Class 9 New Science Book 2026 Ch4 Describing Motion Around Us

Motion is one of the most important concepts in science that helps us understand how objects move around us. In Class 9 New Science Book 2026 Ch4 Describing Motion, students learn about distance, displacement, speed, velocity, acceleration, and different types of motion in a simple and practical way.

This chapter explains motion using real-life examples, activities, graphs, and numerical problems. The solutions provided here will help students understand concepts clearly and prepare well for exams and assignments.

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: Distance between home and shop = 250 m

Movement of father:

1. Home → Shop = 250 m
2. Shop → Home = 250 m
3. Home → Shop = 250 m
4. Shop → Home = 250 m

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

So, the total distance travelled by him is 1000 m.

Since he finally returned home, his initial position and final position are the same.

Therefore, displacement from home = 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.

Solution

Height of each floor = 3 m

Ground floor → Fourth floor
Number of floors crossed = 4

Distance upward
= 4 × 3
= 12 m

Fourth floor → Second floor
Number of floors moved down = 2

Distance downward
= 2 × 3
= 6 m

(i) Total vertical distance travelled

= 12 m + 6 m
= 18 m

(ii) Displacement from the starting point

Starting point = Ground floor
Final point = Second floor

Net upward displacement
= 2 × 3
= 6 m upward

Therefore, displacement = 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?

Solution

Yes, it is possible for the scooter to be accelerating even if the speedometer reading is constant.

Acceleration occurs whenever there is a change in velocity. Velocity changes if:

• speed changes, or
• direction changes.

If the girl turns the scooter along a curved road, the direction of motion changes continuously. Therefore, the scooter is accelerating even though its speed remains constant.

Question 4

A car starts from rest and its velocity reaches 24 m s–1 in 6 s. Find the average acceleration and the distance travelled in these 6 s.

Solution

Given:

Initial velocity, $u = 0$m s$^{-1}$
Final velocity, $v = 24$m s$^{-1}$
Time, $t = 6$s

Average acceleration

Using:

$a=\frac{v-u}{t}$$$a = \frac{24-0}{6}$$$$a = 4 \text{ m s}^{-2}$$

Distance travelled

Using:

$s=ut+\frac{1}{2}at^2$$$s = 0 + \frac{1}{2}(4)(6^2)$$$$s = 2 \times 36$$$$s = 72 \text{ m}$$

Therefore, distance travelled = 72 m.

Question 5

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

Solution

Given:

Initial velocity, $u = 28$m s$^{-1}$−1
Final velocity, $v = 0$m s$^{-1}$
Distance travelled, $s = 98$

Acceleration

Using:

$v^2=u^2+2as$$$0 = 28^2 + 2(a)(98)$$$$0 = 784 + 196a$$$$196a = -784$$$$a = -4 \text{ m s}^{-2}$$

Time taken to stop

Using:

$v=u+at$$$0 = 28 + (-4)t$$$$4t = 28$$$$t = 7 \text{ s}$$

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.

Answer for Graph 1 (Left)

• Answer: No, objects A and B never have equal velocity.
• Justification: Both graphs are straight lines, which means both objects are moving with constant (uniform) velocity. Since the lines have different steepness (slopes) and are never parallel to each other, their velocities can never be equal. The point where they cross only shows they are at the same position at that time, not the same velocity.

Answer for Graph 2 (Right)

Justification: 1. Object A moves with a constant velocity (straight line). 2. Object B’s motion is non-uniform (curved/changing line). Initially, the slope of B is less than A (moving slower). Later, the slope of B becomes steeper than A (moving faster) to catch up. 3. Because the slope of B changes from being smaller than A’s slope to greater than A’s slope, there must be a point in time where the slope of line B becomes parallel to line A. At that exact moment, their velocities are equal.

Answer: Yes, objects A and B will have equal velocity at some instant.

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

Answer:

Both (i) and (ii) are correct.

Short Explanation:

• For (i): Average velocity depends only on the initial and final positions. Since both objects start and end at the same points in 10 seconds, their displacements are equal, making their average velocities equal.
• For (ii): Since they move in a straight line to the same final point, they cover the exact same total distance in the same 10 seconds. Therefore, their average speeds are also equal.

Statements (iii) and (iv) are incorrect because the total distance and time are identical for both objects.

7.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.

Answer:

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

Solution

The motion has three parts:

1. Accelerating motion
2. Uniform motion
3. Retarding (braking) motion

1. Distance during acceleration

Given:

Initial velocity, $u = 0$ m s$^{-1}$
Final velocity, $v = 20$ m s$^{-1}$
Time, $t = 5$ s

Using:

$s=\frac{(u+v)}{2}t$$$s_1 = \frac{(0+20)}{2}\times 5$$$$s_1 = 10 \times 5$$$$s_1 = 50 \text{ m}$$

Distance during acceleration = 50 m

2. Distance during uniform motion

Speed = 20 m s$^{-1}$−1
Time = 10 s

Using:$$\text{Distance} = \text{speed} \times \text{time}$$$$s_2 = 20 \times 10$$$$s_2 = 200 \text{ m}$$

Distance during uniform motion = 200 m

3. Distance during braking

Given:

Initial velocity, $u = 20$ m s$^{-1}$
Final velocity, $v = 0$ m s$^{-1}$
Time, $t = 6$ s

Using:

$s=\frac{(u+v)}{2}t$$$s_3 = \frac{(20+0)}{2}\times 6$$$$s_3 = 10 \times 6$$$$s_3 = 60 \text{ m}$$

Total distance travelled

$$\text{Total distance} = s_1 + s_2 + s_3$$$$= 50 + 200 + 60$$$$= 310 \text{ m}$$

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 m s–2. Will the bus be able to stop before reaching the obstacle?

Solution

Given:

Speed of bus = 36 km h$^{-1}$

Convert into m s$^{-1}$−1:$$36 \times \frac{5}{18} = 10 \text{ m s}^{-1}$$

Initial velocity, $u = 10$ m s$^{-1}$
Reaction time = 0.5 s
Retardation, $a = -2.5$ m s$^{-2}$

Distance of obstacle from bus = 30 m

1. Distance travelled during reaction time

Using:$$\text{Distance} = \text{speed} \times \text{time}$$$$s_1 = 10 \times 0.5$$$$s_1 = 5 \text{ m}$$

2. Braking distance

Using:

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

Final velocity, $v = 0$ m s$^{-1}$$$0 = 10^2 + 2(-2.5)s$$$$0 = 100 – 5s$$$$5s = 100$$$$s_2 = 20 \text{ m}$$

3. Total stopping distance

$$\text{Total stopping distance} = 5 + 20$$$$= 25 \text{ m}$$

Since:$$25 \text{ m} < 30 \text{ m}$$

the bus will stop before reaching the obstacle.

Therefore, the bus stops safely 5 m before 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.

Solution

Rest and motion are relative terms and depend on the reference point.

An object kept on the Earth may appear to be at rest with respect to the Earth and nearby surroundings. For example, a book lying on a table does not change its position relative to the table.

However, since the Earth itself is rotating on its axis and revolving around the Sun, the object is also moving along with the Earth.

Therefore:

• Relative to the Earth, the object is at rest.
• Relative to the Sun or stars, the object is in motion.

Hence, an object on the Earth can be considered both at rest and in motion depending on the frame of reference.

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 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.

Solution

From the graph:

• From $0$0 to $20$20 s, velocity increases from $0$0 to $3$3 m s$^{-1}$−1.
• From $20$20 s to $100$100 s, velocity remains constant at $3$3 m s$^{-1}$−1.
• From $100$100 s to $120$120 s, velocity decreases from $3$3 m s$^{-1}$−1 to $2$2 m s$^{-1}$−1.

(i) Area representing displacement during constant velocity

The cyclist moves with constant velocity from $20$20 s to $100$100 s.

The displacement is represented by the rectangular area under the graph between:

• time = $20$ s to $100$ s
• velocity = $3$ m s$^{-1}$

(ii) Area representing displacement when velocity is decreasing

The velocity decreases from $100$ s to $120$ s.

The displacement is represented by the trapezium-shaped area under the graph between:

• time = $100$100 s to $120$120 s

Total displacement

Total displacement = Total area under the velocity-time graph

1. First triangular area (0–20 s)

Using:$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$$$A_1 = \frac{1}{2} \times 20 \times 3$$$$A_1 = 30 \text{ m}$$

2. Rectangular area (20–100 s)

Using:$$\text{Area} = \text{length} \times \text{breadth}$$$$A_2 = (100-20)\times 3$$$$A_2 = 80 \times 3$$$$A_2 = 240 \text{ m}$$

3. Trapezium area (100–120 s)

Using:$$\text{Area} = \frac{1}{2}(a+b)h$$$$A_3 = \frac{1}{2}(3+2)\times 20$$$$A_3 = \frac{1}{2}\times 5 \times 20$$$$A_3 = 50 \text{ m}$$

Total displacement

$$\text{Displacement} = A_1 + A_2 + A_3$$$$= 30 + 240 + 50$$$$= 320 \text{ m}$$

Average acceleration

Using:

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

Initial velocity, $u = 0$ m s$^{-1}$
Final velocity, $v = 2$ m s$^{-1}$
Total time, $t = 120$ s$$a = \frac{2-0}{120}$$$$a = \frac{1}{60}$$$$a \approx 0.017 \text{ m s}^{-2}$$

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 running distance based on the graph.

Solution

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

From the graph, the approximate values are:

We calculate the area interval by interval using the trapezium formula:$$\text{Area}=\frac{1}{2}(a+b)h$$

where:

• $a$ and $b$ are velocities,
• $h$ is the time interval.

1. From 0 h to 1 h

$$A_1=\frac{1}{2}(6.5+7.5)\times1$$$$A_1=7 \text{ km}$$

2. From 1 h to 2 h

$$A_2=\frac{1}{2}(7.5+7.5)\times1$$$$A_2=7.5 \text{ km}$$

3. From 2 h to 3 h

$$A_3=\frac{1}{2}(7.5+6.5)\times1$$$$A_3=7 \text{ km}$$

4. From 3 h to 4 h

$$A_4=\frac{1}{2}(6.5+5.5)\times1$$$$A_4=6 \text{ km}$$

5. From 4 h to 5 h

$$A_5=\frac{1}{2}(5.5+5.5)\times1$$$$A_5=5.5 \text{ km}$$

Total running distance

$$\text{Distance}=A_1+A_2+A_3+A_4+A_5$$$$=7+7.5+7+6+5.5$$$$=33 \text{ km}$$

Question 14

On entering a state highway, a car continues to move with a constant velocity of 6 m s–1 for 2 minutes and then accelerates with a constant acceleration 1 m s–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.

Solution

Given:

Constant velocity, $u = 6$m s$^{-1}$
Time at constant velocity = 2 minutes = 120 s
Acceleration, $a = 1$m s$^{-2}$
Time of acceleration = 6 s

1. Distance during constant velocity

Using:$$s_1 = vt$$$$s_1 = 6 \times 120$$$$s_1 = 720 \text{ m}$$

2. Distance during accelerated motion

Final velocity after acceleration:

Using:

$v=u+at$$$v = 6 + (1)(6)$$$$v = 12 \text{ m s}^{-1}$$

Distance travelled during acceleration:

Using:

$s=\frac{(u+v)}{2}t$$$s_2 = \frac{(6+12)}{2}\times 6$$$$s_2 = 9 \times 6$$$$s_2 = 54 \text{ m}$$

Total displacement

$$\text{Displacement} = s_1 + s_2$$$$= 720 + 54$$$$= 774 \text{ m}$$

Therefore, 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 m s–1 in 5 s. Car B attains a velocity of 3 m s–1 in 10 s. Plot the velocity-time graphs for both the cars in the same graph. Using the graph, calculate the displacement mentioned in the two time intervals.

Solution

For Car A:

Given:

Initial velocity, $u = 0$
Final velocity, $v = 5$ m s$^{-1}$
Time, $t = 5$ s

Acceleration of Car A:

$a=\frac{v-u}{t}$$$a_A = \frac{5-0}{5}$$$$a_A = 1 \text{ m s}^{-2}$$

Displacement of Car A

Area under velocity-time graph:$$s_A = \frac{1}{2}\times 5 \times 5$$$$s_A = 12.5 \text{ m}$$

Therefore, displacement of Car A = 12.5 m

For Car B:

Given:

Initial velocity, $u = 0$u=0
Final velocity, $v = 3$v=3 m s$^{-1}$−1
Time, $t = 10$t=10 s

Acceleration of Car B:

$a=\frac{v-u}{t}$$$a_B = \frac{3-0}{10}$$$$a_B = 0.3 \text{ m s}^{-2}$$

Displacement of Car B

Area under velocity-time graph:$$s_B = \frac{1}{2}\times 10 \times 3$$$$s_B = 15 \text{ m}$$

Graph plotting points

Car A

Car B

Plot these points on the same velocity-time graph.

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:
(i) distance travelled,
(ii) displacement,
(iii) speed, and
(iv) velocity.

The length of the minute’s hand is 7 cm.

Solution

Radius of circular path:$$r = 7 \text{ cm}$$

Time interval:

From 6 PM to 7:30 PM = 1.5 hours = 90 minutes

The minute hand completes:

• 1 revolution in 60 minutes
• Therefore, in 90 minutes it completes $1.5$ revolutions.

(i) Distance travelled

Circumference of circle:

$C=2\pi r$

$r$

$d = 2r \approx 6.00$

$C = 2\pi r \approx 18.85$$$C = 2 \times \frac{22}{7} \times 7$$$$C = 44 \text{ cm}$$

Distance in 1.5 revolutions:$$\text{Distance} = 1.5 \times 44$$$$= 66 \text{ cm}$$

(ii) Displacement

After 1.5 revolutions, the minute hand reaches the opposite point of the circle.

$$\text{Displacement} = 2r$$$$= 2 \times 7$$$$= 14 \text{ cm}$$

(iii) Speed

$$\text{Speed} = \frac{\text{Distance}}{\text{Time}}$$$$= \frac{66}{90}$$$$= 0.733 \text{ cm min}^{-1}$$

(iv) Velocity

$$\text{Velocity} = \frac{\text{Displacement}}{\text{Time}}$$$$= \frac{14}{90}$$$$= 0.156 \text{ cm min}^{-1}$$

You can access the official NCERT Solutions for Class 10 Mathematics on the NCERT website at the following link:

NCERT Class 10 Mathematics Solutions