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(D) Using the equation of motion $v^{2} = u^{2} + 2as$,where $v$ is the final velocity,$u$ is the initial velocity,$a$ is the acceleration (retardatio

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(D) Using the equation of motion $v^{2} = u^{2} + 2as$,where $v$ is the final velocity,$u$ is the initial velocity,$a$ is the acceleration (retardation),and $s$ is the stopping distance.Since the train stops,$v = 0$,so $0 = u^{2} - 2as$,which gives $s = \frac{u^{2}}{2a}$.Since the retarding force $F = ma$ is constant,the retardation $a$ is constant.Therefore,$s \propto u^{2}$.Given $s_{1} = 80 \ m$ for initial speed $u_{1} = u$.If the speed is doubled,$u_{2} = 2u$.Then,$\frac{s_{2}}{s_{1}} = \left(\frac{u_{2}}{u_{1}}ight)^{2} = \left(\frac{2u}{u}ight)^{2} = 4$.$s_{2} = 4 	imes s_{1} = 4 	imes 80 \ m = 320 \ m$.Thus,the stopping distance becomes four times the original distance.

Column-$I$	Column-$II$
$(1)$ Velocity-time relation	$(a)$ $v = v_0 + at$
$(2)$ Velocity-displacement relation	$(b)$ $S = v_0t + \frac{1}{2}at^2$
	$(c)$ $v^2 = v_0^2 + 2as$

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

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Column-$I$ Column-$II$

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$(2)$ Velocity-displacement relation $(b)$ $S = v_0t + \frac{1}{2}at^2$

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