A man is $48 m$ behind a bus which is at rest. The bus starts accelerating at the rate of $1 m / s^{2}$ , at the same time the man starts running with uniform velocity of $10 m / s$ . What is the minimum time in which the man catches the bus?

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Solution

Given,

Acceleration of bus, $a = 1 m s^{- 2}$

Assume man catches the train in time $t$ .

Displacement by man in time $t$ $s_{1} = v t = 10 t$

Displacement by bus in time t, $s_{2} = u t + \frac{1}{2} a t^{2} = \frac{t^{2}}{2}$

Separation between man and bus is $48 m$

$s_{1} = s_{2} + 48$

$10 t = \frac{t^{2}}{2} + 48$

$t^{2} - 20 t + 96 = 0$

$t = \frac{20 \pm \sqrt{20^{2} - 4 \times 96}}{2}$

$t = 8 a n d 12$

In first attempt he caches the bus

Hence, minimum time he catches the bus is $8 sec$

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A man is 48m behind a bus which is at rest. The bus starts accelerating at the rate of 1m/s2, at the same time the man starts running with uniform velocity of 10m/s. What is the minimum time in which the man catches the bus?

A man is $48 m$ behind a bus which is at rest. The bus starts accelerating at the rate of $1 m / s^{2}$ , at the same time the man starts running with uniform velocity of $10 m / s$ . What is the minimum time in which the man catches the bus?