## Intext Questions 1

#### 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 if it returns to its initial position.

Example: When a racer runs on a running track, his initial and final position is the same. Hence, the displacement is zero.

#### 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**

Given,

Time period of travel = 2 minutes 20 seconds

Convert min into sec

1 min = 60 sec

So 2 min 20 sec = (2 x 60) sec + 20 sec = 120 + 20 = 140 sec

Side of the given square field = 10 m

Hence, the perimeter of a square = 4 x side = 4 x 10 = 40 m

Time taken by the farmer to cover the boundary of 40 m = 40 s

So, in 1 s, the farmer covers a distance of = $\dfrac{40}{40}$ x 1 = 1 m

Hence, distance covered by the farmer in 140 sec = 140 m

The total number of rotations taken by the farmer to cover a distance of 140 m

= $\dfrac{\text{total distance}}{\text{perimeter}}$

= $\dfrac{140}{40}$

= 3.5

If the farmer starts from point A, after 3.5 rounds he will be at point C of the field.

∴ Displacement (from Pythagoras theorem)

$\text{AC} = \sqrt{(\text{AB})^2 + (\text{BC})^2} \\[1em] = \sqrt{(10)^2 + (10)^2} \\[1em] = \sqrt{100 + 100} \\[1em] = \sqrt{200} \\[1em] = 10\sqrt{2} \\[1em] = 10 \times 1.414 \\[1em] = 14.14 \text{ m}$

Hence, **the magnitude of displacement = 14.14 m**

#### 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**

(a) False.

**Reason** — The displacement of an object which travels a certain distance and comes back to its initial position is zero.

(b) False.

**Reason** — The displacement of an object can be equal to, but never greater than the distance travelled.

## Intext Questions 2

#### Question 1

Distinguish between speed and velocity.

**Answer**

Speed | Velocity |
---|---|

The distance travelled per second by a moving object is called its speed. | The distance travelled per second by a moving object in a particular direction is called its velocity. |

It is a scalar quantity. The speed does not tell us the direction of motion. | It is a vector quantity. The velocity tells us the speed as well as the direction of motion. |

The speed is always positive since direction is not taken into consideration. | The velocity can be positive or negative depending upon the direction of motion. |

After one round in a circular path, the average speed is not zero. | After completing each round in a circular path, the average velocity is zero. |

#### Question 2

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

**Answer**

As average speed is the total distance travelled in a time frame and average velocity is the total displacement in a time frame, so the magnitude of average velocity and average speed will be the equal when the total distance travelled is equal to the displacement.

This happens when an object moves along a straight line in the same direction (without turning back).

#### Question 3

What does the odometer of an automobile measure?

**Answer**

An odometer is a device that measures the distance travelled by an automobile based on the perimeter of the wheel as the wheel rotates.

#### Question 4

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

**Answer**

The path of an object in uniform motion is a **straight line**.

#### 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, that is, 3 × 10^{8} ms^{-1}.

**Answer**

Given,

Time taken (t) = 5 min

Convert min into sec

1 min = 60 sec

So, 5 min = 5 x 60 sec = 300 sec

Speed of the signal = 3 × 10^{8} ms^{-1}

We know, speed = $\dfrac{\text{distance}}{\text{time}}$

Substituting the values, we get,

3 × 10^{8} = $\dfrac{\text{distance}}{300}$

Distance covered (d) = 3 × 10^{8} x 300 = 9 x 10^{10} m.

Hence, **the distance of the spaceship from the ground station = 9 x 10 ^{10} m**

## Intext Questions 3

#### Question 1

When will you say a body is in

(i) uniform acceleration?

(ii) non-uniform acceleration?

**Answer**

(i) The acceleration is said to be uniform (or constant) when equal changes in velocity take place in equal intervals of time.

**Example** — The motion of a body under gravity (e.g., free fall of a body)

(ii) If changes in velocity are not same in the same intervals of time, the acceleration is said to be non-uniform.

**Example** — The motion of a vehicle on a crowded road.

#### Question 2

A bus decreases its speed from 80 kmh^{-1} to 60 kmh^{-1} in 5 s. Find the acceleration of the bus.

**Answer**

Given,

Initial velocity (u) = 80 kmh^{-1}

Convert kmh^{-1} to ms^{-1} : multiply by $\dfrac{5}{18}$

80 x $\dfrac{5}{18}$ = 22.22 ms^{-1}

The final velocity (v) = 60 kmh^{-1}

60 x $\dfrac{5}{18}$ = 16.66 ms^{-1}

Time (t) = 5 sec

According to the first equation of motion,

v = u + at

or

a = $\dfrac{\text {v - u}}{\text{t}}$

Substituting we get,

a = $\dfrac{16.66 - 22.22}{5}$ = -1.112 ms^{-2}

∴ **the acceleration of the bus is -1.112 ms ^{-2}.**

The negative sign indicates that the velocity of the bus is decreasing.

#### Question 3

A train starting from a railway station and moving with uniform acceleration attains a speed 40 kmh^{-1} in 10 minutes. Find its acceleration.

**Answer**

Given.

Initial velocity (u) = 0

Final velocity (v) = 40 kmh^{-1}

Convert kmh^{-1} to ms^{-1} : multiply by $\dfrac{5}{18}$

40 x $\dfrac{5}{18}$ = 11.11 ms^{-1}

Time (t) = 10 minute

Convert min into s

1 min = 60 s

10 min = 60 x 10 = 600 s

Acceleration (a) =?

According to the first equation of motion,

v = u + at

or

a = $\dfrac{\text {v - u}}{\text{t}}$

Substituting we get,

a = $\dfrac{11.11-0}{600}$ = 0.0185 ms^{-2}

∴ **the acceleration of the bus is 0.0185 ms ^{-2}.**

## Intext Questions 4

#### Question 1

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

**Answer**

For uniform motion, the distance-time graph is a straight line.

For non-uniform motion, the distance-time graph is a curve.

#### 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**

When distance-time graph is a straight line parallel to the time axis it means that the object is at the same position as time passes. Hence, the object is at rest.

#### 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**

As there is no change in the speed of the object at any point of time, the object is said to be in uniform motion.

#### Question 4

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

**Answer**

As velocity x time = displacement, the area enclosed between the velocity-time sketch and x-axis (i.e., the time axis) gives the displacement of the body.

## Intext Questions 5

#### Question 1

A bus starting from rest moves with a uniform acceleration of 0.1 ms^{-2} for 2 minutes. Find

(a) the speed acquired,

(b) the distance travelled.

**Answer**

(a) Given,

Initial velocity (u) = 0

Acceleration (a) = 0.1 ms^{-2}

Time = 2 minutes

Convert min into sec

1 min = 60 sec

So, 2 min = 2 x 60 sec = 120 s

v = u + at

Substituting we get,

v = 0 + (0.1 x 120) = 12

Hence, **speed acquired = 12 ms ^{-1}**

(b) According to the third equation of motion,

2as = v^{2} - u^{2}

or

s = $\dfrac{\text{{v}}^2 - \text{{u}}^2}{2\text{a}}$

Since a = 0.1 ms^{-2}

v = 12 ms^{-1}

u = 0

t = 120 s

Substituting we get,

s = $\dfrac{{12}^2 - 0}{2 \times 0.1} = \dfrac{144}{0.2}$ = 720 m.

Hence, **the total distance travelled is 720 m**

#### Question 2

A train is travelling at a speed of 90 kmh^{-1}. Brakes are applied so as to produce a uniform acceleration of -0.5 ms^{-2}. Find how far the train will go before it is brought to rest.

**Answer**

Given,

Initial velocity (u) = 90 kmh^{-1}

Convert kmh^{-1} to ms^{-1}

90 x $\dfrac{5}{18}$= 25 ms^{-1}

Terminal velocity (v) = 0

Acceleration (a) = -0.5 ms^{-2}

According to the third equation of motion,

v^{2} – u^{2} = 2as

$\Rightarrow \text{s} = \dfrac{\text{v}^2 - \text{u}^2}{2\text{a}}$

Substituting we get,

s = $\dfrac{0 - {25}^2 }{2 \times 0.5} = \dfrac{625}{1}$ = 625 m.

Hence, **the train travels 625 m before it is brought to rest**

#### Question 3

A trolley, while going down an inclined plane, has an acceleration of 2 cms^{-2}. What will be its velocity 3 s after the start?

**Answer**

Given,

Initial velocity (u) = 0

Acceleration (a) = 2 cms^{-2}

Time (t) = 3s

According to the first equation of motion,

v = u + at

Substituting we get,

v = 0 + (2 x 3) = 6 cms^{-1}

Hence, **velocity of the trolley after 3 seconds will be 6 cms ^{-1}**

#### Question 4

A racing car has a uniform acceleration of 4 ms^{-2}. What distance will it cover in 10 s after start?

**Answer**

Given,

Initial velocity (u) = 0

Acceleration (a) = 4 ms^{-2}

Time period (t) = 10 s

According to the second equation of motion,

S = ut + $\dfrac{1}{2}$at^{2}

Substituting we get,

S = 0 + $\dfrac{1}{2}$ x 4 x 10 x 10

= 200 m

∴ **The car will cover a distance of 200 m**.

#### Question 5

A stone is thrown in a vertically upward direction with a velocity of 5 ms^{-1}. If the acceleration of the stone during its motion is 10 ms^{-2} in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

**Answer**

Given, initial velocity (u) = 5 ms^{-1}

Terminal velocity (v) = 0

Acceleration = -10 ms^{-2} (as it is in upward direction)

According to the third equation of motion,

v^{2} – u^{2} = 2as

$\Rightarrow \text{s} = \dfrac{\text{v}^2 - \text{u}^2}{2\text{a}}$

Substituting we get,

s = $\dfrac{0 - {5}^2 }{2 \times -10} = \dfrac{-25}{-20}$ = 1.25 m.

Hence, **distance covered = 1.25 m**

According to the first equation of motion,

v = u + at

∴ time taken by the stone to reach a position of rest (maximum height) = $\dfrac{\text{v} - \text{u}}{\text{a}}$

Substituting we get,

$\dfrac{0-5}{-10}$ = 0.5 sec

∴ **The stone reaches a maximum height of 1.25 m in 0.5 sec.**

## Exercises

#### 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?

**Answer**

Given,

Diameter of track (d) = 200 m

Circumference of the track = πd = 200π m

So, distance covered in 40 seconds = 200π m

Distance covered in 1 sec = $\dfrac{200π}{40}$

Distance covered in 2 min 20 sec (i.e., 60 + 60 + 20 = 140 sec) = $\dfrac{200π}{40}$ x 140

= $\dfrac{200 \times 22 \times 140}{7 \times 40}$

= 2200 m

Number of rounds completed in 140 sec = $\dfrac{140}{40}$ = 3.5

∴ the final position (with respect to the initial position) is at the opposite end of the circular track. So, the net displacement will be equal to the diameter of the track = 200 m.

Hence, **the net distance covered by the athlete is 2200 m and the total displacement of the athlete is 200 m.**

#### Question 2

Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph's average speeds and velocities in jogging

(a) from A to B and

(b) from A to C?

**Answer**

(a) From A to B

Given,

Distance covered from A to B = 300 m

Time taken to travel from A to B = 2 min and 30 sec = 60 + 60 + 30 = 150 sec

Displacement from A to B = 300 m

Average speed = $\dfrac{\text{total distance travelled}}{\text{total time taken}}$

Average velocity = $\dfrac{\text{total displacement}}{\text{total time taken}}$

Substituting we get,

Average speed = $\dfrac{300}{150}$ = 2 ms^{-1}

Average velocity = $\dfrac{300}{150}$ = 2 ms^{-1}

Hence, **average speed from A to B is 2 ms ^{-1} and average velocity = 2 ms^{-1}**

(b) From A to C

Distance covered from A to C = 300 m + 100 m = 400 m

Time taken from A to C = 2 min 30 sec + 1 min = 60 + 60 + 30 + 60 = 210 sec

Displacement from A to C = 300 m – 100 m = 200 m

Average speed = $\dfrac{\text{total distance travelled}}{\text{total time taken}}$

Average velocity = $\dfrac{\text{total displacement}}{\text{total time taken}}$

Substituting we get,

Average speed = $\dfrac{400}{210}$ = 1.9 ms^{-1}

Average velocity = $\dfrac{200}{210}$ = 0.952 ms^{-1}

Hence, **average speed from A to C is 1.9 ms ^{-1} and average velocity = 0.952 ms^{-1}**

#### Question 3

Abdul, while driving to school, computes the average speed for his trip to be 20 kmh^{-1}. On his return trip along the same route, there is less traffic and the average speed is 30 kmh^{-1}. What is the average speed for Abdul's trip?

**Answer**

Distance travelled to reach the school = distance travelled to reach home = d

Time taken to reach school = t_{1}

Time taken to reach home = t_{2}

Average speed = $\dfrac{\text{total distance travelled}}{\text{total time taken}}$

On going to school :

20 = $\dfrac{\text{d}}{\text{t}_1}$ or

t_{1} = $\dfrac{\text{d}}{20}$

On return trip :

30 = $\dfrac{\text{d}}{\text{t}_2}$ or

t_{2} = $\dfrac{\text{d}}{30}$

Now, the average speed for the entire trip

= $\dfrac{\text{d+d}}{\text{t}_1 + \text{t}_2}$

= $\dfrac{\text{2d}}{\dfrac{\text{d}}{20} + \dfrac{\text{d}}{30}}$

= $\dfrac{\text{2d}}{\dfrac{\text{3d + 2d}}{60}}$

= $\dfrac{\text{2d} \times {60}}{5\text{d}}$

= 24 kmh^{-1}

∴ **Abdul's average speed for the entire trip is 24 kmh ^{-1}**.

#### Question 4

A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms^{-2} for 8.0 s. How far does the boat travel during this time?

**Answer**

Given,

Initial velocity (u) = 0

Acceleration (a) = 3 ms^{-2}

Time period (t) = 8 s

Distance travelled (S) = ?

According to the second equation of motion,

S = ut + $\dfrac{1}{2}$at^{2}

Substituting we get,

S = 0 + $\dfrac{1}{2}$ x 3 x 8 x 8

= 96 m

∴ **The motorboat travels a distance of 96 m**

#### Question 5

A driver of a car travelling at 52 kmh^{-1} applies the brakes. 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?

**Answer**

(a) The initial speed of the graph is 52 kmh^{-1}. After the driver applies the brakes, the speed becomes zero.

The x-axis of the graph represents the time and the y-axis of the graph represents the velocity.

In the graph, point A represents the initial velocity of the car, that is, 52 kmh^{-1}, and point B represents the time when the velocity becomes zero.

The distance travelled by an object is equal to the product of velocity and time.

Therefore, the area under the velocity-time graph represents the distance travelled by the car during the period.

Distance travelled by the car = Area of triangle AOB

(b) The slope of the velocity-time graph is the same throughout. So, the whole graph represents the uniform motion of the car.

#### Question 6

Fig below 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?

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

(c) How far has C travelled when B passes A?

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

**Answer**

(a) As the slope of line B is the greatest, hence, B is travelling at the fastest speed.

(b) No, as the three lines do not intersect at a single point.

(c) As there are 7 unit areas of the graph between 0 and 4 on the Y axis, 1 graph unit equals $\dfrac{4}{7}$ km.

As the initial point of object C is 4 graph units away from the origin, its initial distance from the origin is 4 x $\dfrac{4}{7}$ km = $\dfrac{16}{7}$ km

When B passes A, the distance between the origin and C is 8 km.

∴ total distance travelled by C in this time = 8 – $\dfrac{16}{7}$ km = 5.71 km

(d) Total distance travelled by B when it crosses C = 9 x $\dfrac{4}{7}$ = 5.14 km

#### Question 7

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

**Answer**

Given,

Initial velocity (u) = 0

Distance travelled (s) = 20 m

Acceleration (a) = 10 ms^{-2}

As per the third law of motion,

v^{2} – u^{2} = 2as

Substituting we get,

v^{2} - 0 = 2 x 10 x 20

⇒ v^{2} = 400

⇒ v = $\sqrt{400}$

⇒ v = 20 ms^{-1}

Hence, **ball strikes the ground with a velocity of 20 ms ^{-1}**

As per the first equation of motion,

v = u + at

or

t = $\dfrac{\text{v - u}}{\text{a}}$

Substituting we get,

t = $\dfrac{20-0}{10}$ = 2 seconds

∴ **the ball reaches the ground after 2 seconds.**

#### Question 8

The speed-time graph for a car is shown in Fig. below

(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?

**Answer**

(a) Distance travelled by the car in the first 4 seconds = area under the slope of the speed-time graph.

To find this area, we will use the square counting method.

On time axis (i.e., x axis) :

5 squares = 2 units

∴ 1 square = $\dfrac{2}{5}$ units

On speed axis (i.e., y axis) :

3 squares = 2 units

∴ 1 square = $\dfrac{2}{3}$ units

Area of each square = $\dfrac{2}{5}$ x $\dfrac{2}{3}$ = $\dfrac{4}{15}$ sq units.

We have fully filled squares, more than half filled squares, half filled squares and less than half filled squares in this area.

Area of fully filled square = Area of 1 square = $\dfrac{4}{15}$ sq units

Area of more than half filled square = Area of 1 square = $\dfrac{4}{15}$ sq units

Area of half filled square = $\dfrac{1}{2}$ x Area of 1 square = $\dfrac{1}{2} \times \dfrac{4}{15}$ = $\dfrac{2}{15}$ sq units

Area of less than half filled square = 0

No. of fully filled squares = 56 [Marked in graph with 'F']

Area of fully filled squares = 56 x $\dfrac{4}{15}$ = 14.93 sq units

No. of more than half filled squares = 4 [Marked in graph with 'M']

Area of more than half filled squares = 4 x $\dfrac{4}{15}$ = 1.06 sq units

No. of half filled squares = 3 [Marked in graph with 'H']

Area of half filled squares = 3 x $\dfrac{4}{15}$ = 0.8 sq units

No. of less than half filled squares = 4 [Marked in graph with 'L']

Area of less than half filled squares = 4 x 0 = 0 sq units

∴ Total Area = 14.93 + 1.06 + 0.4 = 16.39 sq units

∴ Total distance covered in first 4 seconds = Area under speed-time graph = 16.4 m

Hence, **the car travels 16.4 m in first 4 seconds**

(b) From graph, the speed of the car does not change from the points (x = 6) to (x = 10). Hence, the car is said to be in uniform motion from the 6th to the 10th second.

#### Question 9

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

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

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

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

**Answer**

(a) It is possible;

When an object is thrown up vertically into the air, it has constant acceleration due to gravity acting on it. However, when it reaches its maximum height, its velocity is zero.

(b) It is possible;

When an object moves on a circular path with uniform speed it is still under acceleration because the velocity changes due to continuous change in the direction of motion.

(c) It is possible;

When an object is accelerating in a circular trajectory, the acceleration is in the perpendicular direction.

#### Question 10

An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

**Answer**

Given,

Radius of the orbit (r) = 42250 km

∴ circumference of the orbit

= 2 x $\dfrac{22}{7}$ x r

= 2 x $\dfrac{22}{7}$ x 42250

= 265571.42 km

Time taken to complete one revolution in the orbit = 24 hours

Convert hours into sec

1 hour = 3600 sec

So, 24 hour = 24 x 3600 sec

Speed = $\dfrac{\text{Distance}}{\text{Time}}$

Substituting we get,

Speed = $\dfrac{265571.42}{24 \times 3600}$ = 3.07 kms^{-1}

∴ **Speed of the artificial satellite = 3.07 kms ^{-1}**.