question_answer

A retarding force is applied to stop a motor car. If the speed of the motor car is doubled, how much more distance will it cover before stopping under the same retarding force?

Answer:

                Work done against retarding force \[=\]Loss in K.E. In first case,     \[F\times s=\frac{1}{2}m{{\upsilon }^{2}}\]                    ...(i) In second case, \[F\times s'=\frac{1}{2}m{{(2\upsilon )}^{2}}\]                ..(ii) From (i) and (ii),   \[s'=\text{ 4s}\] Thus the motor car will cover a distance four times longer than before.