Question & Answer Class 9 Science Exploration Chapter 4 Describing Motion Around Us

Total Distance Travelled $$\text{Total Distance} = 250 + 250 + 250 + 250$$

$$\boxed{{\displaystyle \text{Total Distance}=1000\text{ m}}}$$

Displacement from Home

Displacement is the net change in position between the starting point and the ending point.

• Starting point = Home
• Ending point = Home (he came back home finally)

Since he started and ended at the same point (home): Displacement = Final Position − Initial Position

Displacement = 0 m

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

Answer: Vertical Distance and Displacement

Understanding the Journey

Let’s identify the floors and their heights:

(i) Finding Acceleration

Using the kinematic equation:

$$v^2 = u^2 + 2as$$ 

$$0^2 = (28)^2 + 2 \times a \times 98$$ 

$$0 = 784 + 196a$$ 

$$196a = -784$$ 

$$\boxed{{\displaystyle a=-4{\text{ m s}}^{-2}}}$$ 

$$\boxed{a = -4 \text{ m s}^{-2}}$$​The negative sign indicates that acceleration is opposite to the direction of motion (deceleration/retardation), which makes sense as the motorbike is slowing down.

(ii) Finding Time Taken to Stop

Using the kinematic equation:

$$v=u+at$$$$v = u + at$$ 

$$0=28+(-4)\times t$$ 

$$4t = 28$$ 

$$\boxed{{\displaystyle t=7\text{ s}}}$$ 

​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:

No, A and B never have equal velocity

From the graph (Fig. 4.27):

Both position-time graphs are straight lines starting from the origin, which means both objects move with constant velocity throughout.

The slope of a position-time graph = velocity of the object

Justification:

Since both lines are straight (not curved), the slope of each line is fixed and constant — it never changes at any point during the motion.

The slope of A is always greater than the slope of B throughout the entire motion. The two lines are not parallel — they diverge continuously, meaning A always covers more displacement than B in the same time interval.

Therefore, the velocities of A and B are always different — they never become equal at any instant.

Objects A and B never have equal velocity because their slopes (velocities) are constant but unequal.

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: Correct options are (i) and (iii)

Analysing each option carefully:

From the graph (Fig. 4.28):

• Object A = straight line → constant velocity
• Object B = curved line → changing velocity (starts slow, ends fast)
• Both start at the same initial position (origin) and end at the same final position at t = 10 s

Option (i) CORRECT

“The average velocity of both over 10 s is equal since they have the same initial and final positions.”

$$\text{Average velocity} = \frac{\text{Displacement}}{\text{Time}} = \frac{\text{Final position} – \text{Initial position}}{\text{Time}}$$​

Since both A and B have the same initial and final positions, their displacement is equal. Same displacement ÷ same time = equal average velocity.

Option (ii) INCORRECT

“Average speeds are equal since both cover equal distance in equal time.”

This is wrong. Average speed depends on total distance travelled, not displacement. B follows a curved path and therefore covers a longer total distance than A’s straight path in the same time. Their average speeds are NOT equal.

Option (iii) CORRECT

“Average speed of A is lower than B since A covers a shorter distance than B in 10 seconds.”

$$\text{Average speed} = \frac{\text{Total distance travelled}}{\text{Time}}$$ 

• A travels along a straight line → shorter distance
• B travels along a curve → longer distance

Option (iv)  INCORRECT

“Average speed of A is greater than B since B’s speed is lower in some segments.”

Even though B is slower in the initial segments, it compensates by moving very fast later. The total path length of B is greater than A, so B’s average speed is actually higher, not lower.

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.

Answer: First convert speeds into m/s:

• Initial speed,
  $u = 54 \text{ km/h} = 15 \text{ m/s}$
• Final speed,
  $v = 36 \text{ km/h} = 10 \text{ m/s}$
• Time,
  $t = 36 \text{ s}$ 

Since acceleration is constant, we use:

$$\text{Distance}=\frac{(u+v)}{2}\times t$$

Substitute values:

$$s = \frac{15 + 10}{2} \times 36$$ 

$$s = \frac{25}{2} \times 36 = 12.5 \times 36 = 450 \text{ m}$$ 

Distance travelled = 450 m

9. 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.

Answer: The motion has three parts:

(i) Acceleration phase

Given:u = 0

,

$v=20\text{m/ s, t = 5s}$ 

Use:

$s = ut + \frac{1}{2}at^2$ 

First find acceleration:

$$a = \frac{v-u}{t} = \frac{20}{5} = 4 \, \text{m/s}^2$$

Now distance:

$$s_1=0+\frac{1}{2}\times 4\times 5^2=50\text{m}$$

(ii) Constant speed phase

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

$$s_1 = 0 + \frac{1}{2} \times 4 \times 5^2 = 50 \, \text{m}$$

(iii) Deceleration phase

Given:

$u=20\text{m/s}$,

$v = 0$

v=0,

$t=6$

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

Total distance

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

Answer: 310 m

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?

Step 1: Convert speed

$$36\text{km/h}=10\text{m/s}$$ 

Step 2: Distance during reaction time

$$s_1=vt=10\times 0.5=5\text{m}$$ 

Step 3: Braking distance**

Given:

$u = 10 \, \text{m/s}, v = 0, a = -2.5 \, \text{m/s}^2$ 

Use:

$v^2 = u^2 + 2as$ 

$$0 = 10^2 + 2(-2.5)s$$ 

$$0 = 100 – 5s \Rightarrow s = 20 \, \text{m}$$ 

Step 4: Total stopping distance

$$s=5+20=25\text{m}$$ 

Obstacle distance = 30 m

Conclusion:

Yes, the bus will stop safely because

$$25\text{m}<30\text{m}$$ 

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: The statement “Earth moves around the Sun” is correct. However:

• Motion is relative to a reference point.
• An object kept on Earth does not change its position relative to Earth.

Conclusion:

• The object is at rest with respect to Earth.
• But it is in motion with respect to the Sun.

So, an object can be considered at rest depending on the chosen reference frame.

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.

Answer: Understanding the graph

From the graph:

• 0–20 s: velocity increases from 0 to 3 m/s
• 20–100 s: velocity is constant at 3 m/s
• 100–120 s: velocity decreases from 3 m/s to 2 m/s

(i) Area for constant velocity

This is from 20 s to 100 s
Shape: Rectangle

(ii) Area when velocity is decreasing

This is from 100 s to 120 s
Shape: Trapezium (or sloping region)

Total displacement

Displacement = Area under velocity–time graph

1. From 0 to 20 s (triangle)

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

2. From 20 to 100 s (rectangle)

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

3. From 100 to 120 s (trapezium)

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

Total displacement

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

Average acceleration

$$\text{Average acceleration} = \frac{\text{final velocity} – \text{initial velocity}}{\text{total time}}$$​

$$= \frac{2 – 0}{120} = \frac{2}{120} = \frac{1}{60} \approx 0.017 \, \text{m/s}^2$$ 

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.

Answer:

Step 1: Read approximate values from graph

From the graph (in km/h vs hours):

Step 2: Divide into intervals

(i) From 0 to 2 h

Velocity increases → trapezium

$$\text{Distance}=\frac{(7+7.5)}{2}\times 2=14.5\text{km}$$ 

(ii) From 2 to 4 h

Constant velocity = 7.5 km/h

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

(iii) From 4 to 6 h

Velocity decreases → trapezium

$$\text{Distance}=\frac{(7.5+6.5)}{2}\times 2=14\text{km}$$ 

Step 3: Total distance

$$\text{Total distance}=14.5+15+14=43.5\text{km}$$ 

Final Answer:

Estimated running distance ≈ 43.5 km

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.

Answer:

Step 1: Given data

• Constant velocity
  $= 6 \, \text{m/s}$ 

  $=6\text{m/s}$ for 2 min = 120 s

• Then acceleration
  $a = 1 \, \text{m/s}^2$ 

  6 s

• Initial velocity during acceleration =
  $6 \, \text{m/s}$ 

   

Step 2: Draw understanding of graph

• From 0 to 120 s → horizontal line at 6 m/s (rectangle)
• From 120 to 126 s → velocity increases linearly (trapezium)

Step 3: Distance in first part (rectangle)

$$s_1=v\times t=6\times 120=720\text{m}$$

Step 4: Distance in second part (accelerated motion)

First find final velocity:

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

Now displacement (trapezium area):

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

Step 5: Total displacement

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

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 (Hint: Calculate the acceleration in both cases. Then calculate their velocities at five instants of timeto plot the graph).

Answer:

Step 1: Find Accelerations

Car A:

$a_A = \frac{5-0}{5} = \textbf{1 m s}^{-2}$

Car B:

$a_B = \frac{3-0}{10} = \textbf{0.3 m s}^{-2}$

Step 2: Velocity at 5 instants of time