50,100) and 80% of the best optimal solutions for 10 fixed-multimodal benchmark functions. Also, the best design solutions are obtained by NDWPSO for all 3 classical practical engineering problems.

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Introduction

In the ever-changing society, new optimization problems arise every moment, and they are distributed in various fields, such as automation control 1 , statistical physics 2 , security prevention and temperature prediction 3 , artificial intelligence 4 , and telecommunication technology 5 . Faced with a constant stream of practical engineering optimization problems, traditional solution methods gradually lose their efficiency and convenience, making it more and more expensive to solve the problems. Therefore, researchers have developed many metaheuristic algorithms and successfully applied them to the solution of optimization problems. Among them, Particle swarm optimization (PSO) algorithm 6 is one of the most widely used swarm intelligence algorithms.

However, the basic PSO has a simple operating principle and solves problems with high efficiency and good computational performance, but it suffers from the disadvantages of easily trapping in local optima and premature convergence. To improve the overall performance of the particle swarm algorithm, an improved particle swarm optimization algorithm is proposed by the multiple hybrid strategy in this paper. The improved PSO incorporates the search ideas of other intelligent algorithms (DE, WOA), so the improved algorithm proposed in this paper is named NDWPSO. The main improvement schemes are divided into the following 4 points: Firstly, a strategy of elite opposition-based learning is introduced into the particle population position initialization. A high-quality initialization matrix of population position can improve the convergence speed of the algorithm. Secondly, a dynamic weight methodology is adopted for the acceleration coefficients by combining the iterative map and linearly transformed method. This method utilizes the chaotic nature of the mapping function, the fast convergence capability of the dynamic weighting scheme, and the time-varying property of the acceleration coefficients. Thus, the global search and local search of the algorithm are balanced and the global search speed of the population is improved. Thirdly, a determination mechanism is set up to detect whether the algorithm falls into a local optimum. When the algorithm is “premature”, the population resets 40% of the position information to overcome the local optimum. Finally, the spiral shrinking mechanism combined with the DE/best/2 position mutation is used in the later iteration, which further improves the solution accuracy.

The structure of the paper is given as follows: Sect. “ Particle swarm optimization (PSO) ” describes the principle of the particle swarm algorithm. Section “ Improved particle swarm optimization algorithm ” shows the detailed improvement strategy and a comparison experiment of inertia weight is set up for the proposed NDWPSO. Section “ Experiment and discussion ” includes the experimental and result discussion sections on the performance of the improved algorithm. Section “ Conclusions and future works ” summarizes the main findings of this study.

Literature review

This section reviews some metaheuristic algorithms and other improved PSO algorithms. A simple discussion about recently proposed research studies is given.

Metaheuristic algorithms

A series of metaheuristic algorithms have been proposed in recent years by using various innovative approaches. For instance, Lin et al. 7 proposed a novel artificial bee colony algorithm (ABCLGII) in 2018 and compared ABCLGII with other outstanding ABC variants on 52 frequently used test functions. Abed-alguni et al. 8 proposed an exploratory cuckoo search (ECS) algorithm in 2021 and carried out several experiments to investigate the performance of ECS by 14 benchmark functions. Brajević 9 presented a novel shuffle-based artificial bee colony (SB-ABC) algorithm for solving integer programming and minimax problems in 2021. The experiments are tested on 7 integer programming problems and 10 minimax problems. In 2022, Khan et al. 10 proposed a non-deterministic meta-heuristic algorithm called Non-linear Activated Beetle Antennae Search (NABAS) for a non-convex tax-aware portfolio selection problem. Brajević et al. 11 proposed a hybridization of the sine cosine algorithm (HSCA) in 2022 to solve 15 complex structural and mechanical engineering design optimization problems. Abed-Alguni et al. 12 proposed an improved Salp Swarm Algorithm (ISSA) in 2022 for single-objective continuous optimization problems. A set of 14 standard benchmark functions was used to evaluate the performance of ISSA. In 2023, Nadimi et al. 13 proposed a binary starling murmuration optimization (BSMO) to select the effective features from different important diseases. In the same year, Nadimi et al. 14 systematically reviewed the last 5 years' developments of WOA and made a critical analysis of those WOA variants. In 2024, Fatahi et al. 15 proposed an Improved Binary Quantum-based Avian Navigation Optimizer Algorithm (IBQANA) for the Feature Subset Selection problem in the medical area. Experimental evaluation on 12 medical datasets demonstrates that IBQANA outperforms 7 established algorithms. Abed-alguni et al. 16 proposed an Improved Binary DJaya Algorithm (IBJA) to solve the Feature Selection problem in 2024. The IBJA’s performance was compared against 4 ML classifiers and 10 efficient optimization algorithms.

Improved PSO algorithms

Many researchers have constantly proposed some improved PSO algorithms to solve engineering problems in different fields. For instance, Yeh 17 proposed an improved particle swarm algorithm, which combines a new self-boundary search and a bivariate update mechanism, to solve the reliability redundancy allocation problem (RRAP) problem. Solomon et al. 18 designed a collaborative multi-group particle swarm algorithm with high parallelism that was used to test the adaptability of Graphics Processing Units (GPUs) in distributed computing environments. Mukhopadhyay and Banerjee 19 proposed a chaotic multi-group particle swarm optimization (CMS-PSO) to estimate the unknown parameters of an autonomous chaotic laser system. Duan et al. 20 designed an improved particle swarm algorithm with nonlinear adjustment of inertia weights to improve the coupling accuracy between laser diodes and single-mode fibers. Sun et al. 21 proposed a particle swarm optimization algorithm combined with non-Gaussian stochastic distribution for the optimal design of wind turbine blades. Based on a multiple swarm scheme, Liu et al. 22 proposed an improved particle swarm optimization algorithm to predict the temperatures of steel billets for the reheating furnace. In 2022, Gad 23 analyzed the existing 2140 papers on Swarm Intelligence between 2017 and 2019 and pointed out that the PSO algorithm still needs further research. In general, the improved methods can be classified into four categories:

Adjusting the distribution of algorithm parameters. Feng et al. 24 used a nonlinear adaptive method on inertia weights to balance local and global search and introduced asynchronously varying acceleration coefficients.

Changing the updating formula of the particle swarm position. Both papers 25 and 26 used chaotic mapping functions to update the inertia weight parameters and combined them with a dynamic weighting strategy to update the particle swarm positions. This improved approach enables the particle swarm algorithm to be equipped with fast convergence of performance.

The initialization of the swarm. Alsaidy and Abbood proposed 27 a hybrid task scheduling algorithm that replaced the random initialization of the meta-heuristic algorithm with the heuristic algorithms MCT-PSO and LJFP-PSO.

Combining with other intelligent algorithms: Liu et al. 28 introduced the differential evolution (DE) algorithm into PSO to increase the particle swarm as diversity and reduce the probability of the population falling into local optimum.

Particle swarm optimization (PSO)

The particle swarm optimization algorithm is a population intelligence algorithm for solving continuous and discrete optimization problems. It originated from the social behavior of individuals in bird and fish flocks 6 . The core of the PSO algorithm is that an individual particle identifies potential solutions by flight in a defined constraint space adjusts its exploration direction to approach the global optimal solution based on the shared information among the group, and finally solves the optimization problem. Each particle $i$ includes two attributes: velocity vector ${V}_{i}=\left[{v}_{i1},{v}_{i2},{v}_{i3},{...,v}_{ij},{...,v}_{iD},\right]$ and position vector ${X}_{i}=[{x}_{i1},{x}_{i2},{x}_{i3},...,{x}_{ij},...,{x}_{iD}]$ . The velocity vector is used to modify the motion path of the swarm; the position vector represents a potential solution for the optimization problem. Here, $j=\mathrm{1,2},\dots ,D$ , $D$ represents the dimension of the constraint space. The equations for updating the velocity and position of the particle swarm are shown in Eqs. ( 1 ) and ( 2 ).

Here ${Pbest}_{i}^{k}$ represents the previous optimal position of the particle $i$ , and ${Gbest}$ is the optimal position discovered by the whole population. $i=\mathrm{1,2},\dots ,n$ , $n$ denotes the size of the particle swarm. ${c}_{1}$ and ${c}_{2}$ are the acceleration constants, which are used to adjust the search step of the particle 29 . ${r}_{1}$ and ${r}_{2}$ are two random uniform values distributed in the range $[\mathrm{0,1}]$ , which are used to improve the randomness of the particle search. $\omega$ inertia weight parameter, which is used to adjust the scale of the search range of the particle swarm 30 . The basic PSO sets the inertia weight parameter as a time-varying parameter to balance global exploration and local seeking. The updated equation of the inertia weight parameter is given as follows:

where ${\omega }_{max}$ and ${\omega }_{min}$ represent the upper and lower limits of the range of inertia weight parameter. $k$ and $Mk$ are the current iteration and maximum iteration.

Improved particle swarm optimization algorithm

According to the no free lunch theory 31 , it is known that no algorithm can solve every practical problem with high quality and efficiency for increasingly complex and diverse optimization problems. In this section, several improvement strategies are proposed to improve the search efficiency and overcome this shortcoming of the basic PSO algorithm.

Improvement strategies

The optimization strategies of the improved PSO algorithm are shown as follows:

The inertia weight parameter is updated by an improved chaotic variables method instead of a linear decreasing strategy. Chaotic mapping performs the whole search at a higher speed and is more resistant to falling into local optimal than the probability-dependent random search 32 . However, the population may result in that particles can easily fly out of the global optimum boundary. To ensure that the population can converge to the global optimum, an improved Iterative mapping is adopted and shown as follows:

Here ${\omega }_{k}$ is the inertia weight parameter in the iteration $k$ , $b$ is the control parameter in the range $[\mathrm{0,1}]$ .

The acceleration coefficients are updated by the linear transformation. ${c}_{1}$ and ${c}_{2}$ represent the influential coefficients of the particles by their own and population information, respectively. To improve the search performance of the population, ${c}_{1}$ and ${c}_{2}$ are changed from fixed values to time-varying parameter parameters, that are updated by linear transformation with the number of iterations:

where ${c}_{max}$ and ${c}_{min}$ are the maximum and minimum values of acceleration coefficients, respectively.

The initialization scheme is determined by elite opposition-based learning . The high-quality initial population will accelerate the solution speed of the algorithm and improve the accuracy of the optimal solution. Thus, the elite backward learning strategy 33 is introduced to generate the position matrix of the initial population. Suppose the elite individual of the population is ${X}=[{x}_{1},{x}_{2},{x}_{3},...,{x}_{j},...,{x}_{D}]$ , and the elite opposition-based solution of $X$ is ${X}_{o}=[{x}_{{\text{o}}1},{x}_{{\text{o}}2},{x}_{{\text{o}}3},...,{x}_{oj},...,{x}_{oD}]$ . The formula for the elite opposition-based solution is as follows:

where ${k}_{r}$ is the random value in the range $(\mathrm{0,1})$ . ${ux}_{oij}$ and ${lx}_{oij}$ are dynamic boundaries of the elite opposition-based solution in $j$ dimensional variables. The advantage of dynamic boundary is to reduce the exploration space of particles, which is beneficial to the convergence of the algorithm. When the elite opposition-based solution is out of bounds, the out-of-bounds processing is performed. The equation is given as follows:

After calculating the fitness function values of the elite solution and the elite opposition-based solution, respectively, $n$ high quality solutions were selected to form a new initial population position matrix.

The position updating Eq. ( 2 ) is modified based on the strategy of dynamic weight. To improve the speed of the global search of the population, the strategy of dynamic weight from the artificial bee colony algorithm 34 is introduced to enhance the computational performance. The new position updating equation is shown as follows:

Here $\rho$ is the random value in the range $(\mathrm{0,1})$ . $\psi$ represents the acceleration coefficient and ${\omega }{\prime}$ is the dynamic weight coefficient. The updated equations of the above parameters are as follows:

where $f(i)$ denotes the fitness function value of individual particle $i$ and u is the average of the population fitness function values in the current iteration. The Eqs. ( 11 , 12 ) are introduced into the position updating equation. And they can attract the particle towards positions of the best-so-far solution in the search space.

New local optimal jump-out strategy is added for escaping from the local optimal. When the value of the fitness function for the population optimal particles does not change in M iterations, the algorithm determines that the population falls into a local optimal. The scheme in which the population jumps out of the local optimum is to reset the position information of the 40% of individuals within the population, in other words, to randomly generate the position vector in the search space. M is set to 5% of the maximum number of iterations.

New spiral update search strategy is added after the local optimal jump-out strategy. Since the whale optimization algorithm (WOA) was good at exploring the local search space 35 , the spiral update search strategy in the WOA 36 is introduced to update the position of the particles after the swarm jumps out of local optimal. The equation for the spiral update is as follows:

Here $D=\left|{x}_{i}\left(k\right)-Gbest\right|$ denotes the distance between the particle itself and the global optimal solution so far. $B$ is the constant that defines the shape of the logarithmic spiral. $l$ is the random value in $[-\mathrm{1,1}]$ . $l$ represents the distance between the newly generated particle and the global optimal position, $l=-1$ means the closest distance, while $l=1$ means the farthest distance, and the meaning of this parameter can be directly observed by Fig. 1 .

Spiral updating position.

The DE/best/2 mutation strategy is introduced to form the mutant particle. 4 individuals in the population are randomly selected that differ from the current particle, then the vector difference between them is rescaled, and the difference vector is combined with the global optimal position to form the mutant particle. The equation for mutation of particle position is shown as follows:

where ${x}^{*}$ is the mutated particle, $F$ is the scale factor of mutation, ${r}_{1}$ , ${r}_{2}$ , ${r}_{3}$ , ${r}_{4}$ are random integer values in $(0,n]$ and not equal to $i$ , respectively. Specific particles are selected for mutation with the screening conditions as follows:

where $Cr$ represents the probability of mutation, $rand\left(\mathrm{0,1}\right)$ is a random number in $\left(\mathrm{0,1}\right)$ , and ${i}_{rand}$ is a random integer value in $(0,n]$ .

The improved PSO incorporates the search ideas of other intelligent algorithms (DE, WOA), so the improved algorithm proposed in this paper is named NDWPSO. The pseudo-code for the NDWPSO algorithm is given as follows:

The main procedure of NDWPSO.

Comparing the distribution of inertia weight parameters

There are several improved PSO algorithms (such as CDWPSO 25 , and SDWPSO 26 ) that adopt the dynamic weighted particle position update strategy as their improvement strategy. The updated equations of the CDWPSO and the SDWPSO algorithm for the inertia weight parameters are given as follows:

where ${\text{A}}$ is a value in $(\mathrm{0,1}]$ . ${r}_{max}$ and ${r}_{min}$ are the upper and lower limits of the fluctuation range of the inertia weight parameters, $k$ is the current number of algorithm iterations, and $Mk$ denotes the maximum number of iterations.

Considering that the update method of inertia weight parameters by our proposed NDWPSO is comparable to the CDWPSO, and SDWPSO, a comparison experiment for the distribution of inertia weight parameters is set up in this section. The maximum number of iterations in the experiment is $Mk=500$ . The distributions of CDWPSO, SDWPSO, and NDWPSO inertia weights are shown sequentially in Fig. 2 .

The inertial weight distribution of CDWPSO, SDWPSO, and NDWPSO.

In Fig. 2 , the inertia weight value of CDWPSO is a random value in (0,1]. It may make individual particles fly out of the range in the late iteration of the algorithm. Similarly, the inertia weight value of SDWPSO is a value that tends to zero infinitely, so that the swarm no longer can fly in the search space, making the algorithm extremely easy to fall into the local optimal value. On the other hand, the distribution of the inertia weights of the NDWPSO forms a gentle slope by two curves. Thus, the swarm can faster lock the global optimum range in the early iterations and locate the global optimal more precisely in the late iterations. The reason is that the inertia weight values between two adjacent iterations are inversely proportional to each other. Besides, the time-varying part of the inertial weight within NDWPSO is designed to reduce the chaos characteristic of the parameters. The inertia weight value of NDWPSO avoids the disadvantages of the above two schemes, so its design is more reasonable.

Experiment and discussion

In this section, three experiments are set up to evaluate the performance of NDWPSO: (1) the experiment of 23 classical functions 37 between NDWPSO and three particle swarm algorithms (PSO 6 , CDWPSO 25 , SDWPSO 26 ); (2) the experiment of benchmark test functions between NDWPSO and other intelligent algorithms (Whale Optimization Algorithm (WOA) 36 , Harris Hawk Algorithm (HHO) 38 , Gray Wolf Optimization Algorithm (GWO) 39 , Archimedes Algorithm (AOA) 40 , Equilibrium Optimizer (EO) 41 and Differential Evolution (DE) 42 ); (3) the experiment for solving three real engineering problems (welded beam design 43 , pressure vessel design 44 , and three-bar truss design 38 ). All experiments are run on a computer with Intel i5-11400F GPU, 2.60 GHz, 16 GB RAM, and the code is written with MATLAB R2017b.

The benchmark test functions are 23 classical functions, which consist of indefinite unimodal (F1–F7), indefinite dimensional multimodal functions (F8–F13), and fixed-dimensional multimodal functions (F14–F23). The unimodal benchmark function is used to evaluate the global search performance of different algorithms, while the multimodal benchmark function reflects the ability of the algorithm to escape from the local optimal. The mathematical equations of the benchmark functions are shown and found as Supplementary Tables S1 – S3 online.

Experiments on benchmark functions between NDWPSO, and other PSO variants

The purpose of the experiment is to show the performance advantages of the NDWPSO algorithm. Here, the dimensions and corresponding population sizes of 13 benchmark functions (7 unimodal and 6 multimodal) are set to (30, 40), (50, 70), and (100, 130). The population size of 10 fixed multimodal functions is set to 40. Each algorithm is repeated 30 times independently, and the maximum number of iterations is 200. The performance of the algorithm is measured by the mean and the standard deviation (SD) of the results for different benchmark functions. The parameters of the NDWPSO are set as: ${[{\omega }_{min},\omega }_{max}]=[\mathrm{0.4,0.9}]$ , $\left[{c}_{max},{c}_{min}\right]=\left[\mathrm{2.5,1.5}\right],{V}_{max}=0.1,b={e}^{-50}, M=0.05\times Mk, B=1,F=0.7, Cr=0.9.$ And, $A={\omega }_{max}$ for CDWPSO; ${[r}_{max},{r}_{min}]=[\mathrm{4,0}]$ for SDWPSO.

Besides, the experimental data are retained to two decimal places, but some experimental data will increase the number of retained data to pursue more accuracy in comparison. The best results in each group of experiments will be displayed in bold font. The experimental data is set to 0 if the value is below 10 –323 . The experimental parameter settings in this paper are different from the references (PSO 6 , CDWPSO 25 , SDWPSO 26 , so the final experimental data differ from the ones within the reference.

As shown in Tables 1 and 2 , the NDWPSO algorithm obtains better results for all 49 sets of data than other PSO variants, which include not only 13 indefinite-dimensional benchmark functions and 10 fixed-multimodal benchmark functions. Remarkably, the SDWPSO algorithm obtains the same accuracy of calculation as NDWPSO for both unimodal functions f 1 –f 4 and multimodal functions f 9 –f 11 . The solution accuracy of NDWPSO is higher than that of other PSO variants for fixed-multimodal benchmark functions f 14 -f 23 . The conclusion can be drawn that the NDWPSO has excellent global search capability, local search capability, and the capability for escaping the local optimal.

In addition, the convergence curves of the 23 benchmark functions are shown in Figs. 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 and 19 . The NDWPSO algorithm has a faster convergence speed in the early stage of the search for processing functions f1-f6, f8-f14, f16, f17, and finds the global optimal solution with a smaller number of iterations. In the remaining benchmark function experiments, the NDWPSO algorithm shows no outstanding performance for convergence speed in the early iterations. There are two reasons of no outstanding performance in the early iterations. On one hand, the fixed-multimodal benchmark function has many disturbances and local optimal solutions in the whole search space. on the other hand, the initialization scheme based on elite opposition-based learning is still stochastic, which leads to the initial position far from the global optimal solution. The inertia weight based on chaotic mapping and the strategy of spiral updating can significantly improve the convergence speed and computational accuracy of the algorithm in the late search stage. Finally, the NDWPSO algorithm can find better solutions than other algorithms in the middle and late stages of the search.

Evolution curve of NDWPSO and other PSO algorithms for f1 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f2 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f3 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f4 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f5 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f6 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f7 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f8 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f9 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f10 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f11(Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f12 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f13 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f14, f15, f16.

Evolution curve of NDWPSO and other PSO algorithms for f17, f18, f19.

Evolution curve of NDWPSO and other PSO algorithms for f20, f21, f22.

Evolution curve of NDWPSO and other PSO algorithms for f23.

To evaluate the performance of different PSO algorithms, a statistical test is conducted. Due to the stochastic nature of the meta-heuristics, it is not enough to compare algorithms based on only the mean and standard deviation values. The optimization results cannot be assumed to obey the normal distribution; thus, it is necessary to judge whether the results of the algorithms differ from each other in a statistically significant way. Here, the Wilcoxon non-parametric statistical test 45 is used to obtain a parameter called p -value to verify whether two sets of solutions are different to a statistically significant extent or not. Generally, it is considered that p ≤ 0.5 can be considered as a statistically significant superiority of the results. The p -values calculated in Wilcoxon’s rank-sum test comparing NDWPSO and other PSO algorithms are listed in Table 3 for all benchmark functions. The p -values in Table 3 additionally present the superiority of the NDWPSO because all of the p -values are much smaller than 0.5.

In general, the NDWPSO has the fastest convergence rate when finding the global optimum from Figs. 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 and 19 , and thus we can conclude that the NDWPSO is superior to the other PSO variants during the process of optimization.

Comparison experiments between NDWPSO and other intelligent algorithms

Experiments are conducted to compare NDWPSO with several other intelligent algorithms (WOA, HHO, GWO, AOA, EO and DE). The experimental object is 23 benchmark functions, and the experimental parameters of the NDWPSO algorithm are set the same as in Experiment 4.1. The maximum number of iterations of the experiment is increased to 2000 to fully demonstrate the performance of each algorithm. Each algorithm is repeated 30 times individually. The parameters of the relevant intelligent algorithms in the experiments are set as shown in Table 4 . To ensure the fairness of the algorithm comparison, all parameters are concerning the original parameters in the relevant algorithm literature. The experimental results are shown in Tables 5 , 6 , 7 and 8 and Figs. 20 , 21 , 22 , 23 , 24 , 25 , 26 , 27 , 28 , 29 , 30 , 31 , 32 , 33 , 34 , 35 and 36 .

Evolution curve of NDWPSO and other algorithms for f1 (Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f2 (Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f3(Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f4 (Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f5 (Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f6 (Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f7 (Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f8 (Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f9(Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f10 (Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f11 (Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f12 (Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f13 (Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f14, f15, f16.

Evolution curve of NDWPSO and other algorithms for f17, f18, f19.

Evolution curve of NDWPSO and other algorithms for f20, f21, f22.

Evolution curve of NDWPSO and other algorithms for f23.

The experimental data of NDWPSO and other intelligent algorithms for handling 30, 50, and 100-dimensional benchmark functions ( ${f}_{1}-{f}_{13}$ ) are recorded in Tables 8 , 9 and 10 , respectively. The comparison data of fixed-multimodal benchmark tests ( ${f}_{14}-{f}_{23}$ ) are recorded in Table 11 . According to the data in Tables 5 , 6 and 7 , the NDWPSO algorithm obtains 69.2%, 84.6%, and 84.6% of the best results for the benchmark function ( ${f}_{1}-{f}_{13}$ ) in the search space of three dimensions (Dim = 30, 50, 100), respectively. In Table 8 , the NDWPSO algorithm obtains 80% of the optimal solutions in 10 fixed-multimodal benchmark functions.

The convergence curves of each algorithm are shown in Figs. 20 , 21 , 22 , 23 , 24 , 25 , 26 , 27 , 28 , 29 , 30 , 31 , 32 , 33 , 34 , 35 and 36 . The NDWPSO algorithm demonstrates two convergence behaviors when calculating the benchmark functions in 30, 50, and 100-dimensional search spaces. The first behavior is the fast convergence of NDWPSO with a small number of iterations at the beginning of the search. The reason is that the Iterative-mapping strategy and the position update scheme of dynamic weighting are used in the NDWPSO algorithm. This scheme can quickly target the region in the search space where the global optimum is located, and then precisely lock the optimal solution. When NDWPSO processes the functions ${f}_{1}-{f}_{4}$ , and ${f}_{9}-{f}_{11}$ , the behavior can be reflected in the convergence trend of their corresponding curves. The second behavior is that NDWPSO gradually improves the convergence accuracy and rapidly approaches the global optimal in the middle and late stages of the iteration. The NDWPSO algorithm fails to converge quickly in the early iterations, which is possible to prevent the swarm from falling into a local optimal. The behavior can be demonstrated by the convergence trend of the curves when NDWPSO handles the functions ${f}_{6}$ , ${f}_{12}$ , and ${f}_{13}$ , and it also shows that the NDWPSO algorithm has an excellent ability of local search.

Combining the experimental data with the convergence curves, it is concluded that the NDWPSO algorithm has a faster convergence speed, so the effectiveness and global convergence of the NDWPSO algorithm are more outstanding than other intelligent algorithms.

Experiments on classical engineering problems

Three constrained classical engineering design problems (welded beam design, pressure vessel design 43 , and three-bar truss design 38 ) are used to evaluate the NDWPSO algorithm. The experiments are the NDWPSO algorithm and 5 other intelligent algorithms (WOA 36 , HHO, GWO, AOA, EO 41 ). Each algorithm is provided with the maximum number of iterations and population size ( ${\text{Mk}}=500,\mathrm{ n}=40$ ), and then repeats 30 times, independently. The parameters of the algorithms are set the same as in Table 4 . The experimental results of three engineering design problems are recorded in Tables 9 , 10 and 11 in turn. The result data is the average value of the solved data.

Welded beam design

The target of the welded beam design problem is to find the optimal manufacturing cost for the welded beam with the constraints, as shown in Fig. 37 . The constraints are the thickness of the weld seam ( ${\text{h}}$ ), the length of the clamped bar ( ${\text{l}}$ ), the height of the bar ( ${\text{t}}$ ) and the thickness of the bar ( ${\text{b}}$ ). The mathematical formulation of the optimization problem is given as follows:

Welded beam design.

In Table 9 , the NDWPSO, GWO, and EO algorithms obtain the best optimal cost. Besides, the standard deviation (SD) of t NDWPSO is the lowest, which means it has very good results in solving the welded beam design problem.

Pressure vessel design

Kannan and Kramer 43 proposed the pressure vessel design problem as shown in Fig. 38 to minimize the total cost, including the cost of material, forming, and welding. There are four design optimized objects: the thickness of the shell ${T}_{s}$ ; the thickness of the head ${T}_{h}$ ; the inner radius ${\text{R}}$ ; the length of the cylindrical section without considering the head ${\text{L}}$ . The problem includes the objective function and constraints as follows:

Pressure vessel design.

The results in Table 10 show that the NDWPSO algorithm obtains the lowest optimal cost with the same constraints and has the lowest standard deviation compared with other algorithms, which again proves the good performance of NDWPSO in terms of solution accuracy.

Three-bar truss design

This structural design problem 44 is one of the most widely-used case studies as shown in Fig. 39 . There are two main design parameters: the area of the bar1 and 3 ( ${A}_{1}={A}_{3}$ ) and area of bar 2 ( ${A}_{2}$ ). The objective is to minimize the weight of the truss. This problem is subject to several constraints as well: stress, deflection, and buckling constraints. The problem is formulated as follows:

Three-bar truss design.

From Table 11 , NDWPSO obtains the best design solution in this engineering problem and has the smallest standard deviation of the result data. In summary, the NDWPSO can reveal very competitive results compared to other intelligent algorithms.

Conclusions and future works

An improved algorithm named NDWPSO is proposed to enhance the solving speed and improve the computational accuracy at the same time. The improved NDWPSO algorithm incorporates the search ideas of other intelligent algorithms (DE, WOA). Besides, we also proposed some new hybrid strategies to adjust the distribution of algorithm parameters (such as the inertia weight parameter, the acceleration coefficients, the initialization scheme, the position updating equation, and so on).

23 classical benchmark functions: indefinite unimodal (f1-f7), indefinite multimodal (f8-f13), and fixed-dimensional multimodal(f14-f23) are applied to evaluate the effective line and feasibility of the NDWPSO algorithm. Firstly, NDWPSO is compared with PSO, CDWPSO, and SDWPSO. The simulation results can prove the exploitative, exploratory, and local optima avoidance of NDWPSO. Secondly, the NDWPSO algorithm is compared with 5 other intelligent algorithms (WOA, HHO, GWO, AOA, EO). The NDWPSO algorithm also has better performance than other intelligent algorithms. Finally, 3 classical engineering problems are applied to prove that the NDWPSO algorithm shows superior results compared to other algorithms for the constrained engineering optimization problems.

Although the proposed NDWPSO is superior in many computation aspects, there are still some limitations and further improvements are needed. The NDWPSO performs a limit initialize on each particle by the strategy of “elite opposition-based learning”, it takes more computation time before speed update. Besides, the” local optimal jump-out” strategy also brings some random process. How to reduce the random process and how to improve the limit initialize efficiency are the issues that need to be further discussed. In addition, in future work, researchers will try to apply the NDWPSO algorithm to wider fields to solve more complex and diverse optimization problems.

Data availability

The datasets used and/or analyzed during the current study available from the corresponding author on reasonable request.

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Acknowledgements

This work was supported by Key R&D plan of Shandong Province, China (2021CXGC010207, 2023CXGC01020); First batch of talent research projects of Qilu University of Technology in 2023 (2023RCKY116); Introduction of urgently needed talent projects in Key Supported Regions of Shandong Province; Key Projects of Natural Science Foundation of Shandong Province (ZR2020ME116); the Innovation Ability Improvement Project for Technology-based Small- and Medium-sized Enterprises of Shandong Province (2022TSGC2051, 2023TSGC0024, 2023TSGC0931); National Key R&D Program of China (2019YFB1705002), LiaoNing Revitalization Talents Program (XLYC2002041) and Young Innovative Talents Introduction & Cultivation Program for Colleges and Universities of Shandong Province (Granted by Department of Education of Shandong Province, Sub-Title: Innovative Research Team of High Performance Integrated Device).

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How a Social App Reduces Food Waste and Promotes ‘Feel-Good’ Decluttering: Founder Q&A

"our vision at olio is simple: we want to harness the age-old practice of sharing to solve the very modern problem of food waste.".

Do you ever find yourself feeling guilty when throwing away leftovers or accidentally letting food in your fridge expire? What about when you pass by a full bakery display case on your walk home from work? Do you wonder what will happen to the orphan pastries left behind by the breakfast crowd?

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We hate to be the bearer of bad news, but each year about a third of food produced across the globe goes to waste, according to the World Economic Forum. And food waste generates four to five times the greenhouse gas emissions produced by the aviation industry, per the United Nations Environment Program. In the U.K., for example, a typical household generates over £1,000 ($1,253) of food waste per year while 9 million people face food insecurity.

Tessa Clarke , 47, daughter of a farmer from Yorkshire, England, saw firsthand the effort that goes into growing and harvesting food and decided that there had to be a better alternative. So, in 2015, Clarke and her co-founder Saasha Celestial-One set out to harness the power of sharing with Olio, a mobile app designed to combat food waste by connecting a few communities in North London.

At its launch, neighbors could use Olio to pass on spare food while volunteers coordinated with local businesses and restaurants to redistribute their unsold or unserved food. Olio’s mission has since expanded into fighting all overconsumption, with users now able to post any unneeded items such as furniture, clothing and books to promote what the platform calls “feel-good” decluttering.

So far, 8 million Olio-ers worldwide have shared 160 million portions of food, avoided 181,000 tonnes of CO2 emissions, and saved 23 billion liters of water . Earlier this month, Observer interviewed Clarke about the inspiration driving Olio, the app’s unforeseen social influence, and how her startup encourages people to participate in its mission. The following conversation has been edited for length and clarity.

Observer: What inspired you to start Olio, and how did the idea evolve into what it is today?

Tessa Clarke: I’m a farmer’s daughter and so I grew up with an especially deep dislike of food waste. As a result, when I was moving to the country nine years ago and found myself with some good food that we hadn’t managed to eat, I couldn’t bring myself to throw it away. Instead, I set out onto the streets to try and find someone to give it to. But unfortunately, I failed miserably. That led to the lightbulb moment for O lio: a mobile app that connects neighbors to safely and easily share food.

My co-founder Saasha and I ran a proof-of-concept using a small WhatsApp group in North London. When that worked we decided to take the plunge and build a mobile app. Olio connects people with their neighbors so that they can give away, rather than throw away, their spare food and other household items, and lend and borrow instead of buying brand new.

We also have 100,000 volunteers who collect unsold food from supermarkets, corporate canteens, school canteens and hospital canteens and redistribute it to their local communities via the Olio app. Our largest clients include Tesco, Iceland, and Holland and Barrett.

Can you explain the mission and vision behind Olio and how it aims to address food waste?

Sadly, it’s no exaggeration to say that food waste is one of the largest, but most unrecognized, problems facing humanity today. Globally, one third of all the food we produce each year goes to waste, which is worth over $1 trillion. The environmental impact of this is absolutely devastating: if it were to be a country , food waste would be the third largest source of greenhouse gas emissions after the U.S. and China.

What’s surprising is that in a country such as the U.K., over half of all food waste takes place in the home, with a typical household of four generating over £1,000 of food waste per year. Alongside this, we have approximately 9 million people experiencing food insecurity , of whom 4 million are children .

Our vision at Olio is simple: we want to harness the age-old practice of sharing to solve the very modern problem of food waste. Our research shows that no one enjoys throwing away food; we just do it because we’re no longer connected to our local communities, and so we no longer have anyone to give our spare food to.

What impact has Olio had so far in terms of reducing food waste and fostering community connections?

So far, our community has shared over 160 million portions of food and 11 million household items. This has had an environmental impact equivalent to taking over 616 million car miles off the road and has saved over 23 billion liters of water (because food production is incredibly water -intensive).

Olio’s social impact is just as powerful as its environmental impact: 40 percent of our community say they’ve made friends through the app, two-thirds say sharing has improved their mental health, and three-quarters say sharing has improved their financial well-being. In addition to this, there are countless heartwarming stories shared with us every week of the friendships formed and small acts of kindness taking place on doorsteps across the country.

What strategies does Olio employ to encourage more individuals and businesses to participate in food sharing?

One of the core ways in which Olio has grown is thanks to our Ambassador Program. We now have over 50,000 ambassadors who are passionate about our mission and who want to be able to use Olio in their local communities. We equip our ambassadors with posters, letters, flyers, and content for social media so they can spread the word about Olio on our behalf.

In addition to directly encouraging businesses to participate, we’ve also been campaigning with other organizations such as Feedback , to encourage the U.K. government to pass the regulation they consulted on, which would mandate that large businesses have to publicly report their food waste data. With businesses in the U.K. throwing away the equivalent of over 2 billion meals per year , it’s clear that if we could bring this practice out from behind closed doors, then we can really start to tackle it in earnest and get this precious food redistributed.

How do you balance the social and environmental mission of Olio with the need to generate revenue and sustain the business?

There’s often an assumption that there must be a trade-off, or tension, between having a positive environmental or social impact and making money. However, the most powerful business models are those where impact and revenues grow in lock-step with one another.

How? In the case of Olio, we generate more revenues by getting more listings on the app—because businesses pay us to redistribute their unsold food—and by getting more listings on the app. We also have two smaller revenue streams which are subscriptions and advertising. Building out these capabilities doesn’t directly drive our impact, but they’re absolutely critical to enabling Olio to become self-sustaining and so continue to have an impact over the longer term.

How a Social App Reduces Food Waste and Promotes ‘Feel-Good’ Decluttering: Founder Q&A

SEE ALSO : How Tony Notarberardino’s ‘Chelsea Hotel Portraits’ Captured the End of an Era

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IMAGES

Example problem: Solving for Speed and Velocity
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Solving Problems Calculating the Average Speed of an Object
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Speed word problems
01 -Physics Find the Speed

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Week-4 Tutorial (Semiconductor Devices and Circuits) : NPTEL NOC EE-91, Y2023
Calculating Speed Solution
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Velocity and Speed: Solutions to Problems
Problem 5: If I can walk at an average speed of 5 km/h, how many miles I can walk in two hours? Solution to Problem 5: distance = (average speed) * (time) = 5 km/h * 2 hours = 10 km using the rate of conversion 0.62 miles per km, the distance in miles is given by distance = 10 km * 0.62 miles/km = 6.2 miles Problem 6: A train travels along a straight line at a constant speed of 60 mi/h for a ...
Problems on Calculating Speed
Formula to find out Speed is given by Speed = Distance/Time. Word Problems on Calculating Speed. 1. A man walks 25 km in 6 hours. Find the speed of the man? Solution: Distance traveled = 25 km Time taken to travel = 6 hours Speed of Man = Distance traveled/Time taken = 25km/6hr = 4.16 km/hr Therefore, a man travels at a speed of 4.16 km/hr. 2.
Problems on Calculating Speed
Here we will learn to solve different types of problems on calculating speed. We know, the speed of a moving body is the distance traveled by it in unit time. Formula to find out speed = distance/time. Word problems on calculating speed: 1. A man walks 20 km in 4 hours. Find his speed. Solution: Distance covered = 20 km
8.8 Rate Word Problems: Speed, Distance and Time
Distance, rate and time problems are a standard application of linear equations. When solving these problems, use the relationship rate (speed or velocity) times time equals distance. r⋅t = d r ⋅ t = d. For example, suppose a person were to travel 30 km/h for 4 h. To find the total distance, multiply rate times time or (30km/h) (4h) = 120 km.
Speed, Velocity, and Acceleration Problems
speed = ? and the formula for speed. speed = distance/time = 5000/50 = 100m/s . Problem 3. A motorcycle starting from rest moves with a uniform acceleration until it attains a speed of 108 kilometres per hour after 15 seconds. Find its acceleration. Solution. Data:
PDF Distance, Time, Speed Practice Problems
DISTANCE, TIME, SPEED PRACTICE PROBLEMS YOU MUST SHOW YOUR WORK. You can use a calculator but you must show all of the steps involved in doing the problem. SPEED 1. If a car travels 400m in 20 seconds how fast is it going? ... 11. If you shout into the Grand Canyon, your voice travels at the speed of sound (340 m/s) to the bottom of the canyon ...
Speed and velocity questions (practice)
Solving for time. Displacement from time and velocity example. Instantaneous speed and velocity. Acceleration: At a glance. Acceleration. ... The average speed is 37.5 miles per hour. C. The average speed is 37.5 miles per hour. (Choice D) The car travels at 50 mph for the first half and 30 mph for the second half. D.
Speed and Velocity
The examples so far calculate average speed: how far something travels over a period of time. But speed can change as time goes by. A car can go faster and slower, maybe even stop at lights. So there is also instantaneous speed: the speed at an instant in time. We can try to measure it by using a very short span of time (the shorter the better).
Speed and Velocity
Numbers in, answer out. ∆ s = c ∆t. ∆ s = (3.00 × 108 m/s)(365.25 × 24 × 60 × 60 s) ∆ s = 9.46 × 1015 m. Δ s ≈ 10 petameters. Since both the speed of light and the year have exactly defined values in the International System of Units, the light year can be stated with an unnecessarily large number of significant digits.
2.2 Speed and Velocity
Speed is the rate at which an object changes its location. ... Practice Problems. 9. A pitcher throws a baseball from the pitcher's mound to home plate in 0.46 s. The distance is 18.4 m. What was the average speed of the baseball? ... We can use the formula for average velocity to solve the problem. Solution.
Solved Speed, Velocity, and Acceleration Problems
Speed, velocity, and acceleration problems with detailed answers are provided for high school physics. ... some sample AP Physics 1 problems on acceleration are provided. ... By practicing these problems, you can get better at solving tough physics questions. This can help you do well in your exams.
TIME SPEED AND DISTANCE PROBLEMS
The formula to find time is. Time taken to cover 330 miles distance at the speed of 60 miles per hour is. = ³³⁰⁄₆₀. = 5.5 hours. = 5 hrs 30 minutes. So, if the person is increased by 50%, it will take 5 hrs 30 minutes to cover 330 miles distance. Problem 10 : A person speed at a rate of 40 kms per hour.
Speed Formula
Using Formula for Speed, Speed = Distance/Time = 120000/3600 = 33.3m/sec. Answer: The speed of the train is 33.3 m/s. Example 2: A cyclist covers 20 km in 50 minutes. Use the speed formula to calculate the speed of the cyclist in m/s. Solution: To find: The speed of a cyclist.
Average velocity and speed worked example
So, the average velocity is going to be equal to negative 1/30 meters per second, the negative specifies that on average the velocity is towards the left. If you wanna specify this as a decimal with two significant digits, this is going to be, so this will approximately equal to 0.033. That would be 1/30. Now let's try to tackle average speed.
Speed, time, and distance problems worksheets
Make customizable worksheets about constant (or average) speed, time, and distance for pre-algebra and algebra 1 courses (grades 6-9). Both PDF and html formats are available. You can choose the types of word problems in the worksheet, the number of problems, metric or customary units, the way time is expressed (hours/minutes, fractional hours, or decimal hours), and the amount of workspace ...
Average velocity and speed in one direction: word problems (practice
Grant sprints 50 m to the right with an average velocity of 3.0 m s . How many seconds did Grant sprint? Answer using a coordinate system where rightward is positive. Round the answer to two significant digits. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more.
Average Speed Problems (video lessons, examples and solutions)
Solution: Step 1: The formula for distance is. Distance = Rate × Time. Total distance = 50 × 3 + 60 × 2 = 270. Step 2: Total time = 3 + 2 = 5. Step 3: Using the formula: Answer: The average speed is 54 miles per hour. Be careful! You will get the wrong answer if you add the two speeds and divide the answer by two.
Calculating Speed: Concept, Formula, and Practice Problems
It is crucial to ensure that the data used in calculations, such as distance and speed, are in the same unit of measurement to accurately solve problems. 📏 In the given example, the time taken to cover a distance of 12 meters at a speed of 3 meters per second is 4 seconds.
Average Velocity and Average Speed Problems and Solutions
average speed = s/t = 88/20 = 4,4 m/s (b) in one complete revolution displacement of car is zero. v avr = displacement/time = 0/10 = 0 m/s Problem #5 a train travels from city A to city B with a constant speed of 20 ms-1 and return back to city A with a constant speed of 30 m.s-1. Find its average speed during te entire journey. Answer:
Sample Problems
Problem 1.1. The acceleration of a particle moving along the x-axis is given by the equation: a(t) = (0.300m s3) t +(2.40 m s2) The particle is at position x = +4.60m and is moving in the −x direction at a speed of 12.0m s at time t = 0s. Find the time at which the particle (briefly) comes to rest.
Distance, Time and Speed Word Problems
Time = Distance / Speed. Let us take a look at some simple examples of distance, time and speed problems. Example 1. A boy walks at a speed of 4 kmph. How much time does he take to walk a distance of 20 km? Solution. Time = Distance / speed = 20/4 = 5 hours. Example 2. A cyclist covers a distance of 15 miles in 2 hours.
Kinematic Equations: Sample Problems and Solutions
A useful problem-solving strategy was presented for use with these equations and two examples were given that illustrated the use of the strategy. Then, the application of the kinematic equations and the problem-solving strategy to free-fall motion was discussed and illustrated. In this part of Lesson 6, several sample problems will be presented.
A hybrid particle swarm optimization algorithm for solving ...
A shuffle-based artificial bee colony algorithm for solving integer programming and minimax problems. Mathematics 9 (11), 1211 (2021). Article Google Scholar
Q&A With Founder of Olio, a Social App Aimed at Reducing ...
"Our vision at Olio is simple: we want to harness the age-old practice of sharing to solve the very modern problem of food waste." By Maddie Whitaker • 04/22/24 8:00am Tessa Clarke, founder and ...

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8.8 Rate Word Problems: Speed, Distance and Time

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Speed, Velocity, and Acceleration Problems

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Speed and Velocity

Example: A car travels 50 km in one hour.

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Average vs Instantaneous Speed

Example: Sam uses a stopwatch and measures 1.6 seconds as the car travels between two posts 20 m apart. What is the instantaneous speed ?

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Watch Physics

Calculating Average Velocity

Solving for Displacement when Average Velocity and Time are Known

Solving for Time when Displacement and Average Velocity are Known

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Solved Speed, Velocity, and Acceleration Problems

Speed and velocity Problems:

Acceleration Problems

TIME SPEED AND DISTANCE PROBLEMS

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Examples on Speed Formula

FAQs on Speed Formula

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How To Use Average Speed To Calculate The Distance Traveled?

How To Use Average Speed To Calculate The Time Taken?