[Exploration] Solutions Ch 4 Describing Motion Around Us NCERT Class 9

Q2

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

Given

Height of each floor = 3 m

• Ground floor → Fourth floor
  = $4 \times 3 = 12$
• Fourth floor → Second floor
  = $(4 – 2) \times 3 = 6$

(i) Total vertical distance travelled

$$12 + 6 = 18 \text{ m}$$

Total vertical distance travelled = 18 m. Ans

(ii) Displacement from the starting point

Starting point = Ground floor
Final position = Second floor

Height of second floor from ground floor:$$2 \times 3 = 6 \text{ m}$$
Displacement = 6 m upward. Ans

Q4.

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.

Answer

Given

Initial velocity,

Final velocity,

Time,

(i) Average acceleration

Using,

Average acceleration = 4 m s^⁻² Ans

(ii) Distance travelled

Using,

Distance travelled = 72 m. Ans

Q6.

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

Velocity is obtained from the slope of the position-time graph.

👉In the left graph, the lines representing A and B have different slopes initially, but they intersect at one point. At the point of intersection, the slopes of the two graphs are equal.

Therefore, objects A and B have equal velocity at that instant.

👉In the right graph, the slopes of A and B are different throughout the motion.

Object A moves with constant velocity (straight line with constant slope), while object B changes its slope during motion.

Therefore, in the right graph, A and B never have equal velocity because their slopes are never equal.

Q7.

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

From the graph:

• Both objects start from the same position at $t = 0$
• Both objects reach the same final position at $t = 10$

Therefore,$$\text{Average velocity}=\frac{\text{displacement}}{\text{time}}$$

Since both have the same displacement in the same time interval,

(i) is correct. Ans

The motion is along a straight line in one direction only.
So, distance travelled = displacement.

Both objects cover the same distance in 10 s.
$$\text{Average speed}=\frac{\text{total distance travelled}}{\text{time}}$$

Hence, their average speeds are also equal.

(ii) is correct. Ans

❌ (iii) is incorrect because A and B cover the same distance in 10 s.

❌ (iv) is incorrect because average speed depends on total distance travelled and total time, not on speed in some segments.

Correct options:

(i) and (ii) Ans

Q8.

A truck driver driving at the speed of 54 km/h notices a road sign with a speed limit of 40 km/h for trucks. He slows down to 36 km/hin 36 s. What was the distance travelled by him during this time? Assume the acceleration to be constant while slowing down.

Answer

Given

Initial velocity,$$u = 54 \text{ km h}^{-1}$$

Final velocity,$$v = 36 \text{ km h}^{-1}$$

Time,$$t = 36 \text{ s}$$

Convert velocities into m s^-¹:$$u = 54 \times \frac{1000}{3600} = 15 \text{ m s}^{-1}$$$$v = 36 \times \frac{1000}{3600} = 10 \text{ m s}^{-1}$$

Acceleration

Using,

$a=\frac{v-u}{t}$$$a=\frac{10-15}{36}$$$$a=-\frac{5}{36}\text{ m s}^{-2}$$

Distance travelled

Using,

$s=ut+\frac{1}{2}at^2$$$s=(15)(36)+\frac{1}{2}\left(-\frac{5}{36}\right)(36)^2$$$$s=540-90$$$$s=450 \text{ m}$$

Distance travelled = 450 m. Ans

Q9.

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

Given

First stage: Accelerating motion

Initial velocity,$$u = 0 \text{ m s}^{-1}$$

Final velocity,$$v = 20 \text{ m s}^{-1}$$

Time,$$t = 5 \text{ s}$$

Distance travelled in the first stage

Using,

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

$$s_1 = \frac{0+20}{2}\times 5$$$$s_1 = 10 \times 5$$$$s_1 = 50 \text{ m}$$

Second stage: Constant velocity

Velocity,$$v = 20 \text{ m s}^{-1}$$

Time,$$t = 10 \text{ s}$$

Distance travelled,$$s_2 = vt$$$$s_2 = 20 \times 10$$$$s_2 = 200 \text{ m}$$

Third stage: Retardation motion

Initial velocity,$$u = 20 \text{ m s}^{-1}$$

Final velocity,$$v = 0 \text{ m s}^{-1}$$

Time,$$t = 6 \text{ s}$$

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

$$s = 50 + 200 + 60$$
$$s = 310 \text{ m}$$

Total distance travelled = 310 m. Ans

Q10.

A bus is travelling at 36 km/h 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 a constant acceleration of 2.5 m/s. Will the bus be able to stop before reaching the obstacle?

Answer

Given

Initial velocity,$$u = 36 \text{ km h}^{-1}$$

Convert into m s^-¹:$$u = 36 \times \frac{1000}{3600}$$
$$u = 10 \text{ m s}^{-1}$$

Distance of obstacle from bus,$$30 \text{ m}$$

Reaction time,$$0.5 \text{ s}$$

Retardation after braking,$$a = -2.5 \text{ m s}^{-2}$$

Distance travelled during reaction time

Before applying the brakes, the bus continues moving with the same velocity.

$$s = vt$$
$$s = 10 \times 0.5$$
$$s = 5 \text{ m}$$

So, after reaction time, remaining distance:$$30 – 5 = 25 \text{ m}$$

Distance required to stop after the brakes are applied

Using,

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

For stopping,$$v = 0$$
$$0 = (10)^2 + 2(-2.5)s$$
$$0 = 100 – 5s$$
$$5s = 100$$
$$s = 20 \text{ m}$$

Compare with the available distance

Available distance after reaction:$$25 \text{ m}$$

Distance needed to stop:$$20 \text{ m}$$

Since,$$20 \text{ m} < 25 \text{ m}$$

Yes, the bus will stop before reaching the obstacle. Ans

Distance left before obstacle:$$25 – 20 = 5 \text{ m}$$

The bus stops 5 m before the obstacle.

Q11

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

An object kept on the Earth can be considered to be both at rest and in motion, depending on the reference point.

• With respect to the Earth, the object does not change its position with time. Hence, it is considered to be at rest.
• But since the Earth itself is moving around the Sun, the object also moves along with the Earth. Therefore, with respect to the Sun, the object is in motion.

Thus, motion and rest depend on the chosen reference point. Ans

Q12.

The velocity-time graph from 0s to 120s 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.

Answer

(i) Area representing displacement while moving with constant velocity

• From $t = 20 \text{ s}$ to $t = 100 \text{ s}$, velocity is constant at $3 \text{ m s}^{-1}$
• The rectangular area under the graph between 20 s and 100 s represents this displacement.

(ii) Area representing displacement when the velocity is decreasing

• From $t = 100 \text{ s}$ to $t = 120 \text{ s}$, velocity decreases from $3 \text{ m s}^{-1}$ to $2 \text{ m s}^{-1}$
• The trapezium area under the graph between 100 s and 120 s represents this displacement.

Total displacement in 120 s

Displacement = Total area under the velocity-time graph

1. From 0 s to 20 s

Triangle area:$$\text{Area} = \frac{1}{2} \times 20 \times 3$$
$$= 30 \text{ m}$$

2. From 20 s to 100 s

Rectangle area:$$\text{Area} = 80 \times 3$$
$$= 240 \text{ m}$$

3. From 100 s to 120 s

Trapezium area:$$\text{Area} = \frac{1}{2}(3+2)\times 20$$
$$= \frac{1}{2}\times 5 \times 20$$
$$= 50 \text{ m}$$

Total displacement

$$30 + 240 + 50 = 320 \text{ m}$$

Total displacement = 320 m

Average acceleration in 120 s

Initial velocity:$$u = 0 \text{ m s}^{-1}$$

Final velocity:$$v = 2 \text{ m s}^{-1}$$

Time:$$t = 120 \text{ s}$$

Using,

$a=\frac{v-u}{t}$$$a = \frac{2-0}{120}$$
$$a = \frac{1}{60} \text{ m s}^{-2}$$
$$a \approx 0.017 \text{ m s}^{-2}$$

Average acceleration = 0.017 m s^⁻²

Q13.

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.

Answer

Running distance = Area under the velocity-time graph

Divide the graph into parts.

1. From 0 h to 1 h

Velocity = $7 \text{ km h}^{-1}$

Distance:$$s_1 = 7 \times 1$$

$$s_1 = 7 \text{ km}$$

2. From 1 h to 2 h

Velocity increases from $7$ to $7.5 \text{ km h}^{-1}$

Distance = area of trapezium$$s_2 = \frac{1}{2}(7+7.5)\times 1$$
$$s_2 = 7.25 \text{ km}$$

3. From 2 h to 4 h

Velocity = $7.5 \text{ km h}^{-1}$

Distance:$$s_3 = 7.5 \times 2$$
$$s_3 = 15 \text{ km}$$

4. From 4 h to 5 h

Velocity decreases from $7.5$ to $7 \text{ km h}^{-1}$$$s_4 = \frac{1}{2}(7.5+7)\times 1$$
$$s_4 = 7.25 \text{ km}$$

5. From 5 h to 6 h

Velocity decreases from $7$ to $6.5 \text{ km h}^{-1}$$$s_5 = \frac{1}{2}(7+6.5)\times 1$$
$$s_5 = 6.75 \text{ km}$$

6. From 6 h to 7 h

Velocity = $6.5 \text{ km h}^{-1}$$$s_6 = 6.5 \times 1$$
$$s_6 = 6.5 \text{ km}$$

Total running distance

$$s = 7 + 7.25 + 15 + 7.25 + 6.75 + 6.5$$

$$s = 49.75 \text{ km}$$

Estimated running distance = 49.75 km. Ans

Q14.

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

Velocity for first part:$$v = 6 \text{ m s}^{-1}$$

Time:$$2 \text{ min} = 120 \text{ s}$$

Acceleration in second part:$$a = 1 \text{ m s}^{-2}$$

Time of acceleration:$$t = 6 \text{ s}$$

Draw a velocity-time graph

• From $0$ s to $120$ s :
  velocity remains constant at $6 \text{ m s}^{-1}$
• From $120$ s to $126$ s :
  velocity increases uniformly.

Final velocity after acceleration:

Using,

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

So, the graph consists of:

• a rectangle from $0$ to $120$,
• and a trapezium from $120$ s to $126$ s.

Displacement from the graph

Displacement = Area under the velocity-time graph

Rectangle area

$$A_1 = 120 \times 6$$
$$A_1 = 720 \text{ m}$$

Trapezium area

$$A_2 = \frac{1}{2}(6+12)\times 6$$
$$A_2 = \frac{1}{2}\times 18 \times 6$$
$$A_2 = 54 \text{ m}$$

Total displacement

$$s = 720 + 54$$
$$s = 774 \text{ m}$$

Displacement of the car = 774 m. Ans

Q15.

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 in 5 s. Car B attains a velocity of 3 m/s in 10 s. Plot the velocity-time graphs for both cars in the same graph. Using the graph, calculate the displacement mentioned in the two time intervals (Hint: Calculate the acceleration in both cases. Then calculate their velocities at five instants of time to plot the graph).

Answer

For Car A

Given:$$u = 0 \text{ m s}^{-1}$$
$$v = 5 \text{ m s}^{-1}$$
$$t = 5 \text{ s}$$

Acceleration of Car A

Using,

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

Velocity values for plotting a graph

The graph is a straight line joining $(0,0)$ and $(5,5)$.

Displacement of Car A

Displacement = Area under the velocity-time graph

Triangle area:$$s_A = \frac{1}{2}\times 5 \times 5$$
$$s_A = 12.5 \text{ m}$$

Displacement of Car A = 12.5 m Ans

For Car B

Given:$$u = 0 \text{ m s}^{-1}$$
$$v = 3 \text{ m s}^{-1}$$
$$t = 10 \text{ s}$$

Acceleration of Car B

Using,

$a=\frac{v-u}{t}$$$a_B = \frac{3-0}{10}$$

$$a_B = 0.3 \text{ m s}^{-2}$$

Velocity values for plotting graph

The graph is a straight line joining $(0,0)$ and $(10,3)$

Displacement of Car B

Triangle area:$$s_B = \frac{1}{2}\times 10 \times 3$$
$$s_B = 15 \text{ m}$$

Displacement of Car B = 15 m. Ans

Q16.

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).

Answer

Given

Length of minute hand:$$r = 7 \text{ cm}$$

Time interval:$$6 \text{ PM to } 7:30 \text{ PM} = 90 \text{ min}$$

In 60 min, the minute hand completes 1 revolution.

So, in 90 min, it completes:$$\frac{90}{60} = 1.5 \text{ revolutions}$$

(i) Distance travelled

Distance travelled by tip of minute hand:$$\text{Distance} = 1.5 \times 2\pi r$$
$$= 1.5 \times 2\pi \times 7$$
$$= 21\pi \text{ cm}$$

Using $\pi = \frac{22}{7}$,$$= 66 \text{ cm}$$

Distance travelled = 66 cm. Ans

(ii) Displacement

After 1.5 revolutions, the tip reaches the diametrically opposite point.

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

Displacement = 14 cm. Ans

(iii) Speed

$$\text{Speed} = \frac{\text{distance}}{\text{time}}$$
$$= \frac{66}{90}$$

$$= 0.733 \text{ cm min}^{-1}$$

Speed = 0.733 cm min⁻¹ Ans

(iv) Velocity

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

Velocity = 0.156 cm min^⁻¹. Ans