Question

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?

Answer 1

Answer:

Given,

Acceleration of bus, a=1ms−2

Assume man catches the train in timet.

Displacement by man in time t s1=vt=10t

Displacement by bus in time t, s2=ut+21at2=2t2

Separation between man and bus is 48m

s1=s2+48

10t=2t2+48

t2−20t+96=0

t=220±202−4×96

t=8and12

In first attempt he caches the bus

Hence, minimum time he catches the

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