A bus starts from rest with an acceleration of $1.0 m / s^{2}$ . A man who is $48 m$ behind the bus, starts with a uniform velocity of $10 m / s$ . Find the minimum time after which the man will catch the bus.

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Solution

Let man will catch the bus after time t.Since the distance from man position is same for both at time t
$\Rightarrow$ Distance covered by bus = Distance covered by man -48
$\Rightarrow \frac{1}{2} a t^{2} = v t - 48$

$\Rightarrow \frac{1}{2} (1) t^{2} = 10 t - 48$

$\Rightarrow t^{2} - 20 t + 96 = 0$
$\Rightarrow (t - 12) (t - 8) = 0$
$\Rightarrow t = 8 s$ (minimum value) or $t = 12 s$
So,at $t = 8 s$ ,the man will catch the bus

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A bus starts from rest with an acceleration of 1.0m/s2. A man who is 48m behind the bus, starts with a uniform velocity of 10m/s. Find the minimum time after which the man will catch the bus.

A bus starts from rest with an acceleration of $1.0 m / s^{2}$ . A man who is $48 m$ behind the bus, starts with a uniform velocity of $10 m / s$ . Find the minimum time after which the man will catch the bus.