Starting from a stationary position , Rahul paddles his bicycle to attain a velocity of `6 m//s` in `30 s`. Then he applies brakes such that the velocity of bicycle comes down to `4m//s` in the next `5s` . Calculate the acceleration of the bicycle in both the cases.

Explanation

In the first case :
Initial velocity , u=0,
Final velocity, v=6` m s^(-1)` ,
time , t=30s
From eq. (8.3), we have
`a=(v-u)/(t)`
Substituting the given values of the v,u and t in the above equation , we get
`a=((6 ms ^(-1)-0 m s^(-1)))/(30 s)`
`=0.2 ms^(-2)`
In the second case :
Initial velocity , `u=6 m s^(-1)` ,
time , t=5 s .
Then , `a=((4m s^(-1)- 6 m s^(-1)))/(5 s)`
`=0.4 ms^(-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)`.