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Question

A man is 48 m behind a bus which is at rest. The bus starts accelerating at 1 m/s2 at the same time the man starts running with uniform velocity 10 m/s. What is the minimum time taken by the man to reach the bus?

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Solution

Let after time 't' man catches the bus.
After time 't' bus has covered the distance = 12(1)t2=t22
Therefore the total distance man has to cover = 48+t22
Therefore, vt=s
10t=48+t2220t=96+t2t220t+96=0(t12)(t8)=0t=12 or t=8
Therefore, minimum time in which man catches the bus is 8 seconds.

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