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`color{red} ♦` INTRODUCTION
`color{red} ♦` DESCRIBING MOTION


In everyday life, we see some objects at rest and others in motion.

Birds fly, fish swim, blood flows through veins and arteries and cars move. Atoms, molecules, planets, stars and galaxies are all in motion.

We often perceive an object to be in motion when its position changes with time. However, there are situations where the motion is inferred through indirect evidences.

For example, we infer the motion of air by observing the movement of dust and the movement of leaves and branches of trees.

What causes the phenomena of sunrise, sunset and changing of seasons? Is it due to the motion of the earth? If it is true, why don’t we directly perceive the motion of the earth?

An object may appear to be moving for one person and stationary for some other. For the passengers in a moving bus, the roadside trees appear to be moving backwards.

A person standing on the road-side perceives the bus along with the passengers as moving. However, a passenger inside the bus sees his fellow passengers to be at rest. What do these observations indicate?

Most motions are complex. Some objects may move in a straight line, others may take a circular path. Some may rotate and a few others may vibrate. There may be situations involving a combination of these.

In this chapter, we shall first learn to describe the motion of objects along a straight line. We shall also learn to express such motions through simple equations and graphs. Later, we shall discuss ways of describing circular motion.


We describe the location of an object by specifying a reference point.

Let us assume that a school in a village is `2` km north of the railway station. We have specified the position of the school with respect to the railway station.

In this example, the railway station is the reference point. We could have also chosen other reference points according to our convenience. Therefore, to describe the position of an object we need to specify a reference point called the origin.


The simplest type of motion is the motion along a straight line. We shall first learn to describe this by an example.

Consider the motion of an object moving along a straight path. The object starts its journey from O which is treated as its reference point (Fig. 8.1).

Let A, B and C represent the position of the object at different instants. At first, the object moves through C and B and reaches A. Then it moves back along the same path and reaches C through B.

The total path length covered by the object is `OA + AC`, that is ` 60 km + 35 km = 95 km`. This is the distance covered by the object.

To describe distance we need to specify only the numerical value and not the direction of motion. There are certain quantities which are described by specifying only their numerical values.

The numerical value of a physical quantity is its magnitude. From this example, can you find out the distance of the final position C of the object from the initial position O?

This difference will give you the numerical value of the displacement of the object from O to C through A. The shortest distance measured from the initial to the final position of an object is known as the displacement.

Can the magnitude of the displacement be equal to the distance travelled by an object? Consider the example given in (Fig. 8.1).

For motion of the object from O to A, the distance covered is `60 km` and the magnitude of displacement is also `60 km`.

During its motion from O to A and back to B, the distance covered `= 60 km + 25 km = 85 km` displacement, are used to describe the overall motion of an object and to locate its final position with reference to its initial position at a given time.

Activity ______________ `8.3`

♦ Take a metre scale and a long rope.
♦ Walk from one corner of a basket-ball court to its oppposite corner along its sides.
♦ Measure the distance covered by you and magnitude of the displacement.
♦ What difference would you notice between the two in this case?

Activity ______________ `8.4`

♦ Automobiles are fitted with a device that shows the distance travelled. Such a device is known as an odometer. A car is driven from Bhubaneshwar to New Delhi. The difference between the final reading and the initial reading of the odometer is `1850 km`.
♦Find the magnitude of the displacement between Bhubaneshwar and New Delhi by using the Road Map of India.


Consider an object moving along a straight line. Let it travel `50 km` in the first hour, `50 km` more in the second hour, `50 km` in the third hour and `50 km` in the fourth hour. In this case, the object covers `50 km` in each hour.

As the object covers equal distances in equal intervals of time, it is said to be in uniform motion.

The time interval in this motion may be small or big. In our day-to-day life, we come across motions where objects cover unequal distances in equal intervals of time, for example, when a car is moving on a crowded street or a person is jogging in a park.
These are some instances of non-uniform motion.

Activity ______________ `8.5`

♦ The data regarding the motion of two different objects A and B are given in Table 8.1.

♦ Examine them carefully and state whether the motion of the objects is uniform or non-uniform.


Look at the situations given in Fig. 8.2. If the bowling speed is `143 \ \ km \ \ h^(–1 )` in Fig. 8.2(a) what does it mean?
What do you understand from the signboard in Fig. 8.2(b)?

Different objects may take different amounts of time to cover a given distance. Some of them move fast and some move slowly.

The rate at which objects move can be different. Also, different objects can move at the same rate. One of the ways of measuring the rate of motion of an object is to find out the distance traveled by the object in unit time.

This quantity is referred to as speed. The SI unit of speed is metre per second. This is represented by the symbol `m s^(–1)` or `m//s`. The other units of speed include centimetre per second `(cm \ \ s^(–1))` and kilometre per hour `(km \ \ h^(–1))`.

To specify the speed of an object, we require only its magnitude. The speed of an object need not be constant. In most cases, objects will be in non-uniform motion.

Therefore, we describe the rate of motion of such objects in terms of their average speed. The average speed of an object is obtained by dividing the total distance travelled by the total time taken. That is,

average speed ` = text( Total distance travelled)/text( Total time taken)`
If an object travels a distance s in time t then its speed `v` is,

` v = s/t` ........(8.1)

Let us understand this by an example. A car travels a distance of `100 km` in `2 h`. Its average speed is `50 km \ \ h^(–1)`. The car might not have travelled at `50 km \ \ h^(–1)` all the time. Sometimes it might have travelled faster and sometimes slower than this.


The rate of motion of an object can be more comprehensive if we specify its direction of motion along with its speed. The quantity that specifies both these aspects is called velocity.

Velocity is the speed of an object moving in a definite direction. The velocity of an object can be uniform or variable. It can be changed by changing the object’s speed, direction of motion or both.

When an object is moving along a straight line at a variable speed, we can express the magnitude of its rate of motion in terms of average velocity. It is calculated in the same way as we calculate average speed.

In case the velocity of the object is changing at a uniform rate, then average velocity is given by the arithmetic mean of initial velocity and final velocity for a given period of time. That is,

average velocity ` = text( initial velocity + final velocity)/2`

Mathematically, `v_(av) = (u + v)/2` .........(8.2)

where `v_(a v)` is the average velocity, `u` is the initial velocity and `v` is the final velocity of the object.

Speed and velocity have the same units, that is, `m \ \ s^(–1)` or `m//s`.

Activity ______________ `8.6`

♦ Measure the time it takes you to walk from your house to your bus stop or the school. If you consider that your average walking speed is `4 km \ \ h^(–1)`, estimate the distance of the bus stop or school from your house.

Activity ______________ `8.7`

♦ At a time when it is cloudy, there may be frequent thunder and lightning. The sound of thunder takes some time to reach you after you see the lightning.
♦ Can you answer why this happens?
♦ Measure this time interval using a digital wrist watch or a stop watch.
♦ Calculate the distance of the nearest point of lightning. (Speed of sound in air `= 346 m \ \ s^(-1)` .)

Q 3224791651

An object travels `16 m` in `4 s` and then another `16 m` in `2 s`. What is the average speed of the object?
Class 9 Chapter 8 Example 1

Total distance travelled by the object `= 16 m + 16 m = 32 m`

Total time taken `= 4 s + 2 s = 6 s`

average speed ` = text( Total distance travelled)/text( Total time taken)`

` = (32 m)/(6 s) = 5.33 m \ \ s^(–1)`

Therefore, the average speed of the object is `5.33 m \ \ s^(–1)`.
Q 3234791652

The odometer of a car reads `2000 km` at the start of a trip and `2400 km` at the end of the trip. If the trip took `8 h`, calculate the average speed of the car in `km \ \ h^(–1)` and `m s^(–1)`.

Class 9 Chapter 8 Example 2

Distance covered by the car,

`s = 2400 km – 2000 km = 400 km`

Time elapsed, `t = 8 h`

Average speed of the car is,

` v_(av) = s/t = (400 km)/(8 h) `

`= 50km \ \ h^(-1)`

`= 50 (k m)/h xx (1000 m)/(1 k h) xx ( 1 h)/(3600 s)`

`= 13.9 m \ \ s^(–1)`
The average speed of the car is `50 km\ \ h^(–1)` or `13.9 m s^(–1)`.
Q 3244791653

Usha swims in a `90 m` long pool. She covers `180 m` in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Usha.
Class 9 Chapter 8 Example 3

Total distance covered by Usha in `1` min is `180 m`.
Displacement of Usha in `1` min `= 0 m`

Average speed ` = text( Total distance covered)/text( Total time taken) `

` = (180 m)/(1 min ) = (180 m)/(1 min ) xx (1 min )/(60 s)`

` = 3 m \ \ s^(-1)`

Average velocity ` = text( Displacement)/text( Total time taken)`

` = (0 m)/( 60 s)`

`= 0 m \ \ s^(–1)`
The average speed of Usha is `3 m \ \ s^(–1)`
and her average velocity is `0 m \ \ s^(–1)`.