Tag: Chapter- 7 Motion

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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Chapter – 7 Motion

Page No. 74 – Questions & Answers

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Question 1

An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Answer:
Yes, an object can have zero displacement even after moving through a distance.

Explanation:
Displacement depends on the initial and final positions of an object. If the final position is the same as the initial position, displacement becomes zero.

Example:
If a person walks 10 m forward and then 10 m backward, the distance travelled is 20 m, but since the person returns to the starting point, the displacement is zero.

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Question 2

A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Answer:

• Side of square field = 10 m

• Perimeter of square = 4 × 10 = 40 m

• Time to complete 1 round = 40 s

Total time given = 2 min 20 s = 140 s

Number of rounds completed =
140 ÷ 40 = 3.5 rounds

After 3 full rounds, the farmer comes back to the starting point.
After half round, he reaches the opposite corner of the square.

Magnitude of displacement = diagonal of square

$$\text{Displacement}=\sqrt{{10}^2+{10}^2}=\sqrt{200}=10\sqrt{2}\text{ m}$$

Final Answer:
Magnitude of displacement = $10\sqrt{2}$ m

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Question 3

Which of the following is true for displacement?

(a) It cannot be zero.
(b) Its magnitude is greater than the distance travelled by the object.

Answer:

Neither (a) nor (b) is true.

Explanation:

• Displacement can be zero when the initial and final positions are the same.

• The magnitude of displacement is always less than or equal to the distance travelled, never greater.

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Page No. 76 – Questions & Answers

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Question 1

Distinguish between speed and velocity.

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Question 2

Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

Answer:
The magnitude of average velocity is equal to average speed when the object moves in a straight line without changing direction.

Explanation:
In this case, distance = displacement, so both average speed and average velocity become equal.

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Question 3

What does the odometer of an automobile measure?

Answer:
An odometer measures the total distance travelled by an automobile.

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Question 4

What does the path of an object look like when it is in uniform motion?

Answer:
When an object is in uniform motion, it moves along a straight-line path.

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Question 5

During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station?
(The signal travels at the speed of light = $3\times {10}^8$ m s⁻¹)

Given:
Speed of signal,

$$v=3\times {10}^8{\text{ m s}}^{-1}$$

Time taken,

$$t=5\text{ minutes}=5\times 60=300\text{ s}$$

Distance = Speed × Time

$$s=v\times t$$
$$s=3\times {10}^8\times 300$$
$$s=9\times {10}^{10}\text{ m}$$

Final Answer:
Distance of the spaceship = $9\times {10}^{10}$ m

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Page No. 77 – Questions & Answers

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Question 1

When will you say a body is in
(i) uniform acceleration?
(ii) non-uniform acceleration?

Answer:

(i) Uniform acceleration:
A body is said to be in uniform acceleration when its velocity changes by equal amounts in equal intervals of time, no matter how small the time intervals are.

Example:
A freely falling body under gravity.

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(ii) Non-uniform acceleration:
A body is said to be in non-uniform acceleration when its velocity changes by unequal amounts in equal intervals of time.

Example:
A car moving in heavy traffic where speed changes irregularly.

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Question 2

A bus decreases its speed from 80 km h⁻¹ to 60 km h⁻¹ in 5 s. Find the acceleration of the bus.

Given:
Initial velocity,

$$u=80{\text{ km h}}^{-1}=\frac{80\times 1000}{3600}=22.22{\text{ m s}}^{-1}$$

Final velocity,

$$v=60{\text{ km h}}^{-1}=\frac{60\times 1000}{3600}=16.67{\text{ m s}}^{-1}$$

Time,

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

Formula:

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

Calculation:

$$a=\frac{16.67-22.22}{5}=\frac{-5.55}{5}=-1.11{\text{ m s}}^{-2}$$

Answer:
Acceleration of the bus = –1.11 m s⁻²
(Negative sign indicates deceleration)

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Question 3

A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km h⁻¹ in 10 minutes. Find its acceleration.

Given:
Initial velocity,

$$u=0$$

Final velocity,

$$v=40{\text{ km h}}^{-1}=\frac{40\times 1000}{3600}=11.11{\text{ m s}}^{-1}$$

Time,

$$t=10\text{ minutes}=600\text{ s}$$

Formula:

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

Calculation:

$$a=\frac{11.11-0}{600}=0.0185{\text{ m s}}^{-2}$$

Answer:
Acceleration of the train = 0.0185 m s⁻²

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Page No. 81 – Questions & Answers

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Question 1

What is the nature of the distance–time graphs for uniform and non-uniform motion of an object?

Answer:

• Uniform motion:
  The distance–time graph is a straight line because the object covers equal distances in equal intervals of time.

• Non-uniform motion:
  The distance–time graph is a curved line because the object covers unequal distances in equal intervals of time.

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Question 2

What can you say about the motion of an object whose distance–time graph is a straight line parallel to the time axis?

Answer:
The object is at rest.

Explanation:
A straight line parallel to the time axis means that the distance does not change with time, so the object is not moving.

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Question 3

What can you say about the motion of an object if its speed–time graph is a straight line parallel to the time axis?

Answer:
The object is moving with uniform speed (constant speed).

Explanation:
A straight line parallel to the time axis in a speed–time graph indicates that speed remains constant with time, and acceleration is zero.

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Question 4

What is the quantity which is measured by the area occupied below the velocity–time graph?

Answer:
The area under the velocity–time graph represents the displacement of the object.

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Page No. 82–83 : Questions & Answers

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Question 1

A bus starting from rest moves with a uniform acceleration of 0.1 m s⁻² for 2 minutes. Find
(a) the speed acquired
(b) the distance travelled

Given:

Initial velocity,

$$u=0$$

Acceleration,

$$a=0.1{\text{ m s}}^{-2}$$

Time,

$$t=2\text{ min}=120\text{ s}$$

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(a) Speed acquired

Using equation:

$$v=u+at$$
$$v=0+(0.1\times 120)$$
$$v=12{\text{ m s}}^{-1}$$

Speed acquired = 12 m s⁻¹

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(b) Distance travelled

Using equation:

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

$$s=0+\frac{1}{2}\times 0.1\times (120{)}^2$$
$$s=0.05\times 14400$$
$$s=720\text{ m}$$

Distance travelled = 720 m

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Question 2

A train is travelling at a speed of 90 km h⁻¹. Brakes are applied to produce a uniform acceleration of –0.5 m s⁻². Find how far the train will go before it is brought to rest.

Given:

Initial velocity,

$$u=90{\text{ km h}}^{-1}=25{\text{ m s}}^{-1}$$

Final velocity,

$$v=0$$

Acceleration,

$$a=-0.5{\text{ m s}}^{-2}$$

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Using equation:

$$v^2=u^2+2as$$
$$0=(25{)}^2+2(-0.5)s$$
$$0=625-s$$
$$s=625\text{ m}$$

Distance travelled before stopping = 625 m

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Question 3

A trolley, while going down an inclined plane, has an acceleration of 2 cm s⁻². What will be its velocity 3 s after the start?

Given:

Initial velocity,

$$u=0$$

Acceleration,

$$a=2{\text{ cm s}}^{-2}$$

Time,

$$t=3\text{ s}$$

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Using equation:

$$v=u+at$$
$$v=0+(2\times 3)$$
$$v=6{\text{ cm s}}^{-1}$$

Velocity after 3 s = 6 cm s⁻¹

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Question 4

A racing car has a uniform acceleration of 4 m s⁻². What distance will it cover in 10 s after start?

Given:

$$u=0,a=4{\text{ m s}}^{-2},t=10\text{ s}$$

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Using equation:

$$s=ut+\frac{1}{2}a{t}^2$$
$$s=0+\frac{1}{2}\times 4\times (10{)}^2$$
$$s=2\times 100$$
$$s=200\text{ m}$$

Distance covered = 200 m

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Question 5

A stone is thrown vertically upwards with a velocity of 5 m s⁻¹. If acceleration is 10 m s⁻² downward, find:
(a) maximum height attained
(b) time taken to reach maximum height

Given:

$$u=5{\text{ m s}}^{-1},v=0,a=-10{\text{ m s}}^{-2}$$

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(a) Maximum height

Using equation:

$$v^2=u^2+2as$$
$$0=25-20s$$
$$s=1.25\text{ m}$$

Maximum height = 1.25 m

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(b) Time taken

Using equation:

$$v=u+at$$
$$0=5-10t$$
$$t=0.5\text{ s}$$

Time taken = 0.5 s