Topic Covered



During uniform motion of an object along a straight line, the velocity remains constant with time. In this case, the change in velocity of the object for any time interval is zero.

However, in non-uniform motion, velocity varies with time. It has different values at different instants and at different points of the path. Thus, the change in velocity of the object during any time interval is not zero.
Can we now express the change in velocity of an object?

To answer such a question, we have to introduce another physical quantity called acceleration, which is a measure of the change in the velocity of an object per unit time. That is,

acceleration ` = text( change in velocity) /text( time taken)`

If the velocity of an object changes from an initial value `u` to the final value `v` in time `t`, the acceleration `a` is,

` a = (v - u)/t` .....(8.3)

This kind of motion is known as accelerated motion. The acceleration is taken to be positive if it is in the direction of velocity and negative when it is opposite to the direction of velocity. The SI unit of acceleration is `m \ \ s^(–2)` .

If an object travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time, then the acceleration of the object is said to be uniform.

The motion of a freely falling body is an example of uniformly accelerated motion. On the other hand, an object can travel with non-uniform acceleration if its velocity changes at a non-uniform rate.

For example, if a car travelling along a straight road increases its speed by unequal amounts in equal intervals of time, then the car is said to be moving with non-uniform acceleration.

Activity ______________ `8.8`
♦ In your everyday life you come across a range of motions in which
(a) acceleration is in the direction of motion,
(b) acceleration is against the direction of motion,
(c) acceleration is uniform,
(d) acceleration is non-uniform.
♦ Can you identify one example each of the above type of motion?

Q 3254791654

Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of `6 m \ \s^(–1)` in `30 s`. Then he applies brakes such that the velocity of the bicycle comes down to `4 m \ \ s^(-1)` in the next `5 s`. Calculate the acceleration of the bicycle in both the cases.
Class 9 Chapter 8 Example 4

In the first case:
initial velocity, `u = 0` ;
final velocity, ` v = 6 m \ \ s^(–1) ;`
time, `t = 30 s` .
From Eq. (8.3), we have

` a = (v - u)/t `

Substituting the given values of `u ,v` and `t` in the above equation, we get

` a = ( 6 m \ \ s^(-1) - 0 m \ \ s^(-1) )/(30 s)`

`= 0.2 m \ \ s^(–2)`
In the second case:
initial velocity, `u = 6 m s^(–1);`
final velocity, `v = 4 m s^(–1);`
time, `t = 5 s`

Then, ` a = ( 4 m \ \ s^(-1) - 6 m \ \ s^(-1) )/(5 s)`

`= –0.4 m \ \ s^(–2)` .
The acceleration of the bicycle in the
first case is `0.2 m \ \s^(–2)` and in the second case, it is `–0.4 m \ \s^(–2)`.


Graphs provide a convenient method to present basic information about a variety of events. For example, in the telecast of a one-day cricket match, vertical bar graphs show the run rate of a team in each over.

As you have studied in mathematics, a straight line graph helps in solving a linear equation having two variables.

To describe the motion of an object, we can use line graphs. In this case, line graphs show dependence of one physical quantity, such as distance or velocity, on another quantity, such as time.


The change in the position of an object with time can be represented on the distance-time graph adopting a convenient scale of choice. In this graph, time is taken along the x–axis and distance is taken along the y-axis.

Distance-time graphs can be employed under various conditions where objects move with uniform speed, non-uniform speed, remain at rest etc.

We know that when an object travels equal distances in equal intervals of time, it moves with uniform speed. This shows that the distance travelled by the object is directly proportional to time taken.

Thus, for uniform speed, a graph of distance travelled against time is a straight line, as shown in Fig. 8.3. The portion OB of the graph shows that the distance is increasing at a uniform rate.

`"Note"` that, you can also use the term uniform velocity in place of uniform speed if you take the magnitude of displacement equal to the distance travelled by the object along the y-axis.

We can use the distance-time graph to determine the speed of an object. To do so, consider a small part AB of the distance-time graph shown in Fig 8.3. Draw a line parallel to the x-axis from point

A and another line parallel to the y-axis from point B. These two lines meet each other at point C to form a triangle ABC. Now, on the graph, AC denotes the time interval `(t_2 – t_1)` while BC corresponds to the distance `(s_2 – s_1)`.

We can see from the graph that as the object moves from the point A to B, it covers a distance `(s_2 – s_1)` in time `(t_2 – t_1)`. The speed, v of the object, therefore can be represented as

` v = (s_2 - s_1)/(t_2 - t_1)` (8.4)

We can also plot the distance-time graph for accelerated motion. Table 8.2 shows the distance travelled by a car in a time interval of two seconds.

The distance-time graph for the motion of the car is shown in Fig. 8.4. Note that the shape of this graph is different from the earlier distance-time graph (Fig. 8.3) for uniform motion.

The nature of this graph shows nonlinear variation of the distance travelled by the car with time. Thus, the graph shown in Fig 8.4 represents motion with non-uniform speed.


The variation in velocity with time for an object moving in a straight line can be represented by a velocity-time graph. In this graph, time is represented along the x-axis

and the velocity is represented along the y-axis. If the object moves at uniform velocity, the height of its velocity-time graph will not change with time (Fig. 8.5). It will be a straight line parallel to the x-axis. Fig. 8.5 shows the velocity-time graph for a car moving with uniform velocity of `40 km \ \h^(–1)`.

We know that the product of velocity and time give displacement of an object moving with uniform velocity. The area enclosed by velocity-time graph and the time axis will be equal to the magnitude of the displacement.

To know the distance moved by the car between time `t_1 ` and `t_2` using Fig. 8.5, draw perpendiculars from the points corresponding to the time `t_1` and `t_2` on the graph.

The velocity of `40 km \ \ h^(–1)` is represented by the height AC or BD and the time `(t_2 – t_1)` is represented by the length AB.

So, the distance s moved by the car in time `(t_2 – t_1)` can be expressed as
`s = AC xx CD`
`= [(40 km\ \ h^(–1) ) xx (t_2 – t_1 ) h ]`
`= 40 (t_2– t_1 ) km`
`=` area of the rectangle `ABDC` (shaded in Fig. 8.5).

We can also study about uniformly accelerated motion by plotting its velocity– time graph. Consider a car being driven along a straight road for testing its engine.

Suppose a person sitting next to the driver records its velocity after every ` 5 ` seconds by noting the reading of the speedometer of the car. The velocity of the car, in ` km \ \ h^(–1)` as well as in `m s^(–1)`, at different instants of time is shown in table 8.3.

In this case, the velocity-time graph for the motion of the car is shown in Fig. 8.6. The nature of the graph shows that velocity changes by equal amounts in equal intervals of time.

Thus, for all uniformly accelerated motion, the velocity-time graph is a straight line.

You can also determine the distance moved by the car from its velocity-time graph. The area under the velocity-time graph gives the distance (magnitude of displacement) moved by the car in a given interval of time.

If the car would have been moving with uniform velocity, the distance travelled by it would be represented by the area ABCD under the graph (Fig. 8.6).

Since the magnitude of the velocity of the car is changing due to acceleration, the distance s travelled by the car will be given by the area ABCDE under the velocity-time graph (Fig. 8.6).

That is,
`s =` area `A B C D E`
`=` area of the rectangle `A B C D +` area of the triangle `A D E`
`= A B xx B C + 1/2 ( A D xx D E)`

In the case of non-uniformly accelerated motion, velocity-time graphs can have any shape.

Fig. 8.7(a) shows a velocity-time graph that represents the motion of an object whose velocity is decreasing with time while Fig. 8.7 (b) shows the velocity-time graph representing the non-uniform variation of velocity of the object with time. Try to interpret these graphs.

Activity ______________ 8.9

♦ The times of arrival and departure of a train at three stations A, B and C and the distance of stations B and C from station A are given in table 8.4.

♦ Plot and interpret the distance-time graph for the train assuming that its motion between any two stations is uniform.

Activity _____________`8.10`

♦Feroz and his sister Sania go to school on their bicycles. Both of them start at the same time from their home but take different times to reach the school although they follow the same route. Table 8.5 shows the distance travelled by them in different times

♦ Plot the distance-time graph for their motions on the same scale and interpret.