Class 9th Science Chapter-7 Exercises

Exercise – Motion (Answers)

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Question 1

An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?

Given:

• Diameter of track = 200 m

• Radius, $r=100$ m

• Time for 1 round = 40 s

• Total time = 2 min 20 s = 140 s

Step 1: Number of rounds

$$\text{Number of rounds}=\frac{140}{40}=3.5$$

Step 2: Distance covered

Circumference of track:

$$2\pi r=2\times \pi \times 100=200\pi \text{ m}$$

Distance in 3.5 rounds:

$$=3.5\times 200\pi =700\pi \text{ m}$$
$$=2200\text{ m (approximately)}$$

Step 3: Displacement

After 3.5 rounds, the athlete reaches the diametrically opposite point.

$$\text{Displacement}=\text{Diameter}=200\text{ m}$$

Answer:

• Distance covered = 2200 m

• Displacement = 200 m

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Question 2

Joseph jogs from A to B (300 m) in 2 min 30 s and then jogs back 100 m to point C in 1 min. Find average speed and average velocity for:
(a) A to B
(b) A to C

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(a) From A to B

Given:

• Distance = 300 m

• Time = 2 min 30 s = 150 s

Average speed

$$=\frac{300}{150}=2{\text{ m s}}^{-1}$$

Average velocity

$$=\frac{\text{Displacement}}{\text{Time}}=\frac{300}{150}=2{\text{ m s}}^{-1}$$

Answer (A to B):

• Average speed = 2 m s⁻¹

• Average velocity = 2 m s⁻¹

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(b) From A to C

Total distance travelled

$$=300+100=400\text{ m}$$

Total time

$$=150+60=210\text{ s}$$

Displacement (A to C)

$$=300-100=200\text{ m}$$

Average speed

$$=\frac{400}{210}=1.90{\text{ m s}}^{-1}$$

Average velocity

$$=\frac{200}{210}=0.95{\text{ m s}}^{-1}$$

Answer (A to C):

• Average speed = 1.90 m s⁻¹

• Average velocity = 0.95 m s⁻¹

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Question 3

Abdul’s average speed to school is 20 km h⁻¹ and on return it is 30 km h⁻¹. Find the average speed for the whole trip.

Formula (important):

$$\text{Average speed}=\frac{2v_1{v}_2}{v_1+v_2}$$

Calculation:

$$=\frac{2\times 20\times 30}{20+30}=\frac{1200}{50}=24{\text{ km h}}^{-1}$$

Answer:
Average speed for the whole trip = 24 km h⁻¹

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Question 4

A motorboat starts from rest and accelerates uniformly at 3.0 m s⁻² for 8.0 s. Find the distance travelled.

Given:

$$u=0,a=3.0{\text{ m s}}^{-2},t=8\text{ s}$$

Formula:

$$s=ut+\frac{1}{2}a{t}^2$$

Calculation:

$$s=0+\frac{1}{2}\times 3\times 8^2=1.5\times 64=96\text{ m}$$

Answer:
Distance travelled = 96 m

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Question 5

A driver of a car travelling at 52 km h⁻¹ applies the brakes.
(a) Shade the area on the speed–time graph that represents the distance travelled.
(b) Which part of the graph represents uniform motion?

(a) Answer:

The area under the speed–time graph represents the distance travelled by the car.
Shade the region between the graph line and the time axis.

(b) Answer:

The horizontal straight-line portion of the speed–time graph represents uniform motion, because speed remains constant there.

Exercise – Question 6

Fig. 7.10 shows the distance–time graph of three objects A, B and C. Study the graph and answer the following questions:

(a) Which of the three is travelling the fastest?

Answer:
Object B is travelling the fastest.

Explanation:
In a distance–time graph, the object with the steepest slope (greatest inclination) has the highest speed.
Object B has the steepest line.

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(b) Are all three ever at the same point on the road?

Answer:
Yes, all three objects are at the same point at the same time.

Explanation:
All three distance–time graphs intersect at one point, which indicates that their distances from the origin are the same at that instant.

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How far has C travelled when B passes A?

Answer:
When B passes A, both B and A are at the same distance from the origin.
From the graph, at this point, object C has travelled 6 km.

Distance travelled by C = 6 km

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(d) How far has B travelled by the time it passes C?

Answer:
When B passes C, their graphs intersect again.
From the graph, the distance corresponding to this intersection point is 8 km.

Distance travelled by B = 8 km

Question 7

A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s⁻², with what velocity will it strike the ground? After what time will it strike the ground?

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Given:

• Initial velocity,

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

  (Ball is gently dropped)

• Distance fallen,

  $$s=20\text{ m}$$

• Acceleration due to gravity,

  $$a=10{\text{ m s}}^{-2}$$

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(a) Velocity with which the ball strikes the ground

Using the equation of motion:

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

Substituting the values:

$$v^2=0+2\times 10\times 20$$
$$v^2=400$$
$$v=20{\text{ m s}}^{-1}$$

Velocity at the ground = 20 m s⁻¹ (downward)

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(b) Time taken to strike the ground

Using the equation:

$$v=u+at$$

Substituting the values:

$$20=0+10t$$
$$t=2\text{ s}$$

Time taken = 2 s

Question 8

The speed–time graph for a car is shown in Fig. 7.11.

(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.

(b) Which part of the graph represents uniform motion of the car?

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(a) Distance travelled in the first 4 seconds

Concept Used

Distance travelled = Area under the speed–time graph

In the first 4 seconds, the graph is a straight sloping line starting from the origin, forming a triangle.

From the graph (Fig. 7.11):

• Time = 4 s

• Maximum speed at 4 s = 8 m s⁻¹

Area of triangle

$$\text{Distance}=\frac{1}{2}\times \text{base}\times \text{height}$$
$$=\frac{1}{2}\times 4\times 8$$
$$=16\text{ m}$$

Distance travelled in first 4 seconds = 16 m

Shading instruction:
Shade the triangular area under the graph from 0 to 4 s.

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(b) Which part of the graph represents uniform motion?

Answer:
The horizontal straight-line portion of the speed–time graph represents uniform motion.

Explanation:

• In this part, speed remains constant

• Constant speed ⇒ zero acceleration

• Hence, motion is uniform

Question 9

State which of the following situations are possible and give an example for each:

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(a) An object with a constant acceleration but with zero velocity

Answer:
Yes, this situation is possible.

Example:
A ball thrown vertically upward has zero velocity at the highest point, but it still has a constant acceleration due to gravity (10 m s⁻² downward).

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(b) An object moving with an acceleration but with uniform speed

Answer:
Yes, this situation is possible.

Example:
An object moving in a circular path at constant speed, such as:

• A car moving on a circular track

• The moon revolving around the Earth

Explanation:
Although speed is constant, the direction of motion keeps changing, so the velocity changes, which means the object has acceleration.

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(c) An object moving in a certain direction with an acceleration in the perpendicular direction

Answer:
Yes, this situation is possible.

Example:
An object in uniform circular motion, such as:

• A stone tied to a string and whirled in a circle

Explanation:
The acceleration (centripetal acceleration) is directed towards the centre of the circle, which is perpendicular to the direction of motion at every point.

Question 10

An artificial satellite is moving in a circular orbit of radius 42,250 km. Calculate its speed if it takes 24 hours to revolve around the Earth.

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Given:

Radius of orbit,

$$r=\mathrm{42,250}\text{ km}$$

Time period,

$$T=24\text{ h}$$

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Formula Used (Uniform Circular Motion):

$$\text{Speed}=\frac{\text{Circumference of orbit}}{\text{Time period}}=\frac{2\pi r}{T}$$

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Calculation:

Circumference of orbit:

$$2\pi r=2\times \frac{22}{7}\times \mathrm{42,250}=\mathrm{265,714}\text{ km (approx.)}$$

Speed:

$$v=\frac{\mathrm{265,714}}{24}=\mathrm{11,071}{\text{ km h}}^{-1}$$

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(Optional – in m s⁻¹)

$$\mathrm{11,071}\times \frac{1000}{3600}=3075{\text{ m s}}^{-1}(\text{approx.})$$

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