A car is moving with constant speed `15 m//s`. Suddenly the driver sees an obstruction on the road and takes `0.40 s` to apply the brake, the brake causes a deceleration of `5 m//s^(2)`. Find the distance traveled by the car before it stops.

Explanation

0.40s is the reaction time of the driver. After reaction time, deceleration comes into effect. During reaction time, car moves with contant velocity. Stopping distance will be the sum deceleration.

If `d_(2)` is the distance during deceleration, then
`v^(2)=u^(2)-2as`
`0=(15)^(2)-2xx5xxd_(2)impliesd_(2)=22.5 m`
Stopping distance `=d_(1)+d_(2)=6+22.5=28.5m`