NCERT Exemplar Class 9 Science Solutions chapter 8 Motion

Vishal kumarUpdated on 04 Dec 2025, 11:41 PM IST

Have you ever questioned yourself about how cars, bicycles or even people move between one location and another? Chapter 8: Motion: Class 9 Science outlines the various modes of motion and presents the major concepts of distance, displacement, speed, velocity, acceleration, equations of motion and graphs of motion. Such Class 9 NCERT Exemplar Solutions of Chapter 8 in Science are designed by highly qualified teachers based on the most current CBSE specifications because they can enable the students to develop an adequate conceptual clarity.

This Story also Contains

NCERT Exemplar Class 9 Science Solutions Chapter 8-MCQ
NCERT Exemplar Class 9 Science Solutions Chapter 8-Short Answer
NCERT Exemplar Class 9 Science Solutions Chapter 8-Long Answer
NCERT Exemplar Class 9 Science Solutions Chapter 8 Important Concepts and Formula
Advantages of NCERT Exemplar Class 9 Science Chapter 8 Motion Solutions
NCERT Class 9 Science Exemplar Solutions for Other Chapters
Check the Solutions of Questions Given in the Book

NCERT Exemplar Class 9 Science Solutions chapter 8 Motion

NCERT Exemplar Class 9 Science Solutions Chapter 8 Motion are well structured in terms of answering all the exemplar questions, MCQs, Very Short Answer, Short Answer and Long Answer. By solving these solved exemplar questions, you not only improve your knowledge of motion, but also improve your accuracy in solving numerical problems, and are ready to take on school exams, NTSE and also Olympiads. The NCERT exemplar solutions also encourage critical thinking as they provide explanations, diagrams, and real-life examples that are more meaningful to learn. Use of these NCERT Exemplar Class 9 Solutions Science Chapter 8 Motion regularly gives sound fundamentals and better performance in the exam

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NCERT Exemplar Class 9 Science Solutions Chapter 8-MCQ

The Motion Class 9 NCERT Exemplar MCQs are aimed at enhancing the clarity of the concepts, as well as assessing your knowledge of motion, speed, velocity, and acceleration. These multiple-choice questions are used to allow the students to put concepts into practical use and to increase accuracy in the school exams and competitive tests. Taking these MCQs helps in developing analytical skills and increases confidence to solve more physics problems as well.

Question:1

A particle is moving in a circular path of radius r. The displacement after half a circle would be:
(a) Zero
(b) $\pi r$
(c) 2 r
(d) 2 $\pi r$

Answer:

Displacement is the direct separation length from initial point to final point.
Distance is the total path length travelled by the particle.
As the particle covers a half circle, it will reach at diametrically opposite point.
If someone asks, What is the distance travelled, the answer will be half of the Perimeter.
But displacement will be the direct separation between these two diametric points, and it would be equal to the Diameter.
Hence, the correct answer is C, which is 2r.

Question:2

A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is,
(a) $\frac{u}{g}$
(b) $\frac{u^{2}}{2g}$
(c) $\frac{u^{2}}{g}$
(d) $\frac{u}{2g}$

Answer:

The body is thrown upwards with some initial velocity, and the acceleration due to gravity is constant, which will retard the body.
As the acceleration is constant, we can use the equation of motion.
By using the third equation of motion and putting proper values as-
$v^{2}=u^{2}+2as$
where v = 0 and a= -g
$0^{2}=u^{2}+2(-g)h$
$h=\frac{u^{2}}{2g}$
Hence, the correct answer is option B

Question:3

The numerical ratio of displacement to distance for a moving object is
(a) always less than 1
(b) always equal to 1
(c) always more than 1
(d) equal to or less than 1

Answer:

The shortest distance between the initial point and the final point is called displacement, and the total path length is called distance.
If a particle travels in a straight line without turning back, then its distance and displacement will be equal. In all other cases distance will be more than or equal to the displacement.
Therefore, the ratio of displacement and distance will be equal to or less than 1
Hence, the correct option is D.

Question:4

If the displacement of an object is proportional to the square of time, then the object moves with
(a) uniform velocity
(b) uniform acceleration
(c) increasing acceleration
(d) decreasing acceleration

Answer:

If the object moves with uniform velocity, the displacement increases linearly with time. That means displacement will be proportional to time.
In the case of uniform acceleration, we can use the equation of motion.
$v = u + at$
$s = ut + \frac{1}{2}$
$v^{2}=u^2 + 2as$
If we observe the second equation of motion carefully for zero initial velocity, we will observe that displacement is proportional to the square of time.
$s = \frac{1}{2}at^2$
In case of variable acceleration, whether it is increasing or decreasing, the displacement will not be proportional to the square of time.
Hence, the correct option is B.

Question:5

From the given v – t graph (Figure), 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
V-T graph for uniform velocity

Answer:

If we observe the graph carefully, we can see that at every time, the value of velocity remains constant.
Hence, the acceleration or retardation is zero as the speed is not changing.
The velocity is constant, but it is non-zero, hence the particle must be moving with uniform velocity.
The motion of such type is called uniform motion.
Hence, the correct option is option A

Question:6

Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of 10 ms^–1. It implies that the boy is:
(a) at rest
(b) moving with no acceleration
(c) in accelerated motion
(d) moving with uniform velocity

Answer:

Velocity is a vector quantity, which has direction as well as magnitude.
In a merry-go-round, even though the boy is moving with constant speed, the direction of velocity keeps changing.
Hence, we can say the velocity of the boy is changing.
The quantity, which changes the velocity, is called acceleration.
Therefore, the boy must be in accelerated motion.
Hence, the correct option for this question is C.

Question:7

Area under a v – t graph represents a physical quantity which has the unit
(a) $m^2$
(b) m
(c) $m^3$
(d) $ms^{-1}$

Answer:

The area under any x-y Graph will have units as the product of the unit of x and the unit of y.
That means the area under the Graph of velocity and time will have a unit equal to the Product of the unit of velocity (m/s) and the unit of time (s).
That would be meter.
For the information, the area under the v-t graph gives displacement.
Hence, the correct option is option B

Question:8

Four cars A, B, C and D are moving on a levelled road. Their distance versus time graphs are shown in Figure. 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.
Distance -Time graph

Answer:

If we closely observe the graph, we will notice:
For any time interval distance travelled by car C is maximum followed by D, A and B.
Hence, car C is the fastest car and car B is the slowest car.
In a distance-time graph, if the graph is a straight line, then the angle between the straight line and the time axis represents the slope or velocity of the car.
The larger the angle, implies larger the slope means a larger velocity.
Hence, the correct option is B.

Question:9

Which of the following figures (Figure) represents uniform motion of a moving object correctly?
Distance -Time graph

Answer:

Uniform motion means the body is moving with constant velocity.
Hence, in every second, it will cover the same distance.
Therefore, for uniform motion distance time graph will be a straight line.
In the graph shown in option B, the distances travelled in consecutive seconds are increasing, hence the speed is increasing.
In the graph shown in option C, the distances travelled in consecutive seconds are decreasing, hence the speed is decreasing.
In option D distance is not changing, hence the object is not moving.
Therefore, the correct option is option A.

Question:10

Slope of a velocity – time graph gives
(a) the distance
(b) the displacement
(c) the acceleration
(d) the speed

Answer:

If we observe a velocity-time graph, we can calculate displacement as well as acceleration.
The area under the velocity time graph gives the displacement, and the slope of velocity time graph gives the acceleration.
Qualitatively, the slope of the line represents the angle made by the straight line from the time axis.
The larger the angle, implies larger the slope.
Hence, the correct option is option C

Question: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 straight road
(b) If the car is moving in circular path
(c) The pendulum is moving to and fro
(d) The earth is revolving around the Sun

Answer:

Sol.
The distance is the total path travelled by anybody. The displacement is the shortest distance between the initial and final points of the motion.
Only in the case when the body moves in a straight line without turning, the distance will be equal to the displacement.
In all other cases distance will be more than the displacement.
Hence, the correct option for this question is option A.

NCERT Exemplar Class 9 Science Solutions Chapter 8-Short Answer

Short Answer questions in the NCERT Exemplar Class 9 Science Chapter 8 assist students in explaining things in a clear way and forming step-by-step reasoning. The questions enhance conceptual knowledge and make you write to the point in tests with a concise, accurate answer.

Question: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 you answer.

Answer:

The displacement is the shortest distance between the initial point and the final point in any motion.
The distance is the total path length of the motion.
If anybody starts from a point and comes back to the same point. The displacement will be zero, but the distance will be non-zero.
For example, if you start your scooter from your home and after some time you park the scooter in the same place. The displacement will be zero, but the distance is non-zero.
The reading of scooter meter will show the distance travelled.
Hence, zero displacement does not guarantee zero distance.

Question:13

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

Answer:

If a particle is moving with uniform velocity, it means its acceleration is zero.
Hence, its speed remains constant at all times; therefore, 1st equation looks as:
v = u
The $2^{nd}$ equation of motion looks as:
s = ut
and the 3rd equation looks as:
$v^{2}=u^{2}$

Question: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 Figure. Plot a velocity–time graph for the same.
Dispalcement -Time graph

Answer:

If we observe the displacement time graph, for the part of the graph where the distance is increasing, the graph is a straight line. It means that its slope is constant. Therefore, we can say its velocity will be constant for that time interval.
Later, its distance is decreasing with a constant slope. It means that it is moving with constant velocity in the opposite direction.
Mathematically, the opposite direction of velocity will be negative velocity.

Hence, its v-t graph will look as shown below:

Velocity - Time graph

Question:15

A car starts from rest and moves along the x-axis with constant acceleration $5ms^{-2}$ 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?

Answer:

For the first eight seconds, the car is moving with constant acceleration with zero initial velocity.
Hence, we can use equations of motion to find out the distance and final velocity after that time interval of eight seconds.
$s =ut +\frac{1}{2}at^{2}$
$v = u + at$
where u=0 , a =5 , t=8
By putting values
s = 160 m
v = 40 m/s
Now, for the next four seconds, the car will move with a constant speed of 40 m/s. Hence, in this time interval, it will travel 160 m.
Therefore, for total distance travelled will be 320 m.

Question:16

A motorcyclist drives from A to B with a uniform speed of 30 $km h^{-1}$ and returns back with a speed of 20 $km h^{-1}$. Find its average speed.

Answer:

The average speed is total distance travelled divided by total time taken.
Let the distance between A to B be X kilometres.
As it is moving with uniform speed, the time taken from A to B will be $\frac{X}{30}$.
Similarly time taken from B to A will be $\frac{X}{20}$.
Hence total time of motion will be $\left ( \frac{X}{30} \right ) +\left (\frac{X}{20} \right )= \frac{X}{12}$.
Total distance travelled will be equal to 2X.
Therefore, by dividing the total distance by the total time, we can find the average speed which would be 24 km/h.

Question:17

The velocity time graph (Figure) shows the motion of a cyclist.
Find (i) its acceleration (ii) its velocity and (iii) the distance covered by the cyclist in 15 seconds.
Velocity - Time graph

Answer:

(i) As there is no change in velocity, we can say that its acceleration is zero all the time.
(ii) If we observe the velocity-time graph, we can easily deduce that the velocity is constant and equal to 20 m/s during the complete motion.
(iii) The distance travelled by the cyclist can be calculated by the formula:
$s = v t = (20) (15) m = 300 m$
Hence, the distance travelled will be 300 m.

Question:18

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

Answer:

When a stone is thrown upwards, it will have an initial positive velocity.
This velocity will decrease uniformly due to gravity and become zero at the highest point.
After that, its velocity will increase its magnitude but in a downward direction.
Hence, the correct graph of velocity will be as shown:
Velocity - Time graph

If we try to draw the graph of speed versus time, it would look like as shown:
Speed - Time Graph

NCERT Exemplar Class 9 Science Solutions Chapter 8-Long Answer

Chapter 8 contains the long-answer questions, which make one think more and elaborate on the ideas of motion. Such solutions assist students to know how to give well organized step by step responses to get good marks in exams.

Question:19

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

Answer:

As the objects are dropped from rest, their initial velocity is zero.
The acceleration will be acceleration due to gravity, which will be equal to 10 ms^-2.
The height of the particle is written as:
$D = H-S$
Where H is the initial height, and S is the displacement.
If we write the heights of both the particles by using the second equation of motion: $s = ut + \frac{1}{2}at^2$
$D_{1}=H_{1}- \left ( ut + \frac{1}{2}at^{2} \right )$
$D_{2}=H_{2}- \left ( ut + \frac{1}{2}at^{2} \right )$
Hence,
$D_{1}-D_{2}=H_{1}-H_{2}=50m$
The difference in height at any time is coming out to be 50 m.
So even after two seconds, the difference will be 50 m, and it will not vary with time.

Question: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.

Answer:

We can observe that the distance travelled in consecutive seconds is increasing; hence, we can conclude that the object is accelerating.
Let its initial speed be u, and the acceleration is a.
Hence, by the 2nd equation of motion, we can write:
$s= ut + \frac{1}{2}at^{2}$
$20= u(2) + \frac{1}{2}a(2)^{2}$
$\Rightarrow 20= 2u +2a$
$\Rightarrow 10= u +a$ ..................(i)
and
$180 =u (6)+ \frac{1}{2}a(6)^{2}$
$180 =6u+ 18 a$
$\Rightarrow 30 =u+ 3 a$ ............(ii)
Solving (i) and (ii)
u = 0
$a= 10m /s^{2}$
Since acceleration is the same, the velocity after 7 s from the start will be $v' = 0+(10\times 7 ) = 70 \;ms^{-1}$

Question:21

Using following data, draw time displacement graph for a moving object:
Time (s) 0 2 4 6 8 10 12 14 16
Displacement (m) 0 2 4 4 4 6 4 2 0
Use this graph to find the average velocity for the first 4 s, for the next 4 s and for the last 6 s.

Answer:

By using the data given in the table, if we draw the displacement time graph, it will look as shown:
Displacement - Time graph

Average velocity is displacement divided by time in any time interval.
In first four seconds, it has travelled four meter hence average velocity will be 1 m/s.
For the next four seconds, its position is not changing. This means it is not moving. Hence, its average velocity will be zero.
In the last six seconds, from t = 10 to t = 16 sec, it has changed its position from 6 m to 0 m.
Hence, in this time interval, displacement is 6 m and time is also 6 seconds.
Therefore, the average velocity will be 1 m/s.

Question:22

An electron moving with a velocity of $5 \times 10^4 ms^{-1 }$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?

Answer:

(i) Given that, the Initial velocity of the electron is u =$5 \times 10^4 ms^{-1 }$
Acceleration of the electron is a = $10^4 ms^{-2 }$
We know that the first equation of motion is,
v = u + at
2u = u + at
$t = \frac{u}{a} =\frac{ (5 \times 10^4) }{ (10^4) }= 5 seconds$
(ii) Now we know that,
$s = ut + \frac{1}{2}at^{2}$
$s =( 5 \times 10^{4})5+ \frac{1}{2}(10)^{4}(5)^{2}$
$s = 37.5 \times 10^{4}m$

Question:23

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

Answer:

Distance travelled in the nth second by any particle is equal to the difference of distances travelled in n seconds and n -1 seconds.

For example, the distance travelled in the fourth second is the difference between the distance travelled in four seconds and three seconds.

Hence, by using the second equation of motion, we can find out the distance travelled in the nth second.

$\begin{aligned} & \mathrm{s}_{\mathrm{n}}=\mathrm{un}+\frac{1}{2} \mathrm{an}^2 \\ & \mathrm{~s}_{\mathrm{n}-1}=\mathrm{u}(\mathrm{n}-1)+\frac{1}{2} \mathrm{a}(\mathrm{n}-1)^2 \\ & \Rightarrow \mathrm{~s}_{\mathrm{nth}}=\mathrm{u}+\frac{1}{2} \mathrm{a}(2 \mathrm{n}-1)\end{aligned}$

If the initial velocity is zero,

In that case, the distance travelled in the fourth and fifth seconds will be in the ratio of 7:9.

Question:24

Two stones are thrown vertically upwards simultaneously with their initial velocities $\mathrm{u}_1$ and $\mathrm{u}_2$ respectively. Prove that the heights reached by them would be in the ratio of $u_1^2: u_2^2$

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

Answer:

The height reached by any stone can be calculated by using the third equation of motion.

$v^2=u^2+2as$

We can use the fact that at the highest point the velocity will be zero. The acceleration due to gravity will retard the particles.

Therefore, the height raised for any particle will be given as:

$\\0^2=u^2+2as\\\Rightarrow h=\frac{u^2}{2g}$

Therefore

$\frac{h_1}{h_2}=\frac{u_1^2}{u_2^2}\\$

NCERT Exemplar Class 9 Science Solutions Chapter 8 Important Concepts and Formula

The essential Concepts and Formulas of Chapter 8, Motion, are the central elements of numerical and application problems of physics. They assist students to comprehend the movement of objects, measurement of distances, computation of velocities and accelerated motion. The knowledge of these formulas will provide a strong base in more advanced physics and will be accurate in resolving Exemplar problems.

1. Distance and Displacement

Distance is the total path covered (scalar quantity).
Displacement is the shortest straight-line distance from the initial to the final position (vector quantity).
Displacement $=$ Final position - Initial position

2. Speed

Speed is how fast an object moves (scalar quantity).
Formula:

$
\text { Speed }=\frac{\text { Distance }}{\text { Time }}
$

3. Velocity

Velocity is speed in a given direction (vector quantity).

$
\text { Velocity }=\frac{\text { Displacement }}{\text { Time }}
$

4. Acceleration

Acceleration is the rate of change of velocity (vector quantity).

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

where:
$a=$ acceleration
$v=$ final velocity
$u=$ initial velocity
$t=$ time

5. Uniform and Non-uniform Motion

Uniform motion: Equal distances in equal intervals of time.
Non-uniform motion: Unequal distances in equal time intervals.

6. Equations of Uniformly Accelerated Motion

These are used when acceleration is constant.

First Equation:

$
v=u+a t
$

Second Equation:

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

Third Equation:

$
v^2=u^2+2 a s
$

where:
$s=$ displacement

7. Average Speed

Total distance covered divided by total time taken.

$
\text { Average Speed }=\frac{\text { Total Distance }}{\text { Total Time }}
$

Advantages of NCERT Exemplar Class 9 Science Chapter 8 Motion Solutions

NCERT Exemplar Class 9 Science Chapter 8 is provided with high-quality practice, which enhances clarity in concepts and accuracy in solving problems. These solutions describe the application-based, tricky, and HOTS questions in a simple and systematic manner. They assist students in the interpretation of motion, speed, velocity, and graphs by the use of real-life reasoning. On the whole, they are ideal in establishing robust bases for exams and higher classes.

They reinforce the knowledge of the understanding of core motion concepts using diverse and challenging problems.
They enhance the ability to solve numerical problems because they explain a complex calculation process step by step.
They improve the level of thinking by incorporating higher-order questions that are common in school exams and Olympiads.
They instil confidence by providing well-formulated solutions in a clear way with regard to the CBSE guidelines.
They make students become more accurate in interpreting motion graphs and using formulas properly.
They assist in excellent preparation for examinations because they involve all question types, which are MCQs, short answers, as well as long answers.
They promote self-learning through providing elaborate explanations behind each answer, thus simplifying ideas to revise.

NCERT Class 9 Science Exemplar Solutions for Other Chapters

NCERT Class 9 Science Exemplar Solutions for other chapters provide well-structured, high-quality practice material designed to strengthen conceptual understanding and analytical skills. These solutions assist students in solving application-based and higher-level questions that usually emerge in competitive and school examinations. They simplify the complex topics by providing straightforward explanations and logical ways of learning, thus making it easier to revise.

Chapter 1 Matter in our Surroundings

Chapter 2 Is Matter Around Us Pure?

Chapter 3 Atoms and Molecules

Chapter 4 Structure of the Atom

Chapter 5 The Fundamental Unit of Life

Chapter 6 Tissues

Chapter 7 Diversity in Living Organisms

Chapter 9 Forces and Laws of Motion

Chapter 10 Gravitation

Chapter 11 Work and Energy

Chapter 12 Sound

Chapter 13 Why do We Fall ill?

Chapter 14 Natural Resources

Chapter 15 Improvement in Food Resources

Check the Solutions of Questions Given in the Book

Chapter No.	Chapter Name
Chapter 1	Matter in Our Surroundings
Chapter 2	Is Matter Around Us Pure
Chapter 3	Atoms and Molecules
Chapter 4	Structure of The Atom
Chapter 5	The Fundamental Unit of Life
Chapter 6	Tissues
Chapter 7	Motion
Chapter 8	Force and Laws of Motion
Chapter 9	Gravitation
Chapter 10	Work and Energy
Chapter 11	Sound
Chapter 12	Improvement in Food Resources

Also read - NCERT Solutions for Class 9

Also, Read NCERT Solution Subject Wise

Check NCERT Notes Subject Wise

Also, Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

Q: Are graphs important in understanding motion?

Graphs are important to analyze the behavior of any phenomenon. In motion, graphs are very important and with the help of a graph we can easily find out one variable from another variable such as displacement, distance, or acceleration from the speed-time graph.

Q: “Motion or rest “can we absolutely determine?

No motion or rest is totally relative. Whether a body is in motion or at rest will depend on the observer who is observing that body. A man sitting on the train will find that everybody else as the fellow passengers are at rest but they are moving with respect to the man on the ground.

Q: How to find displacement from the speed-time graph?

The area under the speed-time graph will give displacement or distance of any moving body.

Q: Is Motion an important topic for competitive examination?

Motion is one of the building blocks of physics and is an extremely important topic from the perspective of competitive exams such as JEE Advanced and NEET.

Q: Will I be able to download these solutions and view them in an offline mode?

Certainly, NCERT exemplar Class 9 Science solutions chapter 8 pdf download is the feature provided to students to view/download these solutions for use in offline mode and attempt NCERT exemplar Class 9 science chapter 8. NCERT exemplar Class 9 science solutions chapter 7 are elaborate in nature and judicious use can help the students score well in the examinations.

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