A man is at a distance of 6,m from a bus. The | vedclass

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(C) Let the man run with a constant speed $v$. Let $t$ be the time taken to catch the bus.The distance covered by the bus in time $t$ is $s_b = \frac{

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$6$

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(C) Let the man run with a constant speed $v$. Let $t$ be the time taken to catch the bus.The distance covered by the bus in time $t$ is $s_b = \frac{1}{2} a t^2 = \frac{1}{2} 	imes 3 	imes t^2 = 1.5 t^2$.The total distance the man must cover to reach the bus is $s_m = 6 + s_b = 6 + 1.5 t^2$.Since the man runs with constant speed $v$,the distance he covers is $s_m = v t$.Equating the two expressions for $s_m$: $v t = 6 + 1.5 t^2$,which rearranges to $1.5 t^2 - v t + 6 = 0$.For the man to catch the bus,the time $t$ must be a real value. This requires the discriminant of the quadratic equation to be greater than or equal to zero $(D \ge 0)$.$D = (-v)^2 - 4(1.5)(6) \ge 0$$v^2 - 36 \ge 0$$v^2 \ge 36 \implies v \ge 6\,m s^{-1}$.Thus,the minimum speed required is $6\,m s^{-1}$.

$A$ man is at a distance of $6\,m$ from a bus. The bus begins to move with a constant acceleration of $3\,m s^{-2}$. In order to catch the bus,the minimum speed with which the man should run towards the bus is $.........m s^{-1}$.

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