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(B) Let the time after which the two bodies meet be $t$ seconds.For the first body moving with uniform velocity $v = 8\,m/s$,the displacement $s_1$ in

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$4$

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(B) Let the time after which the two bodies meet be $t$ seconds.For the first body moving with uniform velocity $v = 8\,m/s$,the displacement $s_1$ in time $t$ is given by $s_1 = v 	imes t = 8t$.For the second body starting from rest $(u = 0)$ with uniform acceleration $a = 4\,m/s^2$,the displacement $s_2$ in time $t$ is given by $s_2 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} 	imes 4 	imes t^2 = 2t^2$.Since the two bodies meet,their displacements must be equal: $s_1 = s_2$.Therefore,$8t = 2t^2$.Rearranging the equation: $2t^2 - 8t = 0$,which gives $2t(t - 4) = 0$.Since $t$ cannot be $0$ (as they meet after starting),we have $t = 4\,s$.

$A$ body is moving with a uniform velocity of $8\,m/s$. When the body just crosses another body,the second body starts from rest and moves with a uniform acceleration of $4\,m/s^2$. The time after which the two bodies meet will be $...........\,s$.

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