ACCELERATION
During the 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 of time. That is,
acceleration =\(\frac {Change\,\, in\,\,velocity }{time \,\,taken}\)
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 = \(\frac {v-u }{t}\) ______(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 ms–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 traveling 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.
Source: This topic is taken from NCERT TEXTBOOK
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 for the above type of motion?
Illustration 8.4:
Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 ms–1 in 30s. Then he applies brakes such that the velocity of the bicycle comes down to 4 ms-1 in the next 5s. Calculate the acceleration of the bicycle in both the cases.
Sol:
In the first case:
initial velocity, u = 0 ; final velocity, v = 6 ms–1 ; time, t = 30s .
From Eq. (8.3), we have
a = \(\frac{v-u}{t}\)
Substituting the given values of u,v and t in the above equation, we get
a = \(\frac{6ms^{-1}-0ms^{-1}}{30s}\)
= 0.2 ms–2
In the second case:
initial velocity, u = 6 ms–1; final velocity, v = 4 ms–1; time, t = 5s.
Then, a = \(\frac{4ms^{-1}-6ms^{-1}}{5s}\)
= –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.
Questions
1. When will you say a body is in
(i) uniform acceleration?
(ii) non-uniform acceleration?
2. A bus decreases its speed from 80 km h–1 to 60 km h–1 in 5 s. Find the acceleration of the bus.
3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h–1 in 10 minutes. Find its acceleration.
Source: This topic is taken from NCERT TEXTBOOK