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(D) From the work-energy theorem,the work done by the retarding force $F$ is equal to the change in kinetic energy of the train.$|W| = |\Delta K|$$F \

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Four times

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(D) From the work-energy theorem,the work done by the retarding force $F$ is equal to the change in kinetic energy of the train.$|W| = |\Delta K|$$F \cdot S = \frac{1}{2} m u^2$Since the retarding force $F$ and the mass $m$ are constant,the stopping distance $S$ is directly proportional to the square of the initial velocity $u$,i.e.,$S \propto u^2$.Given that the initial speed is doubled $(u' = 2u)$,the new stopping distance $S'$ will be:$S' \propto (2u)^2 = 4u^2$$S' = 4S$$S' = 4 	imes 80 \ m = 320 \ m$.Therefore,the stopping distance becomes four times the original distance.

The same retarding force is applied to stop a train. The train stops after $80 \ m$. If the speed is doubled,then the distance will be

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