A man is 45 , m behind the bus when the bus s | vedclass

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(C) Let the man catch the bus after time $t$ seconds. The distance covered by the man is $s_m = u t$,where $u$ is the constant velocity of the man.The

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$15$

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(C) Let the man catch the bus after time $t$ seconds. The distance covered by the man is $s_m = u t$,where $u$ is the constant velocity of the man.The distance covered by the bus starting from rest is $s_b = \frac{1}{2} a t^2$,where $a = 2.5 \, m/s^2$.For the man to catch the bus,the total distance covered by the man must be equal to the initial gap plus the distance covered by the bus:$u t = 45 + \frac{1}{2} (2.5) t^2$$u t = 45 + 1.25 t^2$$u = \frac{45}{t} + 1.25 t$To find the minimum velocity $u$,we differentiate $u$ with respect to $t$ and set it to zero:$\frac{du}{dt} = -\frac{45}{t^2} + 1.25 = 0$$1.25 t^2 = 45$$t^2 = \frac{45}{1.25} = 36$$t = 6 \, s$Substituting $t = 6 \, s$ back into the equation for $u$:$u = \frac{45}{6} + 1.25(6) = 7.5 + 7.5 = 15 \, m/s$.

$A$ man is $45 \, m$ behind the bus when the bus starts accelerating from rest with an acceleration of $2.5 \, m/s^2$. With what minimum velocity should the man start running to catch the bus?

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