Lakhmir Singh Physics Solutions for Class 9 Science Chapter 1

Page No 19:

Question 1:

Is displacement a scalar quantity?

ANSWER:

No, displacement is a vector quantity because it has magnitude as well as direction.

Page No 19:

Question 2:

State whether distance is a scalar or a vector quantity.

ANSWER:

Distance is a scalar quantity; it has magnitude but no specific direction.

Page No 19:

Question 3:

Change the speed of 6 m/s into km/h.

ANSWER:

We have to convert 6 m/s into km/h. So,
$6 m / s = \frac{(6) (3600)}{1000} km / h = 21.6 km / hr$

Page No 19:

Question 4:

What name is given to the speed in a specified direction?

ANSWER:

Velocity is the name given to the speed of a body in a particular direction.

Page No 19:

Question 5:

Give two examples of bodies having non-uniform motion.

ANSWER:

Two examples of non-uniform motion:
(a) Motion of a bus on a curved road
(b) Motion of a bee flying randomly in air.

Page No 19:

Question 6:

Name the physical quantity obtained by dividing 'Distance travelled' by 'Time taken' to travel that distance.

ANSWER:

Speed is obtained by dividing 'distance travelled' by 'time taken' to travel that distance.

Page No 19:

Question 7:

What do the following measure in a car?

(a) Speedometer
(b) Odometer

ANSWER:

(a) Speedometer of a car is used to measure the instantaneous speed of the car.
(b) Odometer of a car is used to record and measure the overall distance travelled by the car.

Page No 20:

Question 8:

Name the physical quantity which gives us an idea of how slow or fast a body is moving.

ANSWER:

Speed is the physical quantity that gives the idea of how slow or fast a body is moving.

Page No 20:

Question 9:

Under what conditions can a body travel a certain distance and yet its resultant displacement be zero?

ANSWER:

When the body comes back to its starting point, its resultant displacement is zero. It is because it has covered a certain distance in due course of time; however, there is no difference between the initial and final positions.

Page No 20:

Question 10:

In addition to speed, what else should we know to predict the position of a moving body?

ANSWER:

Besides speed, we should know the direction to predict the position of a moving body.

Page No 20:

Question 11:

When is a body said to have uniform velocity?

ANSWER:

When a body covers equal distances in equal intervals of time in a particular direction, it is said to have uniform velocity.

Page No 20:

Question 12:

Under which condition is the magnitude of average velocity equal to average speed?

ANSWER:

The magnitude of average velocity is equal to average speed when an object covers equal distances in equal intervals of time in a particular direction.

Page No 20:

Question 13:

Which of the two can be zero under certain conditions : average speed of a moving body or average velocity of a moving body?

ANSWER:

Average velocity of a moving body can be zero. This is because the net displacement for a given time interval can be zero.

Page No 20:

Question 14:

Give one example of a situation in which a body has a certain average speed but its average velocity is zero.

ANSWER:

Motion of a boy from his home to shop and back to home is an example of a situation in which a body has a certain average speed but its average velocity is zero.

Page No 20:

Question 15:

What is the acceleration of a body moving with uniform velocity?

ANSWER:

The acceleration of a body moving with uniform velocity is zero.

Page No 20:

Question 16:

What is the other name of negative acceleration?

ANSWER:

Retardation is the other name for negative acceleration.

Page No 20:

Question 17:

Name the physical quantity whose SI unit is :

(a) m/s
(b) m/s2

ANSWER:

(a) Speed or velocity is expressed in m/s.
(b) Acceleration or retardation is expressed in m/s2.

Page No 20:

Question 18:

What type of motion is exhibited by a freely falling body?

ANSWER:

A freely falling body exhibits uniform accelerated motion.

Page No 20:

Question 19:

What is the SI unit of retardation?

ANSWER:

The S.I. unit of retardation is m/s2.

Page No 20:

Question 20:

(a) Displacement is a __________ quantity, whereas distance is a __________ quantity.
(b) The physical quantity that includes both the speed and direction of the motion of a body is called its __________.
(c) A motorcycle has a steady __________ of 3 m/s2. This means that each __________ its __________ increases by __________.
(d) Velocity is the rate of change of __________. It is measured in __________.
(e) Acceleration is the rate of change of __________. It is measured in __________.

ANSWER:

(a) Displacement is a vector quantity, whereas distance is a scalar quantity.
(b) The physical quantity that includes both the speed and direction of the motion of a body is called its velocity.
(c) A motorcycle has a steady acceleration of 3 m/s2. This means that each second its velocity increases by 3 m/s.
(d) Velocity is the rate of change of displacement. It is measured in m/s.
(e) Acceleration is the rate of change of velocity. It is measured in m/s2.

Page No 20:

Question 21:

What type of motion, uniform or non-uniform, is exhibited by a freely falling body? Give reason for you answer.

ANSWER:

A freely falling body exhibits non-uniform motion. Its velocity increases at a constant rate, so it shows uniformly accelerated motion.

Page No 20:

Question 22:

State whether speed is a scalar or a vector quantity. Give reason for your choice.

ANSWER:

Speed is a scalar quantity. This is because it has magnitude, but it does not specify direction. It is the distance travelled by a body per unit time.

Page No 20:

Question 23:

Bus X travels a distance of 360 km in 5 hours whereas bus Y travels a distance of 476 km in hours. Which bus travels faster?

ANSWER:

We should first consider Bus X.
Distance travelled (d1) = 360 km
Time taken (d1) = 5 hr
So, we can calculate the speed as:
$Speed = \frac{Distance}{Time}$
Thus, the speed of Bus X is 72 km/h.
Similarly,
For Bus Y:
Distance travelled
Time taken (t2) = 7 hr
So, we can calculate the speed as:
$Speed = \frac{Distance}{Time}$
Thus, the speed of Bus Y is 68 km/h.
Speed of Bus X is more than that of Bus Y. Hence, Bus X travels faster.

Page No 20:

Question 24:

Arrange the following speeds in increasing order (keeping the least speed first):

(i) An athlete running with a speed of 10 m/s.
(ii) A bicycle moving with a speed of 200 m/min.
(iii) A scooter moving with a speed of 30 km/h.

ANSWER:

We have three different moving bodies. To compare their speeds, we need to express them in similar S.I. units.
(i) An athlete running with a speed of 10 m/s.
(ii) A bicycle moving with a speed of 200 m/min. Converting into S.I. units, the speed will be:
$= \frac{200}{60} = 3.33 m / s$
(iii) A scooter moving with a speed of 30 km/h. Converting into S.I. units, the speed will be:
$= \frac{(30) (1000)}{3600} = 8.33 m / s$
On arranging the speeds in ascending order, we get:
(ii) < (iii) < (i)

Page No 20:

Question 25:

(a) Write the formula for acceleration. Give the meaning of each symbol which occurs in it.
(b) A train starting from Railway Station attains a speed of 21 m/s in one minute. Find its acceleration.

ANSWER:

(a) Acceleration is the change in velocity per unit time; it is a vector quantity.
$Acceleration = \frac{Change in velocity}{Time taken}$
The above expression can also be written as:
$a = \frac{v - u}{t}$
Where,
a = Acceleration
v = Final velocity
u = Initial velocity
t = Time taken

(b) We will find the value of uniform acceleration.
Initial velocity (u) 0 m/s
Final velocity (v) = 21 m/s
Time taken (t) = 60 s
Acceleration:
$a = \frac{v - u}{t}$
Putting the values in the above equation, we get:
$a = \frac{21 - 0}{60} = \begin{matrix} 0.35 m / s^{2} \end{matrix}$

Page No 20:

Question 26:

(a) What term is used to denote the change of velocity with time?
(b) Give one word which means the same as 'moving with a negative acceleration'.
(c) The displacement of a moving object in a given interval of time is zero. Would the distance travelled by the object also be zero? Give reason for your answer.

ANSWER:

(a) Acceleration is used to denote the change in velocity with time.
(b) Retardation is the same as ‘moving with a negative acceleration'.
(c) Displacement is a vector quantity, so it can be zero for two reasons:
1. When the body doesn't move at all: In this case, the distance travelled will also be zero.
2. When the body comes back to its initial position: In this case, the distance travelled is non-zero, but the displacement is zero.

Page No 20:

Question 27:

A snail covers a distance of 100 metres in 50 hours. Calculate the average speed of snail in km/h.

Figure

ANSWER:

Distance (d) = 1000 m = 0.1 km
Time (t) = 50 hr
So, we can calculate the speed as:
$Average speed = \frac{Total distance travelled}{Time}$
Average speed for the entire journey:
$= \frac{0.1}{50} = 0.002 km / hr$

Page No 21:

Question 28:

A tortoise moves a distance of 100 metres in 15 minutes. What is the average speed of tortoise in km/h?

ANSWER:

Distance (d) = 0.1 km
$Time (t) = \frac{15}{60} hr = 0.25 hr$
So, we can calculate the speed as:
$Average speed = \frac{Total distance travelled}{Time}$
Average speed for the entire journey:
$= \frac{0.1}{0.25} = 0.4 km / hr$

Page No 21:

Question 29:

If a sprinter runs a distance of 100 metres in 9.83 seconds, calculate his average speed in km/h.

ANSWER:

Distance (d) = 100 m
Time (t) = 9.83 s
So, we can calculate the speed as:
$Average speed = \frac{Total distance travelled}{Time}$
Average speed for the entire journey:
$= \frac{100}{9.83} = 10.173 m / s$
Now, we can convert it in km/h as:
$= \frac{(10.173) (3600)}{1000} = 36.62 km / hr$

Page No 21:

Question 30:

A motorcyclist drives from place A to B with a uniform speed of 30 km h−1 and returns from place B to A with a uniform speed of 20 km h−1. Find his average speed.

ANSWER:

We have to find the average velocity of the entire journey. For this, we have the following information:
Speed from A to B = (v1) = 30 m/s
Let the distance from A to B be (d).
Also, let the time taken to travel from A to B be (t1).
$Time = \frac{Distance travelled}{Speed}$
We have:
$t_{1} = \frac{d}{30}$
Speed from B to A (v2) = 20 m/s
Let the time taken to travel from B to A be(t2).
Thus, we have:
$t_{2} = \frac{d}{20} Total time of journey : = t_{1} + t_{2} = \frac{d}{30} + \frac{d}{20} = \frac{d}{12}$
Total distance travelled is 2d.
Therefore,
$Average speed = \frac{Total distance travelled}{Time}$
On putting the values to obtain the average speed of the motorcyclist, we get:
$= \frac{(2 d) 12}{d} = 24 km / hr$

Page No 21:

Question 31:

A motorcyclist starts from rest and reaches a speed of 6 m/s after travelling with uniform acceleration for 3 s. What is his acceleration?

ANSWER:

We have to find the value of uniform acceleration.
Initial velocity (u) = 0 m/s
Final velocity (v) = 6 m/s
Time taken (t) = 3s
Acceleration:
$a = \frac{v - u}{t}$
Put the values in the above equation to obtain the value of acceleration.
$a = \frac{6 - 0}{3} = 2 m / s^{2}$

Page No 21:

Question 32:

An aircraft travelling at 600 km/h accelerates steadily at 10 km/h per second. Taking the speed of sound as 1100 km/h at the aircraft's altitude, how long will it take to reach the 'sound barrier'?

ANSWER:

Final velocity, v = 1100 km/h = 1100 × 5/18 = 305.55 m/s
Initial velocity, u = 600 km/h = 600 × 5/18 = 166.66 m/s
Acceleration = 10 km/h per second = 10 × 5/18 = 2.77 m/s2
Time taken by body,
v = u + at
t = (v - u)/a
t = (305.55 -166.66 )/2.77
t = 50.14 sec
t ≃ 50 sec

Page No 21:

Question 33:

If a bus travelling at 20 m/s is subjected to a steady deceleration of 5 m/s2, how long will it take to come to rest?

ANSWER:

We have to find the time taken to reach the given final velocity.
We have:
Initial velocity (u) = 20 m/s
Final velocity (v) = 0 m/s
Acceleration for the entire journey (a) = –5 ms2.
Let the time taken be (t).
We can calculate the time taken using the first equation of motion.
$t = \frac{v - u}{a}$
Time taken by the bus to come to rest:
$= \frac{0 - 20}{- 5} = 4 s$

Page No 21:

Question 34:

(a) What is the difference between 'distance travelled' by a body and its 'displacement'? Explain with the help of diagram.
(b) An ant travels a distance of 8 cm from P to Q and then moves a distance of 6 cm at right angles to PQ. Find its resultant displacement.

ANSWER:

(a)

Distance		Displacement
1.	Distance has only magnitude, with no specified direction.	1.	Displacement has magnitude as well as direction.
2.	It is a scalar quantity.	2.	It is a vector quantity.
3.	Two different distances can be added directly.	3.	We have to follow the vector addition method to add displacements.

One difference in diagrammatic form is as follows:

Here the curved line is the distance traveled and the straight line is the magnitude of the displacement.

(b) We have to find the resultant displacement from the given diagram:

We have:
PQ = 8 cm and QR = 6 cm
Resultant displacement:

P R = \sqrt{P Q^{2} + Q R^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10 cm

The direction of this displacement is from P to R. If θ is the angle made by PR with PQ then,'

This is the angle made by the resultant with PQ.

Page No 21:

Question 35:

Define motion. What do you understand by the terms 'uniform motion' and 'non-uniform motion'? Explain with examples.

ANSWER:

A body is said to be in motion when its position changes continuously with respect to a stationary point taken as the reference point.

Uniform motion: A body is said to be in uniform motion if it travels equal distances in equal intervals of time in a particular direction, no matter how small these time intervals are.
For example, a car running at a constant speed of 10 m/s towards east will cover the equal distance of 10 m every second towards east, so its motion will be uniform.

Non-uniform motion: A body is said to be in non-uniform motion if it travels unequal distances in equal intervals of time.
For example, motion of a freely falling ball from the roof of a tall building.

Page No 21:

Question 36:

(a) Define speed. What is the SI unit of speed?
(b) What is meant by (i) average speed, and (ii) uniform speed?

ANSWER:

(a) Speed of a body is the distance travelled by it per unit time. The S.I. unit of speed is m/s.
(b) (i) Average speed of a body is the total distance travelled by it divided by the total time taken by it to cover the given distance.
$Average speed = \frac{Total distance travelled}{Total time taken}$
(ii) A body has a uniform speed if it travels equal distance in equal intervals of time, no matter how small these time intervals are.

Page No 21:

Question 37:

(a) Define velocity. What is the SI unit of velocity?
(b) What is the difference between speed and velocity?
(c) Convert a speed of 54 km/h into m/s.

ANSWER:

(a) Velocity of a body is the distance travelled by it per unit time in a given direction. The S.I. unit of velocity is m/s. It is a vector quantity.

(b) (i) Speed is a scalar quantity, whereas velocity is a vector quantity.
      (ii) Speed of a body is the distance travelled by it per unit time, whereas the velocity of a body is the distance travelled by it per unit time in a given direction.
      (iii) Speed is always positive, while velocity can be both positive and negative, depending upon the direction.

(c) We have to convert 54 km/h into m/s.
       $54 km / hr = \frac{(54) (1000)}{3600} m / s = 15 m / s$

Page No 21:

Question 38:

(a) What is meant by the term 'acceleration'? State the SI unit of acceleration.
(b) Define the term 'uniform acceleration'. Give one example of a uniformly accelerated motion.

ANSWER:

(a) Acceleration of a body is defined as the rate of change of its velocity with respect to time. It is a vector quantity. The S.I. unit of acceleration is (m/s2) .
(b) A body has uniform acceleration if it travels in a straight line and its velocity increases by equal amounts in equal intervals of time.
For example, a freely falling body has uniform acceleration.

Page No 21:

Question 39:

The distance between Delhi and Agra is 200 km. A train travels the first 100 km at a speed of 50 km/h. How fast must the train travel the next 100 km, so as to average 70 km/h for the whole journey?

ANSWER:

We have the following data to find the speed for the second part of the journey:
Total distance to be travelled by train (D) = 200 km
Average speed required (vavg) = 70 km/hr
Time required for the entire journey (T):
$Time = \frac{Total distance}{Average speed} = \frac{200}{70} hr T = \frac{20}{7} hr$
For the first part of the trip:
Distance covered (d1) = 100 km
Speed for this part of journey (v1) = 50 km/hr
Time taken for the first part of journey:
$Time = \frac{Distance travelled}{Speed} So, t_{1} = \frac{100}{50} hr = 2 hr$
For the second part of the trip,
Distance covered (d2) = 100 km
Time taken for the second part of journey:
$t_{2} = T - t_{1} = (\frac{20}{7} - 2) hr = \frac{6}{7} hr$
Speed of the train for the second part of the journey:
$= \frac{distance travelled}{Time} = \frac{(100) (7)}{6} = 116.67 km / hr$

Page No 21:

Question 40:

A train travels the first 15 km at a uniform speed of 30 km/h; the next 75 km at a uniform speed of 50 km/h; and the last 10 km at a uniform speed of 20 km/h. Calculate the average speed for the entire train journey.

ANSWER:

(i) In the first case, the train travels at a speed of 30 km/h for a distance of 15 km.
We can find the time as:
$Time = \frac{Distance travelled}{Speed} So, t_{1} = \frac{15}{30} hr = 0.5 hr$

(ii) In the second case, the train travels at a speed of 50 km/h for a distance of 75 km.
We can find the time as:
$Time = \frac{Distance travelled}{Speed} So, t_{2} = \frac{75}{50} hr = 1.5 hr$

(iii) In the third case, the train travels at a speed of 20 km/h for a distance of 10 km.
We can find the time as:
$Time = \frac{Distance travelled}{Speed} So, t 3 = \frac{10}{20} hr = 0.5 hr$

Total distance covered:
= (15 + 75 + 10) km
= 100 km
Total time taken = (0.5 + 1.5 + 0.5) km
= 2.5
Therefore, $Average speed = \frac{Total distance travelled}{Time}$
Now, put the values to get the average speed.
$= \frac{100}{2.5} = 40 km / hr$

Page No 21:

Question 41:

A car is moving along a straight road at a steady speed. It travels 150 m in 5 seconds:

(a) What is its average speed?
(b) How far does it travel in 1 second?
(c) How far does it travel in 6 seconds?
(d) How long does it take to travel 240 m?

ANSWER:

(a) We have:
Distance (d) = 150 m
Time (t) = 5s
So, we can calculate average the speed as:
$Average speed = \frac{Total distance travelled}{Time}$
Average speed for the entire journey:
$= \frac{150}{5} = 30 m / s$

(b) We have to calculate the distance travelled in 1s.
Distance = (Speed) (Time)
Distance travelled in one second:
=(30) (1) m
$= 30 m$

(c) We have to calculate the distance travelled in 6 s.
Distance = (Speed) (Time)
Distance travelled in one second:
= (30) (6) m
$= 180 m$

(d) We have:
Distance (d) = 240 m
Speed (v) = 30 m/s
We can find the time as:
$Time = \frac{Distance travelled}{Speed} S o, t = \frac{240}{30} s = 8 s$

Page No 21:

Question 42:

A particle is moving in a circular path of radius r. The displacement after half a circle would be:

(a) 0
(b) πr
(c) 2r
(d) 2πr

ANSWER:

(c) Displacement is the difference between the final and initial position of a body. It is a vector quantity and is independent of the path taken. So, for the movement of half of a circle, the displacement is 2r, where r is the radius of the circular path.

Page No 21:

Question 43:

The numerical ratio of displacement to distance for a moving object is :

(a) always less than 1
(b) equal to 1 or more than 1
(c) always more than 1
(d) equal to 1 or less than 1

ANSWER:

(d), i.e., Equal to 1 or less than 1.
Displacement is always smaller than or equal to displacement.

Page No 21:

Question 44:

A boy is sitting on a merry-go-round which is moving with a constant speed of 10 m s−1. This means that the boy is :

(a) at rest
(b) moving with no acceleration
(c) in accelerated motion
(d) moving with uniform velocity

ANSWER:

(c) Acceleration is the rate of change of velocity, and the velocity of the merry-go-round is changing with respect to time. Thus, it will move in an accelerated motion.

Page No 21:

Question 45:

In which of the following cases of motion, the distance moved and the magnitude of displacement are equal?

(a) if the car is moving on straight road
(b) if the car is moving on circular road
(c) if the pendulum is moving to and fro
(d) if a planet is moving around the sun

ANSWER:

(a) The magnitude of displacement is equal to the distance travelled by a body when it travels in a straight line.

Page No 21:

Question 46:

The speed of a moving object is determined to be 0.06 m/s. This speed is equal to :

(a) 2.16 km/h
(b) 1.08 km/h
(c) 0.216 km/h
(d) 0.0216 km/h

ANSWER:

Page No 22:

Question 47:

A freely falling object travels 4.9 m in 1st second, 14.7 m in 2 nd second, 24.5 m in 3rd second, and so on. This data shows that the motion of a freely falling object is a case of :

(a) uniform motion
(b) uniform acceleration
(c) no acceleration
(d) uniform velocity

ANSWER:

(b) The displacement of the body in equal interval of time is unequal, but acceleration is constant. The acceleration will thus be uniform, so the answer is (b).

Page No 22:

Question 48:

When a car runs on a circular track with a uniform speed, its velocity is said to be changing. This is because :

(a) the car has a uniform acceleration
(b) the direction of car varies continuously
(c) the car travels unequal time intervals.
(d) the car travels equal distances in unequal time intervals

ANSWER:

(d) When a car runs on a circular track, its velocity changes continuously, as its direction keeps changing.

Page No 22:

Question 49:

Which of the following statement is correct regarding velocity and speed of a moving body?

(a) velocity of a moving body is always higher than its speed
(b) speed of a moving body is always higher than its velocity
(c) speed of a moving body is its velocity in a given direction
(d) velocity of a moving body is its speed in a given direction

ANSWER:

(d) Velocity is a vector quantity having a magnitude and a specific direction. So, velocity is nothing but speed in a particular direction.

Page No 22:

Question 50:

Which of the following can sometimes be 'zero' for a moving body?

(i) average velocity
(a) only (i)

(ii) distance travelled
(b) (i) and (ii)

(iii) average speed
(c) (i) and (iv)

(iv) displacement
(d) only (iv)

ANSWER:

(c) Distance is the length of the actual path covered by a moving body. This implies that the distance travelled by a moving body and also its average speed can never be zero at any point of time. However, it is possible for the displacement as well as average velocity for the body to be zero at an instance during the motion of the particle.

Page No 22:

Question 51:

When a car driver travelling at a speed of 10 m/s applies brakes and brings the car to rest in 20 s, then retardation will be:

(a) + 2 m/s2
(b) − 2 m/s2
(c) − 0.5 m/s2
(d) + 0.5 m/s2

ANSWER:

(d) The term “retardation” means negative acceleration.
Initial velocity = 10 m/s
Final velocity = 0 m/s
Time taken = 20 s
$Acceleration = \frac{0 - 10}{20} \Rightarrow Acceleration = - 0.5 m / s^{2} \Rightarrow Retardation = 0.5 m / s^{2}$

Page No 22:

Question 52:

Which of the following could not be a unit of speed?

(a) km/h
(b) s/m
(c) m/s
(d) mm s−1

ANSWER:

(b) Speed is the distance travelled by a moving body per unit time.

Page No 22:

Question 53:

One of the following is not a vector quantity. This one is :

(a) displacement
(b) speed
(c) acceleration
(d) velocity

ANSWER:

(b) Vector quantities have magnitude as well as direction, and they obey the laws of vector addition.

Page No 22:

Question 54:

Which of the following could not be a unit of acceleration?

(a) km/s2
(b) cm s−2
(c) km/s
(d) m/s2

ANSWER:

Page No 22:

Question 55:

A body is moving along a circular path of radius R. What will be the distance travelled and displacement of the body when it completes half a revolution?

ANSWER:

We have to analyse the distance and displacement of a body that has covered half the perimeter of a circle.
Distance travelled in half a rotation of a circular path is equal to the circumference of semi-circle.
Distance travelled = πR
Displacement is calculated from the initial and final positions of a body. It is independent of the path covered. So, displacement is the diameter of the semi-circle.
Hence, displacement is 2R, where R is the radius of the circular path.

Page No 22:

Question 56:

If on a round trip you travel 6 km and then arrive back home :

(a) What distance have you travelled?
(b) What is your final displacement?

ANSWER:

(a) Distance travelled is the actual path covered.
Total distance covered in this case due to going and coming back:
2d = 6 km
d is the distance of one-side journey.

(b) Displacement is calculated from the initial and final positions of a body. It is independent of the path covered. So, displacement in this case is 0 because the initial and final positions are the same.

Page No 22:

Question 57:

A body travels a distance of 3 km towards East, then 4 km towards North and finally 9 km towards East.
(i) What is the total distance travelled?
(ii) What is the resultant displacement?

ANSWER:

1-22-57
(i) Distance travelled is the actual path covered.
Total distance travelled:
= (3 + 4 + 9) km
= 16 km

(ii) The body travels a total distance of 12 km east, which means towards the x-axis, on a Cartesian plane. It travels a distance of 4 km in north direction, which means towards y-axis.
Resultant displacement:
$\sqrt{12^{2} + 4^{2}} = \sqrt{144 + 16} = \sqrt{160} = 12.65 km$

Page No 22:

Question 58:

A boy walks from his classroom to the bookshop along a straight corridor towards North. He covers a distance of 20 m in 25 seconds to reach the bookshop. After buying a book, he travels the same distance in the same time to reach back in the classroom. Find (a) average speed, and (b) average velocity, of the boy.

ANSWER:

(a) Distance travelled is the length of the actual path covered.
Total distance covered in going and coming back:
= (20 + 20) m
= 40 m
Total time taken, = (25 + 25) s
= 50 s
So, we can calculate average speed as:
$Average speed = \frac{Total distance travelled}{Time}$
Average speed for the entire journey:
$= \frac{40}{50} = 0.8 m / s$

(b) Average velocity is zero, as displacement is zero because the boy arrives at the initial point.

Page No 22:

Question 59:

A car travels 100 km at a speed of 60 km/h and returns with a speed of 40 km/h. Calculate the average speed for the whole journey.

ANSWER:

In the first case, the car travels at a speed of 60 km/h for a distance of 100 km.
Thus,
$Time = \frac{Distance travelled}{Speed} So, t_{1} = \frac{100}{60} hr = \frac{5}{3} hr$

In the second case, the car travels at a speed of 40 km/h for a distance of 100 km.
Thus,
$Time = \frac{Distance travelled}{Speed} So, t_{2} = \frac{100}{40} hr = \frac{5}{2} hr$

Total time taken:
$= t_{1} + t_{2} = \frac{5}{3} + \frac{5}{2} hr = \frac{25}{6} hr$

Total distance travelled = 200 km
We can calculate average speed as:
$Average speed = \frac{Total distance travelled}{Time}$
Average speed for the entire journey:
$= \frac{(200) (6)}{25} = 48 km / hr$

Page No 22:

Question 60:

A ball hits a wall horizontally at 6.0 m s−1. It rebounds horizontally at 4.4 m s−1. The ball is in contact with the wall for 0.040 s. What is the acceleration of the ball?

ANSWER:

We have to find the value of uniform acceleration.
We have:
Initial velocity (u) = 6 m/s
Final velocity is in opposite direction to that of initial velocity (v) = –4.4 m/s
Time taken (t) = 0.04 s
Acceleration:
$a = \frac{v - u}{t}$
Put the values in the above equation to get the value of acceleration.
$a = \frac{- 4.4 - 6}{0.04} = \frac{10.4}{0.04} = - 260 m / s^{2}$

Page No 39:

Question 1:

(a) What remains constant in uniform circular motion?
(b) What changes continuously in uniform circular motion?

ANSWER:

(a) Speed remains constant in uniform circular motion.
(b) Direction of motion changes continuously in uniform circular motion.

Page No 39:

Question 2:

State whether the following statement is true or false :
Earth moves round the sun with uniform velocity.

ANSWER:

The statement “Earth moves around the sun in uniform velocity” is false as the direction of motion while travelling in a circular path is always changing and velocity being a vector quantity also changes.

Page No 39:

Question 3:

A body goes round the sun with constant speed in a circular orbit. Is the motion uniform or accelerated?

ANSWER:

The motion is accelerated one. Here the direction of motion of the body changes every instant.

Page No 39:

Question 4:

What conclusion can you draw about the velocity of a body from the displacement-time graph shown below:

Figure

ANSWER:

It represents uniform velocity. In this type of motion, body covers equal distance in equal interval of time in a specific direction.

Page No 39:

Question 5:

Name the quantity which is measured by the area occupied under the velocity-time graph.

ANSWER:

Displacement is the quantity which is measured by the area occupied under the velocity-time graph.

Page No 39:

Question 6:

What does the slope of a speed-time graph indicate?

ANSWER:

The slope of a speed-time graph indicates acceleration.

Page No 39:

Question 7:

What does the slope of a distance-time graph indicate?

ANSWER:

The slope of a distance-time graph indicates speed.

Page No 39:

Question 8:

Give one example of a motion where an object does not change its speed but its direction of motion changes continuously.

ANSWER:

The motion of an artificial satellite around the earth is an example of a motion (uniform circular motion) where an object does not change its speed but its direction of motion changes continuously.

Page No 39:

Question 9:

Name the type of motion in which a body has a constant speed but not constant velocity.

ANSWER:

Uniform circular motion is the type of motion in which a body has a constant speed but not constant velocity.

Page No 39:

Question 10:

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

ANSWER:

If the speed-time graph is a straight line parallel to the time axis then the body is moving with a constant speed.

Page No 40:

Question 11:

What conclusion can you draw about the speed of a body from the following distance-time graph?

Figure

ANSWER:

The body has a uniform speed because the body covers equal distance in equal interval of time.

Page No 40:

Question 12:

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

ANSWER:

The body is not moving because it is not changing its position with respect to any stationary object.

Page No 40:

Question 13:

What conclusion can you draw about the acceleration of a body from the speed-time graph shown below?

Figure

ANSWER:

It represents non-uniform acceleration. In this kind of motion, speed doesn’t vary linearly with time.

Page No 40:

Question 14:

A satellite goes round the earth in a circular orbit with constant speed. Is the motion uniform or accelerated?

Figure

ANSWER:

It is an example of accelerated motion, as direction of motion is changing continuously.

Page No 40:

Question 15:

What type of motion is represented by the tip of the 'seconds' hand' of a watch? Is it uniform or accelerated?

ANSWER:

'Seconds' hand 'of a watch represents uniform circular motion because the speed of the second-hand is always constant and its tip covers a circular path.

Page No 40:

Question 16:

Fill in the following blanks with suitable words :
(a) If a body moves with uniform velocity, its acceleration is _____.
(b) The slope of a distance-time graph indicates _ of a moving object.
(c) The slope of a speed-time graph of a moving body gives its _.
(d) In a speed-time graph, the area enclosed by the speed-time curve and the time axis gives the _ by the body.
(e) It is possible for something to accelerate but not change its speed if it moves in a _______.

ANSWER:

(a) If a body moves with uniform velocity, its acceleration is zero.
(b) The slope of a distance-time graph indicates speed of a moving object.
(c) The slope of a speed-time graph of a moving body gives its acceleration.
(d) In a speed-time graph, the area enclosed by the speed-time curve and the time axis gives the distance traveled by the body.
(e) It is possible for something to accelerate but not change its speed if it moves in a circular path.

Page No 41:

Question 17:

Is the uniform circular motion accelerated? Give reasons for your answer.

ANSWER:

Yes, the uniform circular motion is accelerated because the velocity changes due to continuous change in the direction.

Page No 41:

Question 18:

Write the formula to calculate the speed of a body moving along a circular path. Give the meaning of each symbol which occurs in it.

ANSWER:

The speed of a body moving along a circular path is given by the formula:
$v = \frac{2 π r}{T}$
Where,
(v) - Speed of the object in circular path.
(π) - It is a constant having value close to 3.14
(r) - Radius of circular path
(T) - Time taken for one round of circular path.

Page No 41:

Question 19:

Explain why, the motion of a body which is moving with constant speed in a circular path is said to be accelerated.

ANSWER:

The motion of a body is said to be an accelerated one if the velocity of the body changes with time. In uniform circular motion the magnitude of velocity remains constant but the direction of motion changes at every instant. So, the motion is an accelerated one.

Page No 41:

Question 20:

What is the difference between uniform linear motion and uniform circular motion? Explain with examples.

ANSWER:

In uniform linear motion, the speed and direction of motion is fixed and so, it is not accelerated. Example- A car running on a straight road.
In uniform circular motion, the speed is constant but the direction of motion changes continuously and hence, it is accelerated.
Example- Motion of earth around the sun.

Page No 41:

Question 21:

State an important characteristics of uniform circular motion. Name the force which brings about uniform circular motion.

ANSWER:

An important characteristic of uniform circular motion is that the speed remains constant while the direction of motion changes continuously with time, so it is accelerated.
Centripetal force brings about uniform circular motion. This force is always directed towards the center of the circular path and it is along the radius of the circular path.

Page No 41:

Question 22:

Find the initial velocity of a car which is stopped in 10 seconds by applying brakes. The retardation due to brakes is 2.5 m/s2.

ANSWER:

We have to find the initial velocity.
Final velocity as the car stops after some time, (v) = 0 m/s
Acceleration for the entire journey, (a) = –2.5 m/s2
Time taken is (t) = 10 s
Let the initial velocity be (u)
We can calculate initial velocity using 1st equation of motion as, u = v – at
So initial velocity,
u = 0 – (–2.5)(10)
⇒ u = 25 m/s

Page No 41:

Question 23:

Describe the motion of a body which is accelerating at a constant rate of 10 m s−2. If the body starts from rest, how much distance will it cover in 2 s?

ANSWER:

We have to find the distance travelled by the body. We have the following information given,
So, Initial velocity, (u) = 0 m/s
Acceleration, (a) = 10 m/s2
Time taken, (t) = 2 s
We can calculate the distance travelled by using the 2nd equation of motion,
$s = u t + \frac{1}{2} a t^{2}$
Put the values in above equation to find the distance travelled by the body,
$(s) = 0 (2) + \frac{1}{2} (10) (2)^{2} = 20 m$

Page No 41:

Question 24:

A motorcycle moving with a speed of 5 m/s is subjected to an acceleration of 0.2 m/s2. Calculate the speed of the motorcycle after 10 seconds, an the distance travelled in this time.

ANSWER:

We have to find the distance travelled and final velocity of the body. We have the following information given,
Initial velocity, (u) = 5 m/s
Acceleration, (a) = 0.2 m/s2
Time taken, (t) = 10 s
So, we can find the final velocity using the relation,
v = u + at
So, final velocity,
v = 5 + (0.2)(10)
= 7 m/s
We can calculate the distance travelled by using the 2nd equation of motion,
$s = u t + \frac{1}{2} a t^{2}$
Put the values in above equation to find the distance travelled by the motorcycle,
$(s) = 5 (10) + \frac{1}{2} (0.2) (10)^{2} = (50 + 10) m = 60 m$

eSchool

Lakhmir Singh Physics Solutions for Class 9 Science Chapter 1 - Motion

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