A car accelerates uniformly from ` 18 km//h` to `36 km//h` in 5 second. Calculate (i) the acceleration and (ii) the distance covered by the car in that time .

Explanation

We are given that
u=18 `km h^(-1)=5 m s^(-1)`
`v=36 km h^(-1)=10 m s^(-1)` and
t=5 s.
(i) From Eq. (8.5) we have
`a=(v-u)/(t)`
`=(10 ms^(-1)-5 m s^(-1))/(5 s)`
`=1 m s^(-2)`
(ii) From Eq. (8.6) We have
`s=ut+(1)/(2)at^(2)`
`=5 m s^(-1)xx5 s+(1)/(2)xx 1 m s^(-2)xx(5 s)^(2)`
`=25 m +12.5 m`
=37.5 m
The acceleration of the car is 1 ` m s^(-2)` and the distance covered is 37.5 m.