NCERT Solutions for Class 9 Science Chapter 4 Describing Motion Around Us (2026-27)

Revise, Reflect, Refine (NCERT Textbook Page No. 68)

Question 1. My father went to a shop from home which is located at a distance of 250 m on a straight road. On reaching there, he discovered that he forgot to carry a cloth bag. He came home to take it, went to the shop again, bought provisions and came back home. How much was the total distance travelled by him? What was his displacement from home?

Answer:
The distance between home and the shop is 250 m.

The father travels this distance four times:

Home to shop = 250 m
Shop to home = 250 m
Home to shop again = 250 m
Shop to home again = 250 m

Total distance travelled = 250 + 250 + 250 + 250 = 1000 m

Since he starts from home and finally comes back home, his final position is the same as his starting position.

Displacement = 0 m

Therefore, the total distance travelled is 1000 m, and the displacement from home is 0 m.

Question 2. A student runs from the ground floor to the fourth floor of a school building to collect a book and then comes down to their classroom on the second floor. If the height of each floor is 3 m, find:
(i) the total vertical distance travelled, and
(ii) their displacement from the starting point.

Answer:
Height of each floor = 3 m

The student first moves from the ground floor to the fourth floor.

Distance travelled upward = 4 × 3 = 12 m

Then the student comes down from the fourth floor to the second floor.

Distance travelled downward = 2 × 3 = 6 m

(i) Total vertical distance travelled = 12 + 6 = 18 m

(ii) The student finally reaches the second floor. So, the final position is 2 floors above the ground floor.

Displacement from starting point = 2 × 3 = 6 m upward

Therefore, the total vertical distance travelled is 18 m, and the displacement is 6 m upward.

Question 3. A girl is riding her scooter and finds that its speedometer reading is constant. Is it possible for her scooter to be accelerating, and if so, how?

Answer: Yes, the scooter can be accelerating even if the speedometer reading remains constant.

A speedometer shows only speed, not direction. Acceleration occurs when velocity changes. Since velocity includes both speed and direction, a change in direction also means a change in velocity.

If the scooter is moving along a curved road at constant speed, its direction keeps changing. Therefore, its velocity changes continuously, and the scooter is accelerating even though its speed remains the same.

Question 4. A car starts from rest and its velocity reaches 24 ms-1 in 6 s. Find the average acceleration and the distance travelled in these 6 s.

Answer:
Given:

Initial velocity, u = 0 m/s
Final velocity, v = 24 m/s
Time, t = 6 s

Average acceleration:

a = (v - u) / t
a = (24 - 0) / 6
a = 4 m/s²

Average acceleration = 4 m/s²

Now, distance travelled:

Using the formula,

s = ut + 1/2 at²

s = 0 × 6 + 1/2 × 4 × 6²
s = 0 + 2 × 36
s = 72 m

Therefore, the average acceleration is 4 m/s², and the distance travelled is 72 m.

Question 5. A motorbike moving with initial velocity 28 ms-1 and constant acceleration stops after travelling 98 m. Find the acceleration of the motorbike and the time taken to come to a stop.

Answer:
Given:

Initial velocity, u = 28 m/s
Final velocity, v = 0 m/s
Distance, s = 98 m

Using the equation,

v² = u² + 2as

0² = 28² + 2 × a × 98
0 = 784 + 196a
196a = -784
a = -784 / 196
a = -4 m/s²

The acceleration is -4 m/s². The negative sign shows that the motorbike is slowing down.

Now, using:

v = u + at

0 = 28 + (-4)t
4t = 28
t = 7 s

Therefore, the acceleration of the motorbike is -4 m/s², and the time taken to stop is 7 seconds.

Question 6. Fig. 4.27 shows a position-time graph of two objects A and B that are moving along the parallel tracks in the same direction. Do objects A and B ever have equal velocity? Justify your answer.

Image 1

Answer:
No, objects A and B do not have equal velocity.

In a position-time graph, the slope of the graph represents velocity. A steeper slope means higher velocity, while a gentler slope means lower velocity.

In Fig. 4.27, the lines for objects A and B have different slopes. This shows that their velocities are different. Even if the two lines meet at a point, it only means that both objects are at the same position at that instant. It does not mean that their velocities are equal.

Therefore, objects A and B never have equal velocity because their position-time graphs have different slopes.

Question 7. A graph in Fig. 4.28 shows the change in position with time for two objects, A and B, moving in a straight line from 0 to 10 seconds. Choose the correct option(s).

Image 2

(i) The average velocity of both over the 10s time interval is equal since they have the same initial and final positions.
(ii) The average speeds of both over the 10s time interval are equal since both cover equal distances in equal time.
(iii) The average speed of A over the 10 s time interval is lower than that of B since it covers a shorter distance than B in 10 seconds.
(iv) The average speed of A over the 1st time interval is greater than that of B since B’s speed is lower than A’s in some segments.

Answer:
The correct options are (i) and (iii).

(i) is correct because both objects have the same initial and final positions over the 10 s time interval. Since displacement is the same, their average velocity is also the same.

Average velocity = Displacement / Time

(iii) is correct because object B covers more total distance than object A in the same time interval. Average speed depends on total distance travelled, not just displacement. Since B covers a longer path, its average speed is greater than that of A.

(ii) is incorrect because both objects do not cover equal total distance.

(iv) is not the best overall conclusion because average speed for the complete interval must be judged using the total distance covered in the full 10 seconds.

Question 8. A truck driver driving at the speed of 54 km h-1 notices a road sign with a speed limit of 40 km h-1 (Fig. 4.29) for trucks. He slows down to 36 km h-1 in 36 s. What was the distance travelled by him during this time? Assume the acceleration to be constant while slowing down.

image 3

Answer:
Given:

Initial speed, u = 54 km/h
Final speed, v = 36 km/h
Time, t = 36 s

Convert the speeds into m/s:

u = 54 × 5/18 = 15 m/s
v = 36 × 5/18 = 10 m/s

Since acceleration is constant, the distance travelled can be calculated using the average velocity.

Average velocity = (u + v) / 2
Average velocity = (15 + 10) / 2
Average velocity = 25 / 2 = 12.5 m/s

Distance travelled:

s = Average velocity × time
s = 12.5 × 36
s = 450 m

Therefore, the truck travels 450 m while slowing down.

Question 9. A car starts from rest and accelerates uniformly to 20 ms-1 in 5 seconds. It then travels at 20 ms-1 for 10 seconds and finally applies the brake (with uniform acceleration) to stop in 6 seconds. Find the total distance travelled.

Answer:
The motion has three parts.

Part 1: Uniform acceleration from 0 m/s to 20 m/s in 5 s

u = 0 m/s
v = 20 m/s
t = 5 s

s₁ = [(u + v) / 2] × t
s₁ = [(0 + 20) / 2] × 5
s₁ = 10 × 5 = 50 m

Part 2: Constant velocity of 20 m/s for 10 s

s₂ = v × t
s₂ = 20 × 10
s₂ = 200 m

Part 3: Uniform deceleration from 20 m/s to 0 m/s in 6 s

u = 20 m/s
v = 0 m/s
t = 6 s

s₃ = [(u + v) / 2] × t
s₃ = [(20 + 0) / 2] × 6
s₃ = 10 × 6 = 60 m

Total distance travelled = s₁ + s₂ + s₃
Total distance = 50 + 200 + 60
Total distance = 310 m

Therefore, the total distance travelled by the car is 310 m.

Question 10. A bus is travelling at 36 km h-1 when the driver sees an obstacle 30 m ahead. The driver takes 0.5 seconds to react before pressing the brake. Once the brake is applied, the velocity of the bus reduces with constant acceleration of 2.5 ms2. Will the bus be able to stop before reaching the obstacle?

Answer:
Given:

Speed of bus = 36 km/h
Obstacle distance = 30 m
Reaction time = 0.5 s
Retardation = 2.5 m/s²

Convert speed into m/s:

36 km/h = 36 × 5/18 = 10 m/s

Step 1: Distance covered during reaction time

s₁ = speed × time
s₁ = 10 × 0.5
s₁ = 5 m

Step 2: Braking distance

Initial velocity, u = 10 m/s
Final velocity, v = 0 m/s
Acceleration, a = -2.5 m/s²

Using:

v² = u² + 2as

0² = 10² + 2 × (-2.5) × s₂
0 = 100 - 5s₂
5s₂ = 100
s₂ = 20 m

Total stopping distance = reaction distance + braking distance
Total stopping distance = 5 + 20 = 25 m

The obstacle is 30 m ahead.

Since 25 m is less than 30 m, the bus will stop before reaching the obstacle.

Question 11. A student said, “The Earth moves around the Sun”. In this context, discuss whether an object kept on the Earth can be considered to be at rest.

Answer: Yes, an object kept on the Earth can be considered at rest, but only with respect to the Earth.

Motion and rest depend on the reference point or frame of reference. If we observe the object from the Earth’s surface, its position does not change with respect to the ground. So, it is at rest relative to the Earth.

However, if we observe the same object from the Sun, it is in motion because the Earth is revolving around the Sun and rotating on its axis.

Therefore, an object on Earth can be at rest in one frame of reference and in motion in another frame of reference.

Question 12. The velocity-time graph from 0 s to 120 s for a cyclist is shown in Fig. 4.30. Shade the areas (in different colours) representing the displacement of the cyclist
(i) while the cyclist is moving with constant velocity.
(ii) when the velocity of cyclist is decreasing.
Also, calculate the displacement and average acceleration in the 120 s time interval.

Image 4

Answer:
In a velocity-time graph, the area under the graph gives displacement.

(i) The area representing constant velocity should be shaded between 20 s and 100 s. This part is a rectangle because the cyclist moves with constant velocity.

(ii) The area representing decreasing velocity should be shaded between 100 s and 120 s. This part is a trapezium because the velocity decreases uniformly.

Displacement calculation:

Area from 0 s to 20 s:

This is a triangle.

Area = 1/2 × base × height
Area = 1/2 × 20 × 3
Area = 30 m

Area from 20 s to 100 s:

This is a rectangle.

Area = length × breadth
Area = 80 × 3
Area = 240 m

Area from 100 s to 120 s:

This is a trapezium.

Area = 1/2 × (sum of parallel sides) × height
Area = 1/2 × (3 + 2) × 20
Area = 50 m

Total displacement = 30 + 240 + 50
Total displacement = 320 m

Average acceleration:

Initial velocity = 0 m/s
Final velocity = 2 m/s
Total time = 120 s

Average acceleration = (Final velocity - Initial velocity) / Time
Average acceleration = (2 - 0) / 120
Average acceleration = 1/60 m/s²

Therefore, the total displacement is 320 m, and the average acceleration is 1/60 m/s².

Question 13. A girl is preparing for her first marathon by running on a straight road. She uses a smartwatch to calculate her running speed at different intervals. The graph (Fig. 4.31) depicts her velocity versus time. Estimate the distance she ran based on the graph.

Image 5

Answer:
To estimate the distance, we calculate the area under the velocity-time graph. Since the graph is divided into different time intervals, we calculate the distance in each interval and then add them.

Phase 1: From 0 to 1.5 h
Approximate average velocity = 7.1 km/h

Distance = 7.1 × 1.5 = 10.65 km

Phase 2: From 1.5 h to 3 h
Velocity = 7.5 km/h

Distance = 7.5 × 1.5 = 11.25 km

Phase 3: From 3 h to 5.5 h
Velocity decreases from about 7.5 km/h to 6.5 km/h.
Average velocity ≈ 7.0 km/h

Distance = 7.0 × 2.5 = 17.5 km

Phase 4: From 5.5 h to 7 h
Velocity = 6.5 km/h

Distance = 6.5 × 1.5 = 9.75 km

Total distance = 10.65 + 11.25 + 17.5 + 9.75
Total distance = 49.15 km approximately

Therefore, the girl ran about 49.15 km.

Question 14. On entering a state highway, a car continues to move with a constant velocity of 6 ms-1 for 2 minutes and then accelerates with a constant acceleration 1ms-2 for 6 seconds. Find the displacement of the car on the state highway in the 2 min 6 s time interval by drawing a velocity-time graph for its motion.

Answer:
Given:

Initial constant velocity = 6 m/s
Time for constant velocity = 2 minutes = 120 s
Acceleration = 1 m/s²
Time of acceleration = 6 s

Displacement during constant velocity:

s₁ = v × t
s₁ = 6 × 120
s₁ = 720 m

Displacement during acceleration:

Initial velocity, u = 6 m/s
Acceleration, a = 1 m/s²
Time, t = 6 s

s₂ = ut + 1/2 at²
s₂ = 6 × 6 + 1/2 × 1 × 6²
s₂ = 36 + 18
s₂ = 54 m

Total displacement:

s = s₁ + s₂
s = 720 + 54
s = 774 m

Velocity-time graph explanation:

From 0 s to 120 s, draw a horizontal line at 6 m/s. This shows constant velocity.

From 120 s to 126 s, draw a straight sloping line upward. This shows uniform acceleration from 6 m/s to 12 m/s.

The displacement is equal to the area under the velocity-time graph.

Therefore, the total displacement of the car is 774 m.

Question 15. Two cars A and B start moving with a constant acceleration from rest, in a straight line. Car A attains a velocity of 5 ms-1 in 5 s. Car B attains a velocity of 3 ms-1 in 10 s. Plot the velocity-time graphs for both the cars in the same graph. Using the graph, calculate the displacement in the two time intervals mentioned (Hint: Calculate the acceleration in both cases. Then calculate their velocities at five instants of time to plot the graph).

Answer:
For Car A:

Initial velocity, u = 0 m/s
Final velocity, v = 5 m/s
Time, t = 5 s

Acceleration of Car A:

a = (v - u) / t
a = (5 - 0) / 5
a = 1 m/s²

For Car B:

Initial velocity, u = 0 m/s
Final velocity, v = 3 m/s
Time, t = 10 s

Acceleration of Car B:

a = (v - u) / t
a = (3 - 0) / 10
a = 0.3 m/s²

Velocity values for plotting:

For the velocity-time graph, plot time on the x-axis and velocity on the y-axis. Car A’s graph will be a straight line from (0, 0) to (5, 5). Car B’s graph will be a straight line from (0, 0) to (10, 3).

Displacement of Car A in 5 s:

Area under velocity-time graph = 1/2 × base × height
sA = 1/2 × 5 × 5
sA = 12.5 m

Displacement of Car B in 10 s:

sB = 1/2 × 10 × 3
sB = 15 m

Therefore, Car A travels 12.5 m in 5 s, and Car B travels 15 m in 10 s.

Question 16. Rohan studies science from 6 PM to 7:30 PM at home. Consider the tip of the minute hand of the wall clock. During the given time interval, what is its:
(i) distance travelled,
(ii) displacement,
(iii) speed, and
(iv) velocity.
The length of the minute’s hand is 7 cm (Fig. 4.32).

Image 6

Answer:
Given:

Length of minute hand, r = 7 cm
Time interval = 6:00 PM to 7:30 PM = 90 minutes = 5400 seconds

In 60 minutes, the minute hand completes 1 full revolution.
In 90 minutes, it completes 1.5 revolutions.

(i) Distance travelled:

Distance travelled by the tip = 1.5 × circumference of circle
Distance = 1.5 × 2πr
Distance = 1.5 × 2 × 22/7 × 7
Distance = 1.5 × 44
Distance = 66 cm

(ii) Displacement:

At 6:00 PM, the minute hand is at 12.
At 7:30 PM, the minute hand is at 6.

The displacement is the straight-line distance between 12 and 6, which is equal to the diameter of the circular path.

Displacement = 2r
Displacement = 2 × 7
Displacement = 14 cm

(iii) Speed:

Speed = Distance / Time
Speed = 66 / 90
Speed = 0.733 cm/min

In cm/s:

Speed = 66 / 5400
Speed = 0.0122 cm/s approximately

(iv) Velocity:

Average velocity = Displacement / Time
Average velocity = 14 / 90
Average velocity = 0.155 cm/min

In cm/s:

Average velocity = 14 / 5400
Average velocity = 0.00259 cm/s approximately

Therefore, distance travelled = 66 cm, displacement = 14 cm, speed = 0.733 cm/min, and average velocity = 0.155 cm/min in the direction from 12 to 6.

Think It Over (NCERT Textbook Page No. 48)

Question 1. How much distance should we maintain from the truck ahead to avoid a collision if it suddenly applies brakes?.

Answer: We should maintain a safe distance from the truck ahead so that our vehicle gets time to stop if the truck suddenly applies the brakes.

A common safety rule is to keep at least a 2-second gap from the vehicle in front. At higher speeds, the distance should be even more because the vehicle takes more time and distance to stop.

Question 2. Does this distance depend on the speed with which we are moving?

Answer: Yes, this distance depends on speed.

When a vehicle moves faster, it covers more distance during the driver’s reaction time and also needs a longer braking distance. Therefore, at higher speeds, we must keep a larger gap from the vehicle ahead to avoid a collision.

Think It Over (NCERT Textbook Page No. 48)

Question 3. Isn’t motion in nature wonderful? But how do we study the wide variety of complex motions around us?

Answer: Yes, motion in nature is wonderful. We see motion in flowing rivers, falling leaves, flying birds, moving clouds, rotating planets, and running animals.

To study such complex motion, we simplify it using basic ideas such as position, distance, displacement, speed, velocity, acceleration, time, and graphs. By observing changes in position with time, we can understand and describe different types of motion in a systematic way.

Think It Over (NCERT Textbook Page No. 49)

Question 4. How do we describe the position of an object?

Answer: The position of an object is described by using a reference point or origin.

We mention how far the object is from the reference point and in which direction it is located. For example, if a shop is 250 m east of a house, the house is the reference point, and the position of the shop is described as 250 m towards the east.

Pause and Ponder (NCERT Textbook Page No. 51)

Question 1. In the example of an athlete running back and forth on a straight track (Fig. 4.4), when will the displacement of the athlete be zero? What will be the total distance travelled in that case?

Image 7

Answer:
The displacement of the athlete will be zero when the athlete returns to the starting point.

This is because displacement depends only on the initial and final positions. If both positions are the same, displacement becomes zero.

However, the total distance travelled will not be zero. It will be equal to the total length of the path covered while running back and forth on the track.

Question 2. Fuel used up in a vehicle depends on which of the following? Justify your answer.
(i) Total distance travelled
(ii) Displacement

Answer:
Fuel used up in a vehicle depends on (i) the total distance travelled.

Fuel is consumed according to the actual path covered by the vehicle. Displacement only tells the shortest distance between the starting point and final point, and it does not represent the full path travelled.

For example, if a vehicle goes from home to a shop and returns home, its displacement is zero, but fuel is still used because it has travelled a certain distance.

Question 3. A ball rolls down an inclined track as shown in the figure. Is its motion a straight-line motion? Assuming the starting point of the ball (O) to be the origin, can its motion from O to D be depicted using a horizontal line as shown in the figure? Are the values of total distance travelled and magnitude of displacement from O equal or different at positions A, B, C, and D?

Image 8

Answer: The motion of the ball is not a straight-line motion because the ball moves along an inclined and curved track.

Its motion from O to D cannot be correctly shown using only a horizontal straight line, because the actual path followed by the ball is not horizontal. The horizontal line can only represent the displacement direction in a simplified way, not the actual path.

At positions A, B, C, and D, the total distance travelled and the magnitude of displacement are different. Distance is measured along the actual path of motion, while displacement is the shortest straight-line distance from the starting point O to the current position.

Pause and Ponder (NCERT Textbook Page No. 53)

Question 4. During a family road trip, you drive 200 km north in three hours. Afterwards, you drive 200 km south in two hours. Find the average speed and average velocity for your entire trip.

Answer:
Distance travelled north = 200 km
Distance travelled south = 200 km

Total distance = 200 + 200 = 400 km

Total time = 3 + 2 = 5 hours

Average speed:

Average speed = Total distance / Total time
Average speed = 400 / 5
Average speed = 80 km/h

Since the car returns to the starting point, displacement = 0 km.

Average velocity:

Average velocity = Displacement / Total time
Average velocity = 0 / 5
Average velocity = 0 km/h

Therefore, the average speed is 80 km/h, and the average velocity is 0 km/h.

Question 5. Under what condition(s) is the:
(i) magnitude of average velocity of an object equal to its average speed?
(ii) magnitude of average velocity of an object zero while its average speed is not zero?

Answer:
(i) The magnitude of average velocity is equal to average speed when the object moves along a straight path without changing direction.

In this case, the distance travelled is equal to the magnitude of the displacement.

(ii) The magnitude of average velocity is zero, while average speed is not zero when the object returns to its starting point.

In this case, displacement is zero, so the average velocity is zero. But the object has travelled some distance, so the average speed is not zero.

Activity 4.1: 

Let Us Analyse (NCERT Textbook Page No. 51)

Aim: To analyse the motion of a ball thrown vertically upward and understand the difference between total distance travelled and displacement.

Image 9

fig. 4.5: A ball in vertical motion (two separate lines are shown only for clarity; in reality, the object goes up and falls back in the same straight line)

Table 4.1: Distance Travelled and Displacement of the Ball 

Observations:
The ball moves upward and then comes back down along the same vertical path. Even though two separate lines may be shown in the figure for clarity, the actual motion is one-dimensional.

Distance is the total path length covered by the ball. Displacement is the shortest distance from the starting point O to the current position, along with direction.

At position C, the ball has first moved 140 cm upward and then 60 cm downward. So, the total distance travelled is 200 cm. But its displacement from O is only 80 cm upward.

Conclusion:
This activity shows that displacement can be less than or equal to the total distance travelled. Distance depends on the actual path covered, while displacement depends only on the starting and final positions.

Activity 4.2:
Let Us Calculate (NCERT Textbook Page No. 55)

Aim: To calculate the magnitude of average acceleration of different types of cars using data collected from the internet, and compare their performances.

Table 4.2: The Magnitude of Average Acceleration in a Time Interval

Observations: (Answer may vary)

Table 4.2: The magnitude of average acceleration in a time interval

To calculate acceleration, first convert 100 km/h into m/s.

100 km/h = 100 × 1000 / 3600
100 km/h = 27.78 m/s

Average acceleration:

a = Change in velocity / Time taken

For Rimac Nevera:

a = 27.78 / 1.81
a = 15.35 m/s²

For BMW M3 Competition:

a = 27.78 / 3.5
a = 7.94 m/s²

For Common Hatchback:

a = 27.78 / 10.5
a = 2.65 m/s²

Conclusion:
The car that takes the least time to reach 100 km/h has the highest average acceleration. The Rimac Nevera has the highest acceleration, while the common hatchback has the lowest acceleration among the three examples.

Activity 4.3:
Let Us Plot A Graph (NCERT Textbook Page No. 57)

Aim: To plot a position-time graph for a uniformly moving vehicle using data from a table, and learn how to read and interpret such graphs.

Image 10

Image 11

Image 12

Table 4.3: Positions of Vehicle at Different Instants of Time

Observations:
The data for the position-time graph is:

When these points are plotted on a graph and joined, they form a straight line passing through the origin.

This shows that the vehicle covers equal distances in equal intervals of time. Therefore, the vehicle is moving with uniform motion.

Slope of the graph:

Slope = Change in position / Change in time
Slope = (120 - 0) / (6 - 0)
Slope = 120 / 6
Slope = 20 m/s

Conclusion:
The straight-line position-time graph shows that the vehicle is moving with constant velocity. The slope of the graph gives the velocity of the vehicle, which is 20 m/s.

Activity 4.4:
Let Us Calculate (NCERT Textbook Page No. 59)

Aim: To calculate the average velocity of a moving vehicle from its position-time graph by finding the slope of the line.

Image 13

Observations:
To calculate average velocity from a position-time graph, choose two points on the straight line.

Let the two points be A and B.

From the graph:

Change in position = 80 m - 40 m = 40 m
Change in time = 4 s - 2 s = 2 s

Average velocity:

v = Change in position / Change in time
v = 40 / 2
v = 20 m/s

Conclusion:
The slope of a position-time graph represents velocity. A steeper slope means greater velocity, while a horizontal line means zero velocity. In this case, the average velocity of the moving vehicle is 20 m/s.

Activity 4.5:
Let Us Investigate (NCERT Textbook Page No. 67)

Aim: To investigate the direction of motion of an object moving in a circular path when it is suddenly released, demonstrating Newton’s First Law of Motion (inertia).

Observation:
When the ring is lifted, the marble does not continue moving in a circular path. Instead, it moves in a straight line tangential to the circular path at the point where it is released.

Image 14

Conclusion:
The marble moves in a straight line because an object in motion tends to continue moving in the same direction unless an external force changes its motion. While the marble was inside the ring, the ring provided the force needed for circular motion. Once the ring was removed, that force stopped, and the marble moved in the direction of its motion at that instant.

Class 9 Science Chapter 4 Describing Motion Around Us Solutions

Vedantu provides NCERT Solutions for Class 9 Science Chapter 4, Describing Motion Around Us, from the Exploration textbook for the 2026-27 academic session. This chapter helps students understand how motion is observed, measured, and explained using concepts such as distance, displacement, speed, velocity, acceleration, reference point, and graphical representation of motion.

The solutions include clear answers for exercise questions, in-text questions, numerical problems, graph-based questions, and activity-based learning tasks. Students can use these solutions to understand motion in real-life situations, practise step-by-step calculations, revise important formulas, and prepare confidently for school exams. The downloadable FREE PDF also helps students revise the complete chapter offline whenever required.

CBSE Class 9 Science Chapter 4 Study Materials

Students can use the Chapter 4 study materials below to revise important motion concepts, practise numerical questions, and improve their understanding of graphs, speed, velocity, acceleration, distance, and displacement.

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