A man is standing 40m behindthe bus. The bus starts with acceleration 1m/s^2 also at the same instant the man starts moving with constant speed 9 m/s the minimum time taken by the man to catch the bus will be

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Solution

suppose man catch the bus in time t

distance travelled by bus is

s = ut + 1/2at^2
s = 1/2(1)t^2 = 0.5t^2

the man started running from 40 m behaind when the bus starts. so in the time t man has to run S+40

S+40 = ut = 9t

0.5t^2 + 40 = 9t

t = 8s or 10sec

we need minimum time so the man will catch the bus in 8sec .

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