assignment for class 9 physics motion

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Unit 1: Motion

About this unit.

Motion is all around us, from moving cars to flying aeroplanes. Motion can have different features like speed, direction, acceleration, etc. In this chapter, we will understand these features in detail and see how it can help us predict the future of these moving things.

Distance and displacement

• Distance and displacement introduction (Opens a modal)
• Distance and displacement in one dimension (Opens a modal)

Average speed and average velocity

• Average speed & velocity (with examples) (Opens a modal)
• Calculating average velocity or speed (Opens a modal)
• Average speed for entire journey - solved numerical (Opens a modal)
• Average velocity and speed in one direction: word problems 4 questions Practice
• Average velocity and speed with direction changes: word problems 4 questions Practice

Instantaneous speed and velocity

• Instantaneous speed & velocity (Opens a modal)

Acceleration

• Acceleration (Opens a modal)
• Airbus A380 take-off time (Opens a modal)
• Acceleration and velocity 7 questions Practice

Position time graphs

• Position-time graphs (Opens a modal)
• Calc. velocity from position time graphs (Opens a modal)

Velocity time graphs

• Velocity time graphs (& acceleration) (Opens a modal)
• Calculating displacement from v-t graphs (Opens a modal)

Deriving equations of motion

• Deriving 3 equations of motion (from v-t graph) (Opens a modal)

Problem solving using kinematic equations

• Using equations of motion (1 step numerical) (Opens a modal)
• Using equations of motion (2 steps numerical) (Opens a modal)
• Kinematic equations: numerical calculations 4 questions Practice

Uniform circular motion

• Calc. speed & time in a uniform circular motion - Solved numerical (Opens a modal)

CBSE NCERT Solutions

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Class 9 Physics Assignments

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We have provided below the biggest collection of free CBSE NCERT KVS Assignments for Class 9 Physics . Students and teachers can download and save all free Physics assignments in Pdf for grade 9th. Our expert faculty have covered Class 9 important questions and answers for Physics as per the latest syllabus for the current academic year. All test papers and question banks for Class 9 Physics and CBSE Assignments for Physics Class 9 will be really helpful for standard 9th students to prepare for the class tests and school examinations. Class 9th students can easily free download in Pdf all printable practice worksheets given below.

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Class 9 Motion worksheet – physics word problems

Last updated on December 28th, 2022 at 08:16 am

On this page, you will find a numerical physics worksheet that is based on class 9 Motion numericals .

Also, you can see this similar post with solved numerical: 101 numericals class 9 physics + motion + physicsteacher.in

Class 9 Motion Worksheet

1) In the hare-tortoise race, the hare ran for 2 min at a speed of 7.5 km/h, slept for 56 min, and again ran for 2 min at a speed of 7.5 km/h. Find the average speed of the hare in the race.

2) The maximum speed of a train is 80 km/h. It takes 10 hours to cover a distance of 400 km. Find the ratio of its maximum speed to its average speed.

3) A car moves 100m due east and then 25 m due west. (a) What is the distance covered by the car? (b) What is its displacement?

4) A boy leaves his house at 9.30 a.m. for his school. The school is 2 km away and classes start at 10.00 a.m. If he walks at a speed of 3 km/h for the first kilometer, at what speed should he walk the second kilometer to reach just in time?

More Motion Numericals (with solution)

5) A bus takes 8 hours to cover a distance of 320 km. What is the average speed of the bus?

6) A car moves through 20 km at a speed of 40 km/h, and the next 20 km at a speed of 60 km/h. Calculate its average speed.

7) A person walks along the sides of a square field. Each side is 100 m long. What is the maximum magnitude of displacement of the person at any time interval?

8) An object moves through 10 m in 2 minutes and the next 10 m in 3 minutes. Calculate its average speed.

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• Ohm's Law Problems in Physics - CBSE, ICSE Class 9 and 10

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• Motion Class 9 Notes CBSE Science Chapter 8 (Free PDF Download)
• Revision Notes

CBSE Class 9 Science Chapter 8 - Motion Revision Notes - Free PDF Download

The theories and principles taught in NCERT chapter 8 of motion for class 9 are explained in motion chapter class 9 notes. When the position of a body changes in reference to a stationary object, it is said to be in motion. Understanding higher-level physics subjects requires a comprehension of motion. There are several types of motions, and the phenomenon of motion is regulated by definite principles. As part of the Class 9 science chapter 8 notes, we will go over all of this in great detail.

Vedantu also provides free NCERT Solutions to all the students. You can download Class 9 Maths NCERT Solutions to help you to revise complete syllabus and score more marks in your examinations. 

Download free NCERT Solutions Class 9 Maths to amp up your preparations and to score well in your examinations.

Download CBSE Class 9 Science Revision Notes 2024-25 PDF

Also, check CBSE Class 9 Science revision notes for All chapters:

Access CBSE Class 9 Science Chapter 8 - Motion Notes

Introduction: 

One of the most common phenomena in the physical world is motion. Mechanics is the branch of Physics that deals with the behavior of moving objects.

Mechanics is divided further into two sections: Kinematics and Dynamics.

Kinematics is the study of motion without regard for the cause of motion.

Dynamics is concerned with the source of motion, which is force. 

Motion and Rest: 

An object is said to be in motion if its position in relation to its surroundings changes in a given time.

An object is said to be at rest if its position in relation to its surroundings does not change.

A frame of reference is another object or scene against which we compare the position of an object.

Take a look at the numbers. Figure 1 shows the car to the right of the tree. Figure 2 shows the car to the left of the tree after 2 seconds. The car must have moved from one location to another because the tree does not move. As a result, the tree serves as the frame of reference in this case. 

Types of Motion: 

There are three types of motion: 

Translatory motion 

Rotatory motion 

Vibratory motion 

Translatory Motion: 

A particle in translatory motion moves from one point in space to another. This movement may be in a straight line or in a curved path.

Rectilinear motion is defined as motion along a straight line.

Curvilinear motion is defined as movement along a curved path.

As an example, consider a car driving down a straight road.

Rectilinear Motion: 

Example: A car negotiating a curve 

Curvilinear Motion: 

Rotatory Motion: 

The particles of the body describe concentric circles around the axis of motion in rotatory motion.

Vibrational Motion:

Particles in vibratory motion move back and forth around a fixed point.

Distance and Displacement: 

The distance between termini A and B is 150 kilometers. A bus connects Terminus A and Terminus B. The bus travels a distance of 150 kilometers. The bus returns from terminus B to terminus A along the same route. As a result, the total distance traveled by the bus from A to B and then from B to A is 150 km + 150 km = 300 km. 

A bus traveling from point A to point B and back again.

The distance traveled by a moving object is the length of the path the object takes.

The measure of distance is a scalar quantity. The meter is the SI unit of distance.

The bus's position changed when it moved from Terminus A to Terminus B. The distance between A and B is 150 kilometers. The distance traveled on the return trip is also 150 kilometers.

Displacement is the shortest path covered by a moving object in a specified direction from the point of reference (the initial position of the body).

Note:  

However, the displacement when the bus moves from A B to B B is zero. The meter is the SI unit of displacement.

Displacement is a vector, which means that it is represented by a number with appropriate units and direction.

To emphasize the distinction between displacement and distance, consider a few more examples.

Assume a person moves 3 meters from point A to point B and 4 meters from point B to point C, as shown in the figure. He has travelled a total distance of 7 meters. But is he really 7 meters away from his starting point? No, he is only 5 meters away from his initial position, implying that he is displaced by the shortest distance between his initial and final positions.

Distance and Displacement  

To determine the displacement in this example, we can use Pythagoras' theorem. Consider an object that is changing its position with respect to a fixed point known as the origin 0. 

$\mathrm{x}_{\mathrm{i}}$ and $\mathrm{x}_{\mathrm{f}}$ are the initial and final positions of the object. Then the displacement of the

object $=x_{f}-x_{i}$

Suppose the object is travelling from $+1$ to $+4$, then displacement

$=\mathrm{x}_{\mathrm{f}}-\mathrm{x}_{\mathrm{i}}$

Displacement: Case 1 

Case 2:  

If the object is travelling from -3 to -1, then displacement 

If the object is travelling from $-3$ to $-1$, then displacement

$=x_{f}-x_{i}$

Displacement: Case 2 

Case 3:  

If the object is travelling from +4 to +2, then displacement 

Displacement: Case 3

Case 4:  

If the object follows the path depicted in the figure, the final and initial positions are the same, implying that the displacement is zero.

Displacement: Case 4 

We can conclude from the preceding examples that a body's displacement is positive if its final position is on the right side of its initial position and negative if its final position is on the left side of its initial position. The displacement of a moving object is said to be zero when it returns to its original position. Consider an athlete running in a clockwise direction along a circular track with radius r, beginning at A.

A Circular Track of Radius r 

What is the athlete's total distance traveled when he arrives at point B?

The athlete's total distance traveled when he arrives at point B equals to half of the circumference of the circular track, that is, $\frac{2 \pi r}{2}=\pi r$.

Displacement $=A B=2 r=$ Diameter of circle (the shortest distance between the initial and final positions).

If the athlete arrives at the starting point $A$, the distance covered is equal to the circumference of the circular track, i.e., $2 \pi r .$ However, the displacement is zero because the athlete's initial and final positions are the same.

Difference between Distance and Displacement

Motion: 

Uniform Motion and Non-uniform Motion: 

The distances covered by car A and car B with respect to time is given below:

Car A: 

Car B: 

The car A travels equal distances in equal time intervals, whereas the car B does not travel equal distances in equal time intervals. That is, car A's motion is an example of uniform motion, whereas car B's motion is an example of non-uniform motion.

A body is said to describe uniform motion when it covers equal distances in equal intervals of time.

When a body moves unequal distances in equal time intervals, or vice versa, this is referred to as non-uniform motion.

Speed: 

Ram and Krishna compete in various races over varying distances. Ram covers $1000 \mathrm{~m}$ in 20 minutes and Krishna covers $700 \mathrm{~m}$ in 10 minutes. Who is the fastest?

To determine who is faster, we will calculate the distance they cover in one minute.

Distance covered by Ram in one minute $=\frac{1000 \mathrm{~m}}{20 \mathrm{~min}}=500 \mathrm{~m} / \mathrm{min}$

Distance covered by Krishna in one minute $=\frac{700 \mathrm{~m}}{10 \mathrm{~min}}=70 \mathrm{~m} / \mathrm{min}$

Krishna covered more ground in the same amount of time. We conclude that Krishna is the faster of the two. 

Speed is defined as the distance traveled by a moving object in one unit of time.

$\text { speed }=\frac{\text { distance }}{\text { time }}=\frac{\mathrm{S}}{\mathrm{t}}$

Where S denotes the distance traveled and t denotes the time spent.

The SI unit of speed is millimeters per second (m/s). Speed is defined as a scalar quantity.

Uniform Speed:

The graph depicts the distance traveled by a ball every 2 seconds.

Every 2 seconds, the ball travels 10 meters. At any point between A and E, the ball moves at a speed of 5 m/s. The object is moving at a constant speed.

If an object travels the same distance in the same amount of time, it is said to be moving at a uniform speed.

Surface friction or resistance is ignored in this case.

Variable Speed or Non-Uniform Speed: 

The distance covered varies with time.

Variable Speed:  

For example, when a rubber ball is dropped from a certain height (h 1 ), it bounces up to a height less than the initial one (h 2 ). It keeps bouncing, but the height to which it rises keeps decreasing (h 3 , h 4 ). The ball's distance traveled per unit time decreases. The ball's speed varies from point to point. This type of speed is known as variable speed.

Average Speed and Instantaneous Speed:

When we travel by car, the speed varies depending on the road conditions at the time. The speed is calculated in this case by dividing the total distance traveled by the vehicle by the total time required for the journey. This is known as the average speed. 

The average speed of an object traveling S 1 in time t 1 , S 2 in time t 2 , and S n in time t n is given by,

$\text { Average speed }=\frac{\mathrm{S}_{1}+\mathrm{S}_{2}+\mathrm{S}_{3}+\ldots+\mathrm{S}_{\mathrm{n}}}{\mathrm{t}_{1}+\mathrm{t}_{2}+\mathrm{t}_{3}+\ldots+\mathrm{t}_{\mathrm{n}}}$

When we say that the car travels at an average speed of 60 km/h, we do not mean that it will travel at that speed for the duration of the journey. The actual speed of the car may be less than or greater than the average speed at a given location.

The speed of a moving body at any given point in time is referred to as instantaneous speed.

The diagram below depicts the various routes Shyam can take from his house to school. 

Shyam drives himself to school every day, averaging 60 km/h. Is it possible to find out how long it will take to get to the destination? Yes, you can use the relation to determine the time.

$\text { speed }=\frac{\text { distance }}{\text { time }}$

But you don't know what path he would have taken. As a result, simply providing the speed of a moving object does not allow one to determine the exact position of the object at any given time. As a result, there is a need to define a quantity that has both magnitude and direction.

Starting with A, consider two objects P and Q. Allow them to travel equal distances in equal time intervals, i.e. at the same speed. Can you guess where each of them will be in 20 seconds? P and Q are free to move in any direction. To determine the exact position of P and Q, we must also know their direction of motion.

Pictorial Representation of the Position of the Objects P and Q:

As a result, another physical quantity known as velocity is introduced to provide us with an idea of both speed and direction.

Velocity is defined as the distance traveled in a given direction by a moving object in a given time or speed in a given direction.

$\operatorname{velocity}(v)=\frac{\text { distance travelled in a specified direction }(s)}{\text { time taken }(t)}$

Note: 

Velocity is defined as the distance traveled in a given direction in a given amount of time. Displacement is the distance traveled in a specified direction.

As a result, velocity can be defined as the rate at which displacement changes.

Uniform Velocity and Non-Uniform Velocity:

Assume that two athletes, Ram and Shyam, are running at a constant speed of 5 m/s. Ram moves in a straight line, while Shyam follows a circular path. For a layperson, both Ram and Shyam are moving with uniform velocity, but for a physicist, only Ram is running with uniform velocity because his speed and direction of motion do not change.

In the case of Shyam, who is running on a circular track, the direction of motion changes at every instant because a circle is a polygon with infinite sides, and Shyam must change his direction at every instant.

A body is said to be moving with uniform velocity if it travels the same distance in the same amount of time in the same direction.

A body is said to be moving with variable velocity if it covers unequal distances in equal intervals of time and vice versa in a specified direction, or if its direction of motion changes.

Acceleration:

We are all aware that a car moving down the road does not have a uniform velocity. Either the speed or the direction of travel shifts. We say that a vehicle is accelerating when it is speeding up, i.e. when the speed increases.

Let us look at the change in velocity of a train traveling from Bangalore to Mysore to get an idea of acceleration. The train, which was initially at rest, begins to move; its velocity gradually increases until it reaches a constant velocity after a certain time interval. As the train approaches the next station, its speed gradually decreases until it comes to a halt.

When a train starts from a stop, its speed increases from zero, and we say it is accelerating. After a while, the speed becomes uniform, and we say that the train is moving at a uniform speed, which means that it is not accelerating. However, as the train approaches Mysore, it slows down, indicating that the train is accelerating in the opposite direction. When the train comes to a halt in Mysore, it stops accelerating once more.

As a result, it is clear that the term "acceleration" does not always imply that the speed of a moving body increases; it can also decrease, remain constant, or become zero.

In general, acceleration is defined as the rate at which the velocity of a moving body changes over time.

This change could be a change in the object's speed, direction of motion, or both.

Let us now look up a mathematical formula for calculating acceleration.

If an object moves with an initial velocity 'u' and reaches a final velocity 'v' in time 't,' then the acceleration 'a' produced by the object is

Acceleration = Rate at which velocity changes over time.

$a=\frac{v-u}{t}$

Unit of Acceleration:

The SI unit of acceleration is m/s 2 and it is a vector quantity.

Different Types of Acceleration:  

It is clear from the preceding example that acceleration takes various forms depending on the change in velocity.

Positive acceleration:

When an object's velocity increases, it is said to be moving with positive acceleration.

Positive Acceleration 

Example: a ball rolling downhill on an inclined plane.

Negative acceleration:

When an object's velocity decreases, it is said to be moving with negative acceleration. Negative acceleration can also be referred to as retardation or deceleration.

(1) A ball moving up an inclined plane.

(2) A vertically thrown upwards ball has a negative acceleration as its velocity decreases over time.

Zero Acceleration:  

If the change in velocity is zero, indicating that the object is either at rest or moving at uniform velocity, the object is said to have zero acceleration.

A parked car, for example, or a train moving at a constant speed of 90 km/hr.

Uniform Acceleration:  

The object is said to be moving with uniform acceleration if the change in velocity at equal intervals of time is always the same.

As an example, consider a body falling from a great height towards the earth's surface.

Non-uniform or Variable Acceleration:  

If the change in velocity over equal time intervals is not the same, the object is said to be moving with variable acceleration.

Motion:  

Distance-Time Table and Distance-Time Graph: 

Mr. X is taking a bus from New Delhi to Agra and recording his observations.

According to the table above, the bus travels equal distances at equal times. The bus is moving at a constant speed. In such a case, we can compute the distance traveled by the bus at any given point in time.

Consider an object moving from its initial position x i to its final position x f in time t at a uniform speed v. 

$\text { uniform speed }=\frac{\text { total distance }}{\text { time taken }}$

$v=\frac{x_{f}-x_{i}}{t}$

$x_{f}-x_{i}=v t \cdots \cdots(1)$

The relationship between distance, time, and average speed is given by equation (1). This relationship can be used to generate distance-time tables as well as to determine the position of any moving object at any given time. However, it is a time-consuming and tedious process, especially when we need to determine the position after a long period of time or compare the motion of two objects. In such cases, graphs such as the distance-time graph can be useful. A distance-time graph is a line graph that shows how distance changes over time. A distance-time graph plots time along the x-axis and distance along the y-axis. 

Distance-Time Graph for Non - Uniform Motion  

Let us now look at the nature of a distance-time graph for a non-uniform motion. The distance traveled by a bus every 15 minutes is shown in the table below.

We can deduce from the above table that the motion is non-uniform, i.e. it covers unequal distances in equal time intervals.

Measure time along the x-axis and distance along the y-axis.

Analyze the provided data and select the appropriate scale for time and distance.

Plot the points.

Join the points.

Consider any two points (A, B) on the graph.

Draw perpendicular from A to B to x and y axes.

Join A to C to get a right angled ACB.

The slope of the graph is shown below.

$\mathrm{AB} =\frac{\mathrm{BC}}{\mathrm{AC}}$

$=\frac{\mathrm{S}}{\mathrm{t}}$

$=\mathrm{speed}$

Write the title and scale chosen for the graph.

$\text { speed }=\frac{15-5}{30-15}$

$=\frac{10}{15}$

$=\frac{2}{3}$

$=0.666 \mathrm{~km} / \mathrm{min}$

Consider another two points on the graph, P and Q, and draw a right-angled triangle PRQ.

$\text { slope }=\text { speed }$

$\mathrm{PQ}=\frac{\mathrm{QR}}{\mathrm{PR}}$

$=\frac{35-30}{90-75}$

$=\frac{5}{15}$

$=0.333 \mathrm{Km} / \mathrm{min}$

Complete Graph

Nature of S- t Graph for Non- Uniform Motion and Uses of Graphs

Let us now see the nature of S-t graph for non-uniform motion. 

Nature of s-t Graph for Non-Uniform Motion: 

Figure (a) depicts the S-t graph as the speed of a moving object increases, while Figure (b) depicts the S-t graph as the speed of a moving object decreases. The nature of the S-t graph allows us to determine whether the object is moving at a constant or variable speed.

Uses of Graphical Representation:  

Because it provides a visual representation of two quantities, graphical representation is more informative than tables (e.g., distance vs. time)

A graph provides more information than a table at a glance. Both of the graphs shown here depict increasing speed.

Figure (1) depicts the nature of the variation in speed, indicating that the increase is greater in the beginning up to time t 1 and relatively lower after t 2 .

Similarly, fig (2) depicts how the increase in speed becomes greater after t1. A similar explanation applies to the decreasing speed.

Graphs are simple to read at a glance.

Graph plotting takes less time and is more convenient.

Graphs can be used to determine the position of any moving object at any point in time.

Two moving objects' motions can be easily compared.

Graphs reveal information about the nature of motion.

Velocity-Time Graph:  

The variation of velocity with time can be graphically represented to calculate acceleration in the same way that we calculated speed from the distance-time graph.

Let us now create a velocity-time(v-t) graph using the data below.

Draw time on the x-axis and velocity on the y-axis. 

Analyze the provided data and select the appropriate scale for the x and y axes.

Plot the Given Points.

Join the Points

Consider Any Two Points A and B on the Straight-Line Graph.

Draw Perpendiculars from A and B to x and y-axes.

Join A to C, ACB forms a Right-Angled Triangle.

Slope of the Graph

$\mathrm{AB}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\text { Change in velocity }}{\text { time }}=\text { acceleration }$

Calculations:

$\text { Acceleration }=\frac{30-20}{6-4}$

$=\frac{10}{2}$

$=5 \mathrm{~m} / \mathrm{s}^{2}$

Write the title for the graph.

V - T Graph:

Let us now examine the nature of the v - t graph for various types of motion.

a) Increasing Acceleration:

Uniform Acceleration

Non-Uniform Acceleration

(b) Decreasing Acceleration:

Non-uniform Retardation

Uniform Retardation

Zero Acceleration

Uses of Velocity-time Graphs  

The velocity-time graph can be used to derive the following results. 

The acceleration produced in a body. 

The distance traveled by a moving object. 

The equations of motion.

Speed - Time Graph

To compute the distance traveled by a moving object using a speed-time graph. 

The graph below depicts the speed-time graph of a car traveling at a constant speed of 60 km/h for 5 hours .

Speed-Time Graph of a Car Moving with Uniform Speed 

Distance travelled by the car,

Distance travelled by the car

$(S)=v \times t$

$=60 \times 5$

$=300 \mathrm{~km}$

But $60 \mathrm{~km} / \mathrm{h}=\mathrm{OC}=$ breadth of the rectangle $\mathrm{OABC}$

$5 h=O A=$ length of the rectangle $O A B C$

i.e., the distance covered by the car = length $\times$ breadth $=300 \mathrm{~km}$.

To calculate the distance traveled by a moving object using a speed-time graph, find the area enclosed by the speed-time graph and the time axis. In the case of non-uniform motion, the distance covered by the object increases in steps as the object's speed increases. During the time intervals $0-t_{1}, t_{1}-t_{2}, t_{2}-t_{3}, \ldots \ldots$, the speed remains constant.

The motion of an object moving at a variable speed is depicted in the figure below.

Speed - Time Graph for an Object Moving with Variable Speed

Calculation of Distance:

The object's total distance traveled during the time interval. 

0- t 6 = Area of rectangle 1 + area of rectangle 2 + …… + area of rectangle 6. 

Motion 

Equations of Motion  

Time, speed, distance covered, and acceleration are the variables in a uniformly accelerated rectilinear motion. These quantities have simple relationships. These relationships are expressed using equations known as equations of motion. 

The equations of motion are: 

(1) $v=u+a t$

(2) $S=u t+\dfrac{1}{2} a t^{2}$

(3) $v^{2}-u^{2}=2 a S$

Derivation of the First Equation of Motion

Consider a particle moving in a straight line with a constant acceleration ‘a’. Let the particle be at A at t=0, and u be its initial velocity, and v be its final velocity at t=t.

$a=\dfrac{v-u}{t}$ 

$v-u=a t$ 

I Equation of Motion

Second Equation of Motion

$\text { Average velocity } =\dfrac{\text { total distance travelled }}{\text { total time taken }}$

$=\dfrac{S}{t} \ldots \ldots(1)$

Average velocity can alsobe written as

$\frac{u+v}{2} \cdots \cdots(2)$

From equation (1) and (2),

$\frac{S}{t}=\frac{u+v}{2} \cdots \cdots(3)$

The first equation of motion is $v=u+a t$. Substituting the value of $v$ in equation (3), we get

$\frac{S}{t}=\frac{u+u+a t}{2}=\frac{(u+u+a t) t}{2}$

$\quad=\frac{(2 u+a t) t}{2}$

$S=u t+\frac{1}{2} a t^{2}$

II equation of motion

Third Equation of Motion

The first equation of motion is

$v-u=a t \ldots(1)$

Average velocity $=\frac{S}{t} \cdots \cdots(2)$

Average velocity $=\frac{u+v}{t} \cdots \cdot(3)$

From equation (2) and equation (3) we get,

$\frac{u+v}{t}=\frac{S}{t} \cdots \ldots(4)$

Multiplying equation (1) and equation (4) we get,

$(v-u)(v+u)=a t \times \frac{2 S}{t}$

$(v-u)(v+u)=2 a S$

$\left(v^{2}-u^{2}\right)=2 a S$

III equation of motion

Derivations of Equations of Motion (Graphically) 

First Equation of Motion

Consider an object moving in a straight line with a uniform velocity u. When its initial velocity is u, give it a uniform acceleration an at time t = 0. The object's velocity increases as a result of the acceleration to v (final velocity) in time t, and S is the distance covered by the object in time t.

The graph depicts the velocity-time graph of the object's motion.

The acceleration of a moving object is given by the slope of the v - t graph.

Thus, acceleration = slope

$\mathrm{AB}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{v-u}{t-0}$

I equation of motion

Second Equation of Motion  

Let u be an object's initial velocity and 'a' be the acceleration produced in the body. The area enclosed by the velocity-time graph for the time interval 0 to t gives the distance travelled S in time t.

Graphical Derivation of Second Equation

Distance travelled S = area of the trapezium ABDO

Distance travelled $S=$ area of the trapezium ABDO

$=$ area of reactangle $A C D O+$ area of $\triangle A B C$

$=\mathrm{OD} \times \mathrm{OA}+\frac{1}{2} \mathrm{BC} \times \mathrm{AC}$

$=\mathrm{t} \times \mathrm{u}+\frac{1}{2}(\mathrm{v}-\mathrm{u}) \times \mathrm{t}$

$=\mathrm{ut}+\frac{1}{2}(v-u) \times t$

$v=u+$ at $\mid$ equation of motion; $v-u=a t$

$S=u t+\frac{1}{2} a t \times t$

$=u t+\frac{1}{2} a t^{2}$

Third Equation of Motion  

Let 'u' be an object's initial velocity and a be the acceleration produced in the body. The area enclosed by the v - t graph gives the distance travelled 'S' in time 't'.

Graphical Derivation of Third Equation

$\mathrm{S}=$ area of trapezium $\mathrm{OABD}$

$=\frac{1}{2}\left(b_{1}+b_{2}\right) h$

$=\frac{1}{2}(\mathrm{OA}+\mathrm{BD}) \mathrm{AC}$

$=\frac{1}{2}(u+v) t \cdots \cdots(1)$

But we know that $a=\frac{v-u}{t}$ or $t=\frac{v-u}{a}$

Substituting the value of $t$ in equation (1) we get,

$S=\frac{1}{2} \frac{(u+v)(v-u)}{a}$

$=\frac{1}{2} \frac{(v+u)(v-u)}{a}$

$2 a s=(v+u)(v-u)$

$(v+u)(v-u)=2 a s$

$v^{2}-u^{2}=2 a s$

III Equation of Motion 

Circular Motion 

We classified motion along a circular track as non-uniform motion in the example discussed under the topic uniform and non-uniform motion. Let's take a look at why circular motion is considered non-uniform motion. The figure depicts an athlete running at a constant speed on a hexagonal track.

Athlete Running on a Regular Hexagonal Track 

The athlete runs at a constant speed along the track segments AB, BC, CD, DE, EF, and FA, and at the turns, he quickly changes direction to stay on the track without changing his speed. Similarly, if the track had been a regular octagon, the athlete would have had to change directions eight times in order to stay on the track.

Athlete Running on a Regular Octagonal Track 

The athlete must turn more frequently as the number of sides of the track increases. If we increase the number of sides indefinitely, the track will take on the shape of a circle. As a result, because a circle is a polygon with infinite sides, motion along a circular path is classified as non-uniform motion.

Athlete Running on a Circular Track 

Thus, an object moving at uniform speed along a circular track is an example of non-uniform motion because the object's direction of motion changes at every instant of time. 

Examples of Uniform Circular Motion 

(1) A car negotiating a curve at a constant speed.

(2) An athlete spinning a hammer in a circle before throwing it.

(3) An aircraft looping the loop.

Expression for Linear Velocity

If an athlete takes t seconds to complete one circular path of radius r, then the velocity v is given by the relation,

$\mathrm{v}=\frac{\text { distance travelled }}{\text { time }}$

Distance travelled = circumference of the circle

$=2\pi\mathrm{r}$

Linear velocity $=\frac{\text { 2r }}{\mathrm{t}}$

Class 9 Science Chapter 8 Revision Notes - Free PDF Download

Contents of class 9 science motion notes.

The Class 9 science notes chapter 8 contains detailed discussions on the topic of motion. It tells you about the basic definition of motion and rest. Then the chapter discusses the different types of motion. Next comes distance and displacement. 

With all this knowledge in your hand, we thereby venture to find out the speed and velocities of different objects. We also look into acceleration and its types. 

The notes also contain worked-out examples of numerical problems and other questions to give us a clear idea of the type of questions we can expect in the examination.

Subtopics Covered in Class 9 Chapter 8 Science Notes

The subtopics covered in class 9 science ch 8 notes are:

Motion and Rest: The change in position of an object with time and with respect to a stationary object is moving. If a body undergoes no change in position with time and with respect to a stationary object then it is said to be at rest

Types of Motion: Motion is divided into the following types:

Translation Motion: Motion along a straight line or curved path

Vibrational Motion: Motion of a particle to and from fixed positions.

Rotatory Motion: The circular motion of an object around the axis of rotation.

Distance and Displacement: Distance is defined as the total length of the path traveled by an object. Displacement on the other hand is the net effective change in position. In this section, we shall also look into various examples of numerical problems that aim at finding the distance and displacement of various objects.

Uniform and Non-Uniform Motion: Motion is said to be uniform when the rate of change of position remains the same across intervals multiple times with respect to the same object. Non-uniform motion sees different rates of change of position with respect to the object of reference.

Speed: Uniform And Variable: Speed is defined as the rate of change of position. It can either be uniform or variable.

Velocity: The velocity of an object is defined as the rate of change of displacement.

Acceleration: Acceleration is the rate of change of velocity with respect to time.

Basics of Plotting a Graph: Graphs are basically illustrations that are used to show the rate of change of a character. A graph is made by the intersection of two mutually perpendicular lines. Each of these lines is called the axis. 

How to work out time vs distance or displacement

Laws of Motion: Motion is governed by laws of physics known as Newton’s law of motion or simply the laws of motion. 

Ch 8 science class 9 notes are very important from the point of view of examinations as the chapter of motion has a net weightage of 9. Apart from that this chapter is especially important for candidates who want to pursue science in higher classes.

Why Choose Notes of ch 8 Science Class 9 by Vedantu?

Vedantu's class 9th science chapter 8 notes are the greatest study companion available. Physics may be a difficult topic to master, especially on your own. Vedantu's Motion class 9 notes are designed just for students. Despite the fact that it is developed by specialists, the language used is relatively easy. The subject of motion is both wide and vital. It is critical for students to understand as much as they can about motion without becoming overburdened. This is why Vedantu's notes give the ideal blend of study material that is both easy and dense with knowledge. Some advantages of utilising Vedantu include:

It is easy to access

It is easy to practice from

It is easy to check your answers and compare them with the solved ones

It is free to download from the official website of Vedantu

• It has solved numerical problems in it to help improve your understanding

The chapter 'Motion' may provide pupils with a wealth of information. All of the major topics are discussed here. The students are required to understand the important coverage so that they can prepare accordingly, this can be done by studying these notes from the chapter 'Motion'.

The students can also download the free pdf which is mentioned in this article. 

FAQs on Motion Class 9 Notes CBSE Science Chapter 8 (Free PDF Download)

1. When can an object have a zero displacement according to what you have studied in Notes of the Chapter Motion?

If an item goes along a certain path and then returns to its original position, the associated displacement of that object is zero. The length of the path will be travelled, but because the beginning and ending points of the movement are the same, the displacement will be zero.

2. Explain the type of motions according to what you have studied in Notes of the Chapter Motion.

Motion is the term given to the change in position of an object with respect to time. There are three types of motions:

Rotatory Motion: This takes place in a circular path when the object moves around the axis of rotation.

Vibrational Motion: This occurs when the object moves to and from a fixed position.

Translation Motion: This is caused by an object moving along a curved path or a straight line.

3. What are the subtopics covered in Class 9 Science Chapter 8 notes?

The CBSE Class 9 Science Chapter 8 - Motion Revision Notes provide well-explained details on each topic that has been covered in the NCERT and is a part of the latest syllabus provided by CBSE for Class 9 Science. These subtopics include the following:

Motion and Rest

Types of Motion

Distance and Displacement

Uniform and Non-uniform Motion

Acceleration

Basics of Plotting a Graph

Newton’s Laws of Motion

4. What are the benefits of referring to Vedantu’s Class 9 Science Chapter 8 notes?

Referring to CBSE Class 9 Science Chapter 8 - Motion Revision Notes by Vedantu can provide students with several benefits like:

Providing one of the best study materials for preparation

Save time by providing notes that are ready for studying

Help in increasing clarity about the different concepts taught in chapter 8

Help in understanding and solving numerical that are an important part of this chapter

Free of cost PDF download for studying offline

5. What are the Laws of Motion discussed in Chapter 8 Class 9 Science?

Newton established three laws of motion. According to the first law of motion, an object at rest or in uniform motion will remain in that condition until an external force is applied to it. Newton's second law states that the acceleration of an item is proportional to the force applied to it and its mass. According to the third and final law, every action has an equal and opposite response. Students may learn more by visiting the Vedantu website or app.

STUDY MATERIALS FOR CLASS 9

• NCERT Exemplar
• Science Exemplar Class 9

NCERT Exemplar Class 9 Science Solutions for Chapter 8 - Motion

Ncert exemplar solutions class 9 science chapter 8 – free pdf download.

NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion are provided to help the students of Class 9 who wish to have a good foundation on the subject, which will help them to prepare for their upcoming annual exams as well as various competitive exams. One such basic chapter that helps students in their future studies is Motion. Here, we have provided the NCERT Exemplar for Class 9 Science Chapter 8 PDF so that students can access and download the study material from anywhere in the world at any time convenient for them.

NCERT Exemplar Class 9 Motion offers answers to the questions provided in the NCERT Exemplar book. This Exemplar has MCQs, numerical problems, short answer questions and long answer questions. Motion is one of the most important topics in Physics. In Class 9, students will learn the basic concepts of Physics like motion in a straight line, uniform and non-uniform motion, velocity and rate of change of velocity. Enhance your learning by solving the NCERT Exemplar Science Class 9 Chapter 8 Motion. To download these solutions, visit the below link.

NCERT Exemplar for Class 9 Science Chapter 8 – Motion

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Access Answers to NCERT Exemplar Class 9 Science Chapter 8 – Motion

Multiple-choice questions.

1. A particle is moving in a circular path of radius r. The displacement after half a circle would be:

The answer is (c) 2 r.

Explanation:

After half revolution

Distance travelled = X circumference= πr

Path length

Displacement = Final position- Initial Position

It comes out to be the diameter of the circle = 2 r.

2. A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is,

(b) u 2 /2g

The answer is (b) u 2 /2g.

V 2 = u 2 +2 as

0 = u² -2gH

3. The numerical ratio of displacement to the distance for a moving object is

(a) always less than 1

(b) always equal to 1

(c) always more than 1

(d) equal or less than 1

The answer is (d) equal or less than 1

The shortest distance between the initial and the endpoint is called displacement. Distance is the total path length.

Displacement is vector, and it may be positive or negative, whereas Distance is scalar, and it can never be negative.

The distance can be equal to or greater than displacement, which means the ratio of displacement to distance is always equal to or less than 1.

4. If the displacement of an object is proportional to square of time, then the object moves with

(a) uniform velocity

(b) uniform acceleration

(c) increasing acceleration

(d) decreasing acceleration

The answer is (b) uniform acceleration

Velocity is measured in distance/second, and acceleration is measured in distance second 2 . Hence uniform acceleration is the right answer.

5. From the given v – t graph (Fig. 8.1), it can be inferred that the object is

(a) in uniform motion

(b) at rest

(c) in non-uniform motion

(d) moving with uniform acceleration

The answer is (a) in uniform motion

From the above-given graph, it is clear that the velocity of the object remains constant throughout hence the object is in uniform motion.

6. Suppose a boy is enjoying a ride on a merry-go-round which is moving at a constant speed of 10 m/s. It implies that the boy is

(a) at rest

(b) moving with no acceleration

(c) in accelerated motion

(d) moving with uniform velocity

The answer is (c) in accelerated motion

The boy is moving in a circular motion, and circular motion is an accelerated motion; hence C) is the right answer.

7. Area under a v – t graph represents a physical quantity which has the unit

The answer is (b) m

The area given in the graph represents Displacement, and its unit is meter. Hence, the answer is (b) m.

8. Four cars, A, B, C and D, are moving on a levelled road. Their distance versus time graphs are shown in Fig. 8.2. Choose the correct statement

(a) Car A is faster than car D.

(b) Car B is the slowest.

(c) Car D is faster than car C.

(d) Car C is the slowest.

The answer is (b) Car B is the slowest.

The graph shows that Car B covers less distance in a given time than A, C and D cars hence it is the slowest.

9. Which of the following figures (Fig. 8.3) represents the uniform motion of a moving object correctly?

The answer is (a)

Distance in graph a) is uniformly increasing with time hence it represents uniform motion.

10. Slope of a velocity–time graph gives

(a) the distance

(b) the displacement

(c) the acceleration

(d) the speed

The answer is (c) the acceleration

11. In which of the following cases of motions the distance moved and the magnitude of displacement are equal?

(a) If the car is moving on a straight road

(b) If the car is moving in a circular path

(c) The pendulum is moving to and fro

(d) The earth is revolving around the Sun

The answer is (a) If the car is moving on a straight road

In other cases given here, displacement can be less than distance; hence option (a) If the car is moving on a straight road, is the right answer.

Short Answer Questions

12. The displacement of a moving object in a given interval of time is zero. Would the distance travelled by the object also be zero? Justify your answer.

Displacement zero does not mean zero distance. The distance can be zero when moving an object back to the place it started. Displacement is either equal to or less than distance, but the distance is always greater than one, and it cannot be a negative value.

13. How will the equations of motion for an object moving with a uniform velocity change?

If the object is moving with a uniform velocity, then v = µ and a = 0. In this scenario equation for distance is given below.

S= ut and V 2 – µ 2 =0

14. A girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement–time graph is shown in Fig.8.4. Plot a velocity-time graph for the same.

15. A car starts from rest and moves along the x-axis with a constant acceleration of 5 m/s 2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest?

Car Starts from rest hence Initial velocity u=o acceleration a=5 m/s 2 and time t=8s

v = 0+5×8

v = 40ms -1

From second equation

s = ut + \(\begin{array}{l}\frac{1}{2}\end{array} \) at 2

s = 0x8 + \(\begin{array}{l}\frac{1}{2}\end{array} \) x5x(8) 2

s = \(\begin{array}{l}\frac{1}{2}\end{array} \) x5x(8) 2

s = \(\begin{array}{l}\frac{1}{2}\end{array} \) x5x64

s = 5×32 =160 is the distance covered in 8 seconds.

Therefore, the total distance covered in 12 seconds  is 160+160=320m

16. A motorcyclist drives from A to B with a uniform speed of 30 km/h and returns back with a speed of 20 km h –1 . Find its average speed.

Let the distance from A to B is D kms.

Distance for the entire journey is 2D kms.

The time taken to go from A to B is D/30 hr, and that of B to A is D/20 hr. So, the total time taken T is

T = (D/30) + (D/20). By solving, we will get,

T = D/12 hrs.

Average speed = Total distance/Total time.

Av.speed = 2D ÷ D/12

=> 2D x 12/D = 24 km/h.

Hence Average speed of the motorcycle is 24 km/h.

17. The velocity-time graph (Fig. 8.5) shows the motion of a cyclist. Find (i) its acceleration, (ii) its velocity, and (iii) the distance covered by the cyclist in 15 seconds

(i) As velocity is constant, acceleration is 0 m/s 2

(ii) Here, the velocity is constant, hence v=20m/s

(iii) s = v x t

18. Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height.

The velocity versus time graph of a stone thrown upwards vertically is as given below:

19. An object is dropped from rest at a height of 150 m, and simultaneously another object is dropped from rest at a height of 100 m. What is the difference in their heights after 2 s if both the objects drop with the same accelerations? How does the difference in heights vary with time?

When two objects fall with the same acceleration simultaneously, after 2 seconds, the difference in their heights will not change, and it remains 50 m.

Therefore the height of the first object after 2 seconds is 130 m.

In the same way, the height of the second object is

Therefore, the height of the second object after 2 seconds is 80 m.

So, the difference is the same, i.e. 50 m.

This concludes that the difference in the height of the two objects does not depend on time and will always be the same.

20. An object starting from rest travels 20 m in first 2 s and 160 m in next 4 s. What will be the velocity after 7 s from the start?

Here Object starts from rest hence initial velocity u=0 t =2s and s=20 m

According to the second equation of motion s= ut+at 2

S = 0+ \(\begin{array}{l}\frac{1}{2}\end{array} \) ax2 2

20 = 2+ \(\begin{array}{l}\frac{1}{2}\end{array} \) ax2 2 = 2a

According to the first equation of motion velocity after 7 s from the start

V = 0+10×7

21. Using the following data, draw time-displacement graph for a moving object:

Use this graph to find the average velocity for the first 4 s, for the next 4 s and for the last 6 s.

Average velocity for the first 4s = \(\begin{array}{l}\frac{change in displacement}{Total time taken}\end{array} \) = (4-0)/(4-0)=4/4 = 1ms -1

Average velocity of next 4 s = V = \(\begin{array}{l}\frac{4-4}{8-4}\end{array} \) =0

Average velocity for last 6 s = \(\begin{array}{l}\frac{(0-6)m}{(16-10)s}\end{array} \) = \(\begin{array}{l}\frac{-6}{6}\end{array} \) = 1 ms -1

22. An electron moving with a velocity of 5 × 10 4 m/s enters into a uniform electric field and acquires a uniform acceleration of 10 4  ms -2  in the direction of its initial motion.

(i) Calculate the time in which the electron would acquire a velocity double of its initial velocity.

(ii) How much distance would the electron cover in this time?

Given initial velocity, u = 5 × 10 4  m/s  and acceleration, a = 10 4 ms -2

(i) final velocity = v = 2 u = 2 × 5 ×10 4  m/s =10 × 10 4  m/s

To find t,      use  v  =  at  or t = u – u / a = (5 × 10 4 )/10 4

(ii)   Using s = ut + \(\begin{array}{l}\frac{1}{2}\end{array} \) at 2  = (5 ×10 4 ) × 5 + \(\begin{array}{l}\frac{1}{2}\end{array} \) (10 ) × (5) 2

= 25 ×10 4 + 25 /2 ×10 4

= 37.5×10 4 m

23. Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between the 4th and 5th seconds.

Assume that air resistance is nil.

We can directly contain it by using Newton’s equations of motion or from the below-mentioned method:

Thus, the area under the v-t curve and the x-axis where the slope of the curve is the instantaneous acceleration.

In this case, acceleration g is constant, and due to the free-fall condition, the initial velocity is zero. Therefore the v-t curve is a straight line with a slope equal to g equal to 9.81 m/s  passing through the origin.

On dividing the total area under the curve into the interval of unit seconds, then we initially obtain a triangle followed by trapeziums of increasing height.

The ratio of the area of the first triangle to the second triangle to the third triangle is equal to the ratio of displacement in the first, second and third second. We get ratio equal to 1:3:5:7:9…  and so on.

For the 4th & 5th second, it is 7:9.

24. Two stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2, respectively. Prove that the heights reached by them would be in the ratio of u1² :u2²

(Assume upward acceleration is –g and downward acceleration is +g ).

Important Topics of NCERT Exemplar for Class 9 Science Chapter 8 Motion

• Introduction
• Distance and displacement
• Uniform and non-uniform motion
• Equations of motion
• Uniform circular motion

We, at BYJU’S, provide NCERT Exemplar chapter-wise to help students in their studies. All the solutions in this study material are explained in simple language and in an easy format for students to understand better and learn effectively. These NCERT Exemplars for Class 9 Science , Chapter 8 Motion provide complete information regarding the topic, along with the definitions and examples. These NCERT Exemplars are available in PDF, and students can directly download these materials or can practise online by visiting BYJU’S website or by downloading BYJU’S – The Learning App. Also, they can access other CBSE study materials, such as notes , sample papers, videos and animations by accessing them.

Frequently Asked Questions on NCERT Exemplar Solutions for Class 9 Science Chapter 8

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