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(A) According to the work-energy theorem,the work done by the retarding force is equal to the change in kinetic energy of the body.For the first case,

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$s_{2} = 4s_{1}$

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(A) According to the work-energy theorem,the work done by the retarding force is equal to the change in kinetic energy of the body.For the first case,the body comes to rest from velocity $u$ after covering distance $s_{1}$:$F s_{1} = \frac{1}{2} m u^{2} \quad (i)$For the second case,the body comes to rest from velocity $2u$ after covering distance $s_{2}$:$F s_{2} = \frac{1}{2} m (2u)^{2} = \frac{1}{2} m (4u^{2}) = 4 \left( \frac{1}{2} m u^{2} \right) \quad (ii)$Dividing equation $(ii)$ by equation $(i)$:$\frac{F s_{2}}{F s_{1}} = \frac{4 (\frac{1}{2} m u^{2})}{\frac{1}{2} m u^{2}}$$\frac{s_{2}}{s_{1}} = 4$$s_{2} = 4s_{1}$

$A$ body of mass $m$ is travelling with a velocity $u$. When a constant retarding force $F$ is applied,it comes to rest after travelling a distance $s_{1}$. If the initial velocity is $2u$,with the same force $F$,the distance travelled before it comes to rest is $s_{2}$. Then,

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