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.
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