A car moving at a speed ' V ' is stopped in a certain distance when the breaks produce a deceleration 'a'. If the speed of the car is 'nv', what must be the deceleration of the car to stop it in the same distance and in the same time?

Step-by-step solution

A car stops from speed (v ) under deceleration (a ). Using (v=a t and s= 1 2 v t ) (the distance-time relations during uniform deceleration). Now the new speed is (n v ). Let required deceleration be ( A ). For same stopping time: (n v=A t A=n a ) For same stopping distance: (s= 1 2 v t= 1 2 (n v) t^ ) But time is same, so: ( 1 2 v t= 1 2 n v t a A ) Or use direct formula for distance: ( v^2 2 a = (n v)^2 2 A ) Thus, ( v^2 a = n^2 v^2 A A=n^2 a . ) (n^2 a )