NCERT Solutions Class 9 Exploration Chapter 4: Describing Motion Around Us 2026-27

Class 9 Science Chapter 4, Describing Motion Around Us, introduces students to the fundamental concepts used to represent motion in terms of distance, displacement, speed, velocity, acceleration, motion graphs, and uniform circular motion. The chapter also explains how mathematical equations and graphical representations can be used to study the motion of objects, with real-world examples and activities. 

Download NCERT Solutions for Class 9 Science Chapter 4 PDF

Download expert-led NCERT Solutions for Class 9 Science Chapter 4 of the Exploration textbook in PDF format. These include carefully curated answers to all NCERT questions, including InText, Revise, Reflect, and Refine, numerical, graph and activity-based questions. 

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The NCERT Class 9 Science Ch 4 Solutions provide detailed, step-by-step answers to each question exercise-wise in the new textbook Exploration. These solutions are designed for the current CBSE Class 9 syllabus 2026-27, making them a reliable source for homework, revision, and exam preparation. The answers help students build their conceptual understanding, practice numerical-based questions, and prepare for school exams confidently. Also, they can utilise these solutions to understand both theoretical and practical concepts, verify answers, revise key points, and prepare for tests.

NCERT Topics Covered in Class 9 Science Ch4

Chapter 4 Describing Motion Around Us of the Class 9 Exploration book teaches students how to describe the motion of an object using physical quantities, numbers, equations, and graphs. The chapter covers:

• Motion and rest
• Reference point and position of an object
• Motion in a straight line
• Distance and displacement
• Speed and Velocity
• Average speed and average velocity
• Acceleration and average acceleration
• Uniform and non-uniform motion
• Interpreting motion through graphs
• Position-time graphs
• Velocity-time graphs
• Equations of motion
• Uniform circular motion
• Numerical problems based on motion
• Real-life applications of motion and its measurement

NCERT Solutions for Class 9 Science Chapter 4 2026-27 - Question Answers

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:

1. Distance between home and shop = 250 m

Distance travelled by father:

• From home to the shop = 250 m
• From shop to home = 250 m
• From home to the shop again = 250 m
• From shop to the home after shopping = 250 m

Total distance travelled:

=250 + 250 + 250 + 250
=1000 m or 1 km

Displacement is the shortest distance between the initial and final positions.

Since father started from home and finally returned home, his initial and final positions are the same. Therefore, 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:

(A) the total vertical distance travelled, and

(B) their displacement from the starting point.

Answer:

(A) Given that the height of each floor = 3 m

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

Vertical distance travelled upward,

4 × 3 = 12 m

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

Vertical distance travelled downward,

(4−2) × 3 = 6 m

Therefore,

Total vertical distance travelled = 12 + 6 = 18 m

(B) Displacement is the shortest distance between the initial and final positions.

The student starts from the ground floor and ends on the second floor.

Displacement = 2 × 3 = 6 m upward

Therefore, the displacement from the starting point 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, it is possible for the scooter to be accelerating even if the speedometer reading is constant. A speedometer shows only the speed of the scooter, not its direction. Acceleration occurs whenever there is a change in velocity, and velocity depends on both speed and direction. If the girl is riding the scooter along a curved path or taking a turn, the direction of motion changes continuously even though the speed remains constant. Due to this change in direction, the velocity changes and the scooter is said to be accelerating. Thus, a body moving with constant speed can still have acceleration if its direction of motion changes.

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.

Given:

Initial velocity, u = 0 ms^–1

Final velocity, v = 24 ms^–1

Time taken, t = 6 s

Average acceleration can be calculated by

using the formula:

a = v - u / t

a = 24 - 0 / 6

a = 4 ms^–2

Therefore, the average acceleration is 4 ms^–2

Distance travelled can be calculated by assuming uniform acceleration as:

s = ut + ½ at²

s = (0 × 6) + ½ × 4 × 6 sq²

s = 36 × 2

s = 72 m

Therefore, the distance travelled in 6 s 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:

Initial velocity, u = 28 ms^–1

Final velocity, v = 0 ms^–1

Distance travelled, s = 98 m

Acceleration of the motorbike can be calculated as:

v² = u² + 2as

0² = (28)² + 2 × a × 98

0 = 784 + 196a

196a = –784

a = –4 ms^–2

Therefore, the acceleration of the motorbike is –4 ms⁻². The negative sign shows that the motorbike is slowing down.

Time taken to come to a stop can be calculated using the equation,

v = u + at

0 = 28 + (–4)t

4t = 28

t = 7 s

Therefore, the time taken by the motorbike to stop is 7 s.

Question 6: The given figure 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.

Answer:

No, objects A and B cannot have equal velocity at any time.

In a position-time graph, the slope of the graph represents the velocity of the object. Greater the slope, greater is the velocity.

In the given graph, the line representing object A is steeper than the line representing object B. This means object A covers more displacement in the same interval of time. Therefore, the velocity of object A is greater than the velocity of object B.

Although the two graphs meet at one point, it only shows that the two objects have the same position at that instant. Since their slopes are different, their velocities are not equal.

Question 7: The given graph 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).

(a) The average velocity of both over the 10 s time interval is equal since they have the same initial and final positions.

(b) The average speeds of both over the 10 s time interval are equal since both cover equal distance in equal time.

(c) 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.

(d) The average speed of A over the 10 s time interval is greater than that of B since B's speed is lower than A's in some segments.

Answer:

Both (a) and (b) are correct.

Explanation: Both objects A and B start from the same position and reach the same final position at t = 10. Therefore, their displacements are equal. Since

Average velocity = Displacement / Time

their average velocities are equal. Hence, (a) is correct.

Also, the graph shows that both objects move only in the positive direction, so the distance travelled equals the displacement for each object. Since they cover the same distance in the same time interval, their average speeds are also equal. Hence, (b) is correct.

Therefore, the correct options are (a) and (b).

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 as shown in the figure 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.

Initial velocity,

u = 54 km h^–1

u = 54 × 1000 / 3600 

= 15 ms^–1

Final velocity,

v = 36 km h^–1

v = 36 × 1000 / 3600

= 10 ms^–1

Time taken,

t = 36 s

Using the equation,

s = (u + v / 2) × t

s = (15 + 10 / 2) × 36  

s = 25/2 × 36

s = 12.5 × 36

s = 450 m

Therefore, the distance travelled by the truck during this time is 450 m.

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 of the car takes place in three stages.

First stage: Car accelerates from rest

Given,

u = 0 ms^–1

v = 20 ms^–1

t = 5 s

Distance travelled during acceleration,

s₁ = (( u + v) / 2) × t

s₁ = ((0 + 20) / 2) × 5  

s₁ = 10 × 5

s₁ = 50 m

Second stage: Car moves with constant velocity

Velocity (v) = 20 ms^–1

Time (t) = 10 s

Distance travelled,

s₂ = vt

s₂ = 20 × 10

s₂ = 200 m

Third stage: Car slows down and stops

Given,

u = 20 ms^–1

v = 0 ms^–1

t = 6 s

Distance travelled during braking,

s₃ = (( u + v) / 2) × t

s₃ = (( 20 + 0) / 2) × 6

s₃ = 10 × 6

s₃ = 60 m

Therefore,

Total distance travelled = s₁ + s₂ + s₃

= 50 + 200 + 60

= 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 ms⁻². Will the bus be able to stop before reaching the obstacle?

Answer: 

Initial velocity of the bus, 

u = 36 km h⁻¹ 

u = 36 × 1000 / 3600 = 10 ms⁻¹ 

Distance of obstacle from the bus = 30 m 

Reaction time of driver, t = 0.5s 

Retardation after applying brake, a = –2.5 ms⁻² 

During the reaction time, the driver does not apply brakes, so the bus continues to move with the same velocity. 

Distance travelled during reaction time, 

s₁ = vt 

s₁ = 10 × 0.5 

s₁ = 5 m 

Now, distance left before reaching the obstacle, 

30 – 5 = 25 m 

Then, stopping distance after brakes are applied. 

v² = u² + 2as 

Since the bus stops, 

v = 0 

0 = (10) 2 + 2(–2.5)s 

0 = 100 – 5 s 

5s = 100 

s = 20 m 

Thus, the bus requires 20 m to stop after applying the brakes. 

Total distance travelled before stopping, 

= 5 + 20 = 25m 

Since the obstacle is 30 m ahead, the bus stops 5 m before the obstacle. Therefore, the bus will be able to 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:

Whether an object is at rest or in motion depends upon the reference point chosen for observation.

An object kept on the Earth appears to be at rest with respect to its surroundings on the Earth because its position does not change relative to nearby objects. For example, a book lying on a table remains at the same position with respect to the table and the room.

However, the Earth is continuously rotating about its axis and revolving around the Sun. Therefore, every object present on the Earth also moves along with it. Hence, with respect to the Sun or an external observer in space, the object is in motion.

Thus, an object kept on the Earth can be considered at rest with respect to the Earth, but not with respect to the Sun.

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

Answer:

(A) In the given velocity-time graph, shade the rectangular area under the graph from 20 s to 100 s in one colour. This area represents the displacement while the cyclist is moving with constant velocity.

(B) Shade the trapezium-shaped area under the graph from 100 s to 120 s in another colour. This area represents the displacement while the cyclist is moving with decreasing velocity.

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

Displacement = ½ × 20 × 3 × 80 × 3 + ½ (3 + 2) × 20

           = 30 + 240 + 50

           = 320 m

Now,

u = 0 ms^–1, v = 2 ms^–1, t = 120 s

Using, a = v - u / t

a = 2 / 120

a = 0.017 ms^–2

The displacement of the cyclist is 320 m and the average acceleration is 0.017 ms^–2.

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 given graph depicts her velocity versus time. Estimate the distance she ran based on the graph.

‍

From the graph:

0 to 1 h: velocity increases from 7 to 7.5 km h^–1

1 to 3 h: velocity remains 7.5 km h^–1

3 to 6 h: velocity decreases from 7.5 to 6.5 km h^–1

So, From 0 h to 1.6 h,

Distance = 7 × 0.6 = 4.2 km

From 0.6 h to 1.6 h,

Distance = ((7 + 7.5) / 2) × 1 = 7.25 km

From 1.6 h to 3.0 h,

Distance = 7.5 × 1.4 = 10.5 km

From 3.0 h to 4.6 h,

Distance = ((7 + 7.5) / 2) × 1.6 = 11.6 km

From 4.6 h to 5.6 h,

Distance = ((7 + 6.5) / 2) × 1 = 13.5 km

From 5.6 h to 6.6 h,

Distance = 6.5 × 1 = 6.5 km

Therefore,

Total distance = 4.2 + 7.25 + 10.5 + 11.6 + 13.5 + 6.5

                         = 53.55 km

Distance travelled ≈ 53.55 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 1 ms⁻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:

The car moves with a constant velocity of 6 ms−1 for 2 minutes and then accelerates for 6 seconds.

The required velocity-time graph:

We know that

v = u + at

v = 6 + (1 × 6)

v = 12 ms^–1

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

Displacement during first 120 s:

s₁ = 6 × 120

s₁ = 720 m

Displacement during next 6 s:

s₂ = ((6 + 12)/2) × 6  

s₂ = 9 × 6

s₂ = 54 m

Therefore,

Total displacement = 720 + 54

                                  = 774 m

The displacement of the car in 2 min 6 s 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:

u = 0, v = 5 ms^–1, t = 5 s

Acceleration of car A,

a = v - u/ t

a = 5 - 0/ 5

a = 1 ms^–2

For car B:

u = 0, v = 3 ms^–1, t = 10 s

Acceleration of car B,

a = 3 - 0/ 10

a = 0.3 ms^–2

To plot the velocity-time graph:

(1) Draw time on the X-axis and velocity on the Y-axis.

(2) For car A, join the points (0, 0) and (5, 5).

(3) For car B, join the points (0, 0) and (10, 3).

Therefore, the displacement of car A is 12.5m and the displacement of car B is 15 m.

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

For car A:

Displacement = ½ × 5 × 5

          = 12.5 m

For car B:

Displacement =  ½ × 10 × 3

           = 15 m

Question 16: Rohan studies science from 6 PM to 7:30 PM at home. Consider the tip of the minute's hand of the wall clock. During the given time interval, what is its:

(A) distance travelled,

(B) displacement,

(C) speed, and

(D) velocity.

The length of the minute's hand is 7 cm as shown in the given figure:

Time interval = 6 PM to 7:30 PM = 1.5 hours = 90 min = 5400 s;

Length of minute’s hand (radius) R = 7 cm

In 90 minutes, the minute’s hand completes 1.5

revolutions (60 min = 1 revolution).

Circumference = 2πR = 2 × 22/ 7 × 7 = 44 cm

(A) Distance travelled:

d = 1.5 × 2πR = 1.5 × 44 = 66 cm

(B) Displacement:

After 1.5 revolutions, the tip ends up diametrically opposite to its starting position ( ½ revolution extra).

Displacement = 2R = 2 × 7 = 14 cm

(C) Speed:

Speed = Distance/ Time

= 66 cm/ 5400s » 0.0122 cm/s

(D) Velocity:

Velocity = Displacement/ Time

= 14cm/ 5400s » 2.59 ×10^–3 cm/s

Directed along the diameter (from initial to final position of the tip).

Think It Over [NCERT Textbook Pg. No. 48]

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

[Pg. No. 48]

Answer:

1. To avoid a collision if the truck ahead suddenly applies brakes, we should maintain a sufficient distance from it. This safe distance gives us enough time to react and stop our vehicle safely. Maintaining a proper distance helps prevent accidents and ensures safer driving on the road.

Question 2: Does this distance depend upon the speed with which we are moving? 

[Pg. No. 48]

Answer:

Yes, this distance depends upon the speed with which we are moving. If we are moving at a higher speed, we need to maintain a larger distance from the truck ahead because it takes a longer distance to stop the vehicle. At lower speeds, a smaller distance may be sufficient.

Pause and Ponder [NCERT Textbook Pg. No. 51]

Question 3: In the example of an athlete running back and forth on a straight track, when will the displacement of the athlete be zero? What will be the total distance travelled in that case?  

[Pg. No. 51]

Answer:

An athlete starts from O (0 m) at  t=0 s. Reaches A (100 m) at t=10 s. Then runs back and reaches B (40 m) at t=16 s. Displacement becomes zero when the athlete returns to the starting point O. Total distance travelled =  100+100=200 m. Hence, when the athlete returns to the starting point, the displacement becomes zero and the total distance travelled is 200 m.

Question 4: Fuel used up in a vehicle depends on which of the following?
(A) Total distance travelled
(B) Displacement
Justify your answer. 

[Pg. No. 51]

Answer:

Fuel used up in a vehicle depends on the total distance travelled and not on displacement, because fuel is consumed continuously as the vehicle moves along the actual path covered. Even if the displacement is zero, fuel is still used if some distance has been travelled.

Question 5: A ball rolls down an inclined track as shown in given 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?   

[Pg. No. 51]

Answer:

Yes, the motion of the ball is a straight-line motion because it moves along a straight inclined track from O to D. Taking O as the origin, the motion can be represented on a horizontal line as shown, since only the distance from the starting point is considered. At positions A, B, C and D, the total distance travelled and the magnitude of displacement are equal because the ball moves only in one direction along a straight path without changing direction.

Question 6: 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. 

[Pg. No. 53]

Answer:

Total distance travelled = 200 + 200 = 400 km

Total time taken = 3 + 2 = 5 h

Average speed = Total distance/ Total time 

   = 400/ 5 

   = 80 km/h

Since the final position is the same as the

starting position, total displacement = 0.

Average velocity = Total displacement/ Total time

      = 0/ 5 = 0 km/h

Hence, since the family returns to the starting point, the net displacement is zero. Therefore, the average speed is 80 km/h and the average velocity is 0 km/h.

Question 7: Under what condition(s) is the:
(A) magnitude of average velocity of an object equal to its average speed?
(B) magnitude of average velocity of an object zero while its average speed is not zero? 

[Pg. No. 53]

Answer:

(A) The magnitude of average velocity is equal to the average speed when an object moves in a straight line without changing its direction.

(B) The magnitude of average velocity is zero while its average speed is not zero when the object returns to its starting point after covering some distance. In this case, displacement is zero, but distance travelled is not zero.

How to Study Class 9 Exploration Chapter 4 Effectively?

Start by learning the effect of change in position over time. Then move to the concepts of reference point, distance and displacement - learning these first makes the concepts of speed and velocity easier to understand.

Carefully distinguish between distance and displacement. Distance depends on the actual path taken, and displacement describes the change in position from initial to end point.

Next, revise speed, velocity and acceleration and focus on what each physical quantity tells about motion. Practise identifying whether an object is moving uniformly or non-uniformly from the information provided in a question.

Another key component of this chapter is graphs. Understand how to read and interpret position-time and velocity-time graphs, rather than rote-learn their shapes.

Lastly, practise numerical problems using the equations of motion. Write the given quantities, choose the right equation, plug in the values of the quantities carefully, and don’t forget to add units to your answer.

Essential Tips to Focus on Before Appearing for Exams

• Read each section of the NCERT Class 9 Science textbook and attempt the questions without referring to the answer. 
• Refer to the NCERT solution to confirm if your method, working procedure, and answer are correct or not.
• In numerical questions, always record the value being used, the value required and the appropriate formula and then substitute numbers. 
• Be careful of units, particularly if a question is using the speed in km/h, but your calculations need km/s.
• With graph-based questions, practice determining what information is being given in the graph and what the slope or rate of change means. 
• Once you have finished the textbook questions, revising the chapter from Prasant Kirad CBSE Class 9 Science Notes can help boost your preparation.

FAQs on NCERT Solutions for Class 9 Science Chapter 4

Q1. What do I learn in Chapter 4 of Class 9 Science?

Ans. Chapter 4 Describing Motion Around Us explains to you the concepts of distance, displacement, speed, velocity, acceleration and how to describe and measure motion. It also covers motion graphs, equations of motion and uniform circular motion.

Q2. Which are the most important topics of Class 9 Science Chapter 4?

Ans. While the entire chapter holds equal importance, certain topics deserve special attention: motion and rest, position, distance and displacement, speed and velocity, acceleration, uniform and non-uniform motion, position-time graph, velocity-time graph, and equations of motion.

Q3. Do these NCERT Solutions align with the New textbook Exploration?

Ans. Yes. These NCERT solutions for Class 9 Science Chapter 4 are prepared by Educart experts based on the new exploration book for the CBSE 2026–27 academic syllabus.

Q4. Are numericals important for Class 9 science exams?

Ans. Yes. Numerical problems are an important part of this chapter since students need to use concepts like speed, velocity, acceleration and equations of motion to calculate unknown quantities.

Q5. Are these solutions for the in-text questions and activities?

Ans. Yes. The solutions include prompts and questions throughout the chapter, along with the Think It Over, Pause and Ponder,  Revise, Reflect, Refine, and Activity-based questions.

Q6. What are the strategies to prepare Class 9 Science ch4 effectively?

Ans. First, clear your basic concepts from the NCERT textbook, such as distance, displacement, speed, velocity and acceleration. Then practice motion graphs and equations of motion, followed by textbook numericals and application-based questions. Use NCERT Solutions to check your approach and a Class 9 Science Question Bank 2026–27 for additional practice.

Q7. What is the difference between distance and displacement?

Ans. Distance is the length of the path covered by an object, while displacement is the net change in the position of an object from the start to the end of the journey, and it is a vector quantity. Distance is a scalar quantity; on the other hand, displacement is a vector quantity, which has both direction and magnitude.

Q8. Do NCERT Class 9 Science Ch 4 Solutions help in exam preparation?

Ans. Yes. They assist you in grasping the key idea, validate numerical calculations, work through graph- and diagram-based problems, and prepare you to tackle textbook problems.