# LAKHMIR SINGH AND MANJIT KAUR Solutions for Class 9 Physics Chapter 1 - Motion

## Chapter 1 - Motion Exercise 19

Is displacement a scalar quantity?

No, displacement is a vector quantity.

State whether distance is a scalar or a vector quantity?

Distance is a scalar quantity.

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

6 m/s

= 6 x (3600/1000) km/hr = 21.6km/hr

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

Speed of a body in a specified direction is called velocity.

Give two examples of bodies having non-uniform motion.

(a) Motion of a bus on a road

(b) Motion of a racing horse

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

Speed is defined as the distance travelled per unit time.

What do the following measure in a car?

(a) Speedometer (b) Odometer

(a) The speedometer of a car measures instantaneous speed of the car.

(b) Odometer is a device used to record the distance travelled by the car.

## Chapter 1 - Motion Exercise 20

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

In addition to speed, we should know the direction in which the body is moving.

When is a body said to have uniform velocity?

When a body covers equal distances in equal intervals of time in a particular direction however small or big the time interval may be, the object is said to have uniform velocity.

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

When the object moves in a single straight line, the magnitude of average velocity equal to average speed.

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

Average velocity of a moving body can be zero.

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

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

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

When a body is moving with uniform velocity, its acceleration is zero.

What is the other name of negative acceleration?

Negative acceleration is also called retardation.

Name the physical quantity whose S.I. unit is :

(a) m/s

(b) m/s^{2}

(a) Speed (or Velocity)

(b) Acceleration

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

Uniformly accelerated motion

What is the S.I. unit of retardation?

S.I. unit of retardation is m/s^{2}.

Fill in the following blanks with suitable words:

(a) Displacement is a _______ quantity whereas distance is a ________quantity.

(b) The physical quantity which gives both, the speed and direction of motion of a body is called its _________.

(c) A motorcycle has a steady ________ of 3m/s^{2}. This means that every _______ 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________.

(a) vector, scalar

(b) velocity

(c) acceleration, second, velocity, 3m/s

(d) displacement, m/s

(e) velocity, m/s^{2}

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

A freely falling body has non-uniform motion because it covers smaller distances in the initial '1 second' intervals and larger distances in the later '1 second' intervals, i.e., it covers unequal distances in equal intervals of time.

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

Speed is a scalar quantity as it has magnitude only, it has no specified direction.

Bus X travels a distance of 360km in 5 hours whereas bus Y travels a distance of 476 km in 7 hours. Which bus travels faster?

For bus X,

Speed= Distance/Time

Speed=360/5=72km/h

For bus Y,

Speed= Distance/Time

Speed=476/7=68 km/h

Speed of bus X is more than that of bus Y. Hence, bus X travels faster.

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.

Speed of athelete = 10 m/s

Speed of bicycle = 200 m/min = 200/60 m/s = 3.33 m/s

Speed of scooter = 30 km/h = 30000/3600 m/s = 8.33 m/s

3.33 m/s < 8.33 m/s <10 m/s

i.e. 200 m/min < 30 km/h < 10 m/s

(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 21m/s in one minute. Find its acceleration.

(a)Acceleration=_{}

_{}

(b) u = 0 m/s

v = 21 m/s

Time, t = 1 min = 60 sec

_{}

(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.

(a) Acceleration

(b) Retardation

(c) No, because if a body takes a round trip such that its final position is same as the starting position, then the displacement of the body is zero but the distance travelled is non-zero.

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

Average speed = Total distance travelled/ Total time taken

Total distance travelled = 100m = 0.1 km; Total time taken = 50 hr

Average speed= 0.1/50=0.002km/h

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

Speed gives an idea of how slow or fast a body is moving.

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

When the body comes back to its starting point, it has zero resultant displacement but covers a certain non-zero distance.

## Chapter 1 - Motion Exercise 21

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

Total distance=100m =0.1 km

Total time taken=15 minutes= 15/60=0.25 hour

Average speed = Total distance travelled/ Total time taken

=0.1/0.25= 0.4km/h

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

Total distance travelled=100m

Total time taken = 9.83 sec

Average speed = Total distance travelled/ Total time taken

=100/9.83 =10.172m/s

Averge speed in km/h:

10.172 x(3600/1000)=36.62 km/h

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

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

Initial velocity= 0m/s

Final velocity=6m/s

Time=3 sec

Acceleration=_{}

=_{}

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

Initial velocity, u = 600 km/h

Final velocity, v = 1100 km/h

Acceleration = 10 km/h/s = 600 km/h^{2}

From relation, a = (v-u)/t

t = (v-u)/a

t = (1100-600)/600 = 500/600 = 5/6 hr = 50 sec

If a bus travelling at 20m/s is subjected to a steady deceleration of 5m/s^{2}, how long will it take to come to rest?

Deceleration, a=-5m/s^{2}

Initial velocity, u=20m/s

Final velocity, v=0m/s

t=?

_{}

(a)What is the difference between 'distance travelled' by a body and its displacement? Explain with the help of a 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.

(a) Distance travelled is the actual length of the indirect path covered by the body whereas displacement refers to the straight line path between the initial and final positions. For e.g. In the figure given below, a body moves from point A to point B and then from point B to point C. Here, the distance travelled by the body is AB + BC and displacement is AC.

(b)

PQ=8cm

QR=6 cm

Resultant Displacement PR = _{}

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

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

A body has uniform motion if it travels equal distances in equal intervals of time, no matter how small these time intervals may be. For example: a car running at a constant speed of 10m/s, will cover equal distance of 10m every second, so its motion will be uniform.

Non-uniform motion: A body has a non-uniform motion if it travels unequal distances in equal intervals of time. For example: dropping a ball from the roof of a tall building .

(a) Define speed. What is the SI unit of speed?

(b) What is meant by (i) average speed, and (ii) uniform speed?

(a) Speed of a body is the distance travelled by it per unit time. The SI unit of speed is m/s.

(b) (i) Average speed of a body is the total distance travelled divided by the total time taken to cover this distance.

_{}

(ii) Uniform speed refers to the constant speed of a moving body. A body has a uniform speed if it travels equal distance in equal intervals of time, no matter how small these time intervals may be.

(a) Define velocity. What is SI unit of velocity?

(b) What is the difference between speed and velocity?

(c) Convert a speed of 54km/h into m/s.

(a) Velocity of a body is the distance travelled by it per unit time in a given direction. SI unit of velocity is m/s.

( b) (i)Speed is a scalar quantity whereas velocity is a vector quantity.

(ii) Speed of a body is distance travelled by it per unit time whereas velocity of a body is the distance travelled by it per unit time in a given direction.

(iii)Speed is always positive whereas velocity can be both positive as well as negative.

(c) Speed = 54km/h = 54 x (1000/3600) = 15m/s

(a) What is meant by the term acceleration? State the SI unit of acceleration.

(b) Define the term 'uniform acceleration'. Give one example of uniformly accelerated motion.

(a) Acceleration of a body is defined as the rate of change of its velocity with time. SI unit of acceleration is m/s^{2}.

(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: Motion of a freely falling body.

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

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

A car is moving along a straight road at a steady speed. It travels 150m 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 240m?

## Chapter 1 - Motion Exercise 22

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 ?

Distance travelled in half a rotation of a circular path is equal to the circumference of semi-circle, i.e., =R.

Displacement= diameter of circle= 2R

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

(i) What distance have you travelled?

(ii) What is your final displacement?

(i) Distance travelled = 6 km

(ii) Displacement = zero (since final position is same as initial position)

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?

(i) Total distance travelled= 3 + 4 +9=16 km

(ii) The body travels a total distance of 12 km in east direction i.e. towards x-axis.

And it travels a distance of 4 km in North direction, i.e. towards y-axis.

Hence, resultant displacement is

=_{}

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.

(a) Total distance covered in going to the bookshop and coming back to the classroom = 20 + 20 = 40m

Total time taken= 25 + 25 = 50 sec

_{ }

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.

In the first case, car travels at a speed of 60 km/h for a distance of 100 km

In the second case, car travels at a speed of 40 km/h for a distance of 100 km

_{ }

Total distance travelled = 200 km

A ball hits a wall horizontally at 6m/s. It rebounds horizontally at 4.4m/s. The ball is in contact with the wall for 0.040 s. What is the acceleration of the ball?

Initial velocity, u=6m/s

Final velocity ,v=-4.4m/s (the ball rebounds in opposite direction)

Time, t = 0.040 s

## Chapter 1 - Motion Exercise 39

(a) What remains constant in uniform circular motion?

(b) What changes continuously in uniform circular motion?

(a) Speed

(b) Direction of motion

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

The Speed of the body is constant or uniform.

State whether the following statement is true or false:

Earth moves round the sun with uniform velocity.

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

The motion is accelerated.

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

It represents uniform velocity.

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

Distance travelled by the moving body .

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

The slope of a speed-time graph indicates acceleration.

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

The slope of a distance-time graph indicates speed.

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

Motion of moon around the earth.

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

Uniform circular motion.

## Chapter 1 - Motion Exercise 40

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

The body has uniform speed.

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

The body is not moving. It is stationary.

What conclusion can you draw about the acceleration of a body from the following Speed - time graph shown below:

It represents non-uniform acceleration.

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

It is accelerated motion as the velocity is changing continuously.

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

The tip of the 'seconds' hand' of a watch represents uniform circular motion. It is an accelerated motion.

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 body.

(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 _______

(a) zero

(b) speed

(c) acceleration

(d) distance travelled

(e) circular path

## Chapter 1 - Motion Exercise 41

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

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

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.

The speed of a body moving along a circular path is given by the formula:

_{}

where, v= speed

_{ }=3.14 ( it is a constant)

t= time taken for one round of circular path

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

The motion of a body which is moving with constant speed in a circular path is said to be accelerated because its velocity changes continuously due to the continuous change in the direction of motion.

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

Uniform linear motion is uniform motion along a linear path or a straight line. The direction of motion is fixed. So, it is not accelerated. For e.g.: a car running with uniform speed of 10km/hr on a straight road.

Uniform circular motion is uniform motion along a circular path. The direction of motion changes continuously. So, it is accelerated. For e.g.: motion of earth around the sun.

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

An important characteristic of uniform circular motion is that the direction of motion in it changes continuously with time, so it is accelerated.

Centripetal force brings about uniform circular motion.

Find the initial velocity of a car which is stopped in 10 s by applying brakes. The retardation due to brakes is 2.5 m/s^{2}.

Initial velocity, u=?

Final velocity, v=0m/s (car is stopped)

Retardation, a=-2.5 m/s^{2}

Time, t=10s

v=u + at

0=u +(-2.5)x 10

u=25m/s

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

The velocity of this body is increasing at a rate of '10 metres per second' every second.

Initial velocity, u=0m/s

Time, t=2s

Acceleration,
a=10m/s^{2}

Using , _{}

_{}

A motorcycle moving with a speed of 5m/s is subjected to an acceleration of 0.2 m/s^{2}. Calculate the speed of the motorcycle after 10s, and the distance travelled in this time.

Initial velocity, u=5m/s

Final velocity, v=?

Acceleration, a=0.2m/s^{2}

Time, t=10 sec

Using , v=u + at

v=5 + 0.2 x 10

v=5 + 2 =7 m/s

Now distance travelled in time is calculated;

Using, _{}

_{}

A bus running at a speed of 18km/h is stopped in 2.5 seconds by applying brakes. Calculate the retardation produced.

Initial velocity, u=18km/h

Final velocity, v=0m/s

Time, t=2.5 sec

Acceleration, a=?

Using , v= u + at

_{}

So, retardation is 2m/s^{2}.

A train starting from rest moves with a uniform acceleration of 0.2m/s^{2} for 5 minutes. Calculate the speed acquired and the distance travelled in this time.

Initial velocity, u=0m/s

Final velocity, v=?

Acceleration, a=0.2 m/s^{2}

Time, t=5min= 5 x60=300 sec

Using, v = u + at

v =0 + 0.2 x 300=60m/s

And the distance travelled is

_{}

_{ }

Name the two quantities, the slope of whose graph gives:

(a) speed, and

(b) acceleration

(a) Distance and Time

(b) Speed (or velocity) and Time

A cheetah starts
from rest, and accelerates at 2m/s^{2} for 10s. Calculate:

(a) The final velocity

(b) The distance travelled.

Initial velocity, u=0m/s

Final velocity, v=?

Acceleration, a=2m/s^{2}

Time, t=10s

(a) Using, v = u + at

v=0 + 2 x 10= 20 m/s

(b) Distance travelled is:

_{}

A train travelling
at 20m/s accelerates at 0.5m/s^{2} for 30 s. How far will it travel
in this time?

Initial velocity, u=20m/s

Time, t=30 s

Acceleration,
a=0.5m/s^{2}

Distance travelled is:

_{}

A cyclist is travelling at 15m/s. She applies brakes so that she does not collide with a wall 18m away. What deceleration must she have?

Initial velocity, u=15m/s

Final velocity, v=0m/s

Distance, s=18m

Acceleration, a=?

_{}

So, deceleration is 6.25 m/s^{2}.

Draw a velocity-time graph to show the following motion:

A car accelerates uniformly from rest for 5 sec; then it travels at a steady velocity for 5s.

The velocity-time graph for part of a train journey is a horizontal straight line. What does this tell you about

(a) the train's velocity

(b) about its acceleration?

(a) The train has a uniform velocity.

(b) There is no acceleration.

(a) Explain the meaning of the following equation of motion:

v = u + at

where symbols have their usual meanings.

(b) A body starting from rest travels with uniform acceleration. If it travels 100m in 5s, what is the value of acceleration?

(a) v=u + at is the first equation of motion. It gives the velocity acquired by a body in time t when the body has initial velocity u and uniform acceleration a.

(b) Initial velocity, u=0m/s

Time, t=5 s

Distance, s=100m

Acceleration, a=?

_{}

(a) Derive the formula: v= u + at, where the symbols have usual meanings.

(b) A bus was moving with a speed of 54km/h. On applying brakes it stopped in 8 sec. calculate the acceleration.

(a) Consider a body having initial velocity 'u'. Suppose it is subjected to a uniform acceleration 'a' so that after time't' its final velocity becomes 'v'. Now, from the definition of acceleration we know that:

_{}

(b) Initial velocity, u=54km/h= 15m/s

Final velocity, v=0m/s

Time, t=8s

Acceleration, a=?

_{}

(a) Derive the formula: _{}, where the symbols have usual meanings.

(b) A train starting from stationary position and moving with uniform acceleration attains a speed of 36km per hour in 10 minutes. Find its acceleration.

(a) Suppose a body has an initial velocity 'u' and a uniform acceleration' a' for time 't' so that its final velocity becomes 'v'. Let the distance travelled by the body in this time be 's'. The distance travelled by a moving body in time 't' can be found out by considering its average velocity. Since the initial velocity of the body is 'u' and its final velocity is 'v', the average velocity is given by:

_{ }

(b) Initial velocity, u=0m/s

Final velocity, v=36km/h=10m/s

Time, t=10min =10 x 60=600 sec

Acceleration, a=?

_{}

(a) Write the three equations of uniformly accelerated motion. Give the meaning of each symbol which occurs in them.

(b) A car acquires a velocity of 72 km/h in 10 sec starting from rest. Find (i) the acceleration, (ii) average velocity, and (iii) the distance travelled in this time.

(a) What is meant by uniform circular motion? Give two examples of uniform circular motion.

(b) The tip of second's hand of a clock takes 60 sec to move once on the circular dial of the clock. If the radius of the dial of the clock be 10.5cm, calculate the speed of the tip of the seconds' hand of the clock. (Given _{}=22/7)

(a) When a body moves in a circular path with uniform speed (constant speed), its motion is called uniform circular motion. For e.g.

(i) Artificial satellites move in uniform circular motion around the earth.

(ii) Motion of a cyclist on a circular track.

(b) The speed of a body moving along a circular path is given by the formula:

_{}

Given, t=60 sec

Radius, r=10.5cm=0.105 m

_{}

## Chapter 1 - Motion Exercise 42

Show by means of graphical method that:

v= u + at

where the symbols have their usual meanings.

Consider the velocity-time graph of a body shown in figure.

The body has an initial velocity u at a point A and then its velocity changes at a uniform rate from A to B in time t. In other words, there is a uniform acceleration a from A to B, and after time t its final velocity becomes v which is equal to BC in the graph. The time t is represented by OC. To complete the figure, we draw the perpendicular CB from point C, and draw AD parallel to OC. BE is the perpendicular from point B to OE.

Now, Initial velocity of the body, u= OA -----(1)

And, Final velocity of the body, v=BC ------(2)

But from the graph BC =BD + DC

Therefore, v=BD + DC ------(3)

Again DC = OA

So, v =BD + OA

Now, from equation (1), OA =u

So, v=BD + u ------(4)

We should find out the value of BD now. We know the slope of a velocity-time graph is equal to the acceleration, a.

Thus, Acceleration, a= slope of line AB

_{or a = BD/AD}

But AD =OC= t, so putting t in place of AD in the above relation, we get:

_{}

or BD=at

Now, putting this value of BD in equation(4), we get:

v= u+ at

Show by using the graphical method that:

_{}

where the symbols have their usual meanings.

Consider the velocity-time graph of a body shown in figure. The body has an initial velocity u at a point A and then its velocity changes at a uniform rate from A to B in time t. In other words, there is a uniform acceleration a from A to B, and after time t its final velocity becomes v which is equal to BC in the graph. The time t is represented by OC.

Suppose the body travels a distance s in time t. In the figure, the distance travelled by the body is given by the area of the space between the velocity-time graph AB and the time axis OC, which is equal to the area of the figure OABC. Thus:

Distance travelled = Area of figure OABC

= Area of rectangle OADC + area of triangle ABD

Now, we will find out the area of rectangle OADC and area of triangle ABD.

(i) Area of rectangle OADC =OA x OC

= u x t

=ut

(ii) Area of triangle ABD= _{(1/2)}x Area of rectangle AEBD

=_{(1/2) }x AD x BD

=_{(1/2)} x t x at

=_{(1/2)}at^{2}

Distance travelled, s = Area of rectangle OADC + area of triangle ABD

_{}

Derive the following equation of motion by graphical method:

_{}

where the symbols have their usual meanings.

Consider the velocity-time graph of a body shown in figure. The body has an initial velocity u at a point A and then its velocity changes at a uniform rate from A to B in time t. In other words, there is a uniform acceleration a from A to B, and after time t its final velocity becomes v which is equal to BC in the graph. The time t is represented by OC. To complete the figure, we draw the perpendicular CB from point C, and draw AD parallel to OC. BE is the perpendicular from point B to OE.

The distance travelled s by a body in time t is given by the area of the figure OABC which is a trapezium.

Distance travelled, s= Area of trapezium OABC

_{}

Now, OA + CB = u + v and OC =t Putting these values in the above relation, we get:

_{}

Eliminate t from the above equation. This can be done by obtaining the value of t from the first equation of motion.

Thus, v = u + at (first equation of motion)

And, at = v

## Chapter 1 - Motion Exercise 43

The graph given alongside shows the positions of a body at different times. Calculate the speed of the body as it moves from:

(i) A to B,

(ii) B to C, and

(iii) C to D

(i) The distance covered from A to B is ( 3-0) =3 cm

Time taken to cover the distance from A to B =(5 -2) =3s

_{}

_{}

(ii) The speed of the body as it moves from B to C is zero.

(iii) The distance covered from C to D is (7-3)=4 cm

Time taken to cover the distance from C to D = (9-7)=2s

_{}

What can you say about the motion of a body if:

(a) its displacement-time graph is a straight line?

(b) its velocity-time graph is a straight line?

(a) The body has a uniform velocity if its displacement-time graph is a straight line.

(b) The body has a uniform acceleration if its velocity-time graph is a straight line.

A body with an initial velocity x moves with uniform acceleration y. Plot its velocity-time graph.