A car travelling at a speed of 30km/h is brought to rest in a distance of 8m by applying brakes. If the same car is moving at a speed of 60km/h then it can be brought to rest with same brakes in?? a)64m b)32m c)16m d)4m

Hello! So, in this question, two cases are given.In the first case, speed and stopping distance is given. So, we find out the retardation/maximum braking force.Stopping distance(s)=-(u^2)/2aBut, the speed is in kmph and has to converted into m/sWe get 30 kmph=8.3 m/sSubstituing in stopping distance formula we get8(distance given)=-(8.3)^2/2aSolving for a, we get a=-4.3 m/s^2Now, in the second case, we have the maximum braking force and the speed and we have to find stppping distance.Convert the speed to m/s which gives 60 kmph = 16.6 m/s.Substitute in stoppong distance formula, we get-(16.6)^2/-4.3=sAfter solving, we get s=32 mSo answer is (b)32m.Hope this helped.

In the question,it is given that the car stops with same braking means retardation remains same in both cases so caculating retardation from first and put it into second case so that distance can be found.