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A bus starts from rest with an acceleration of 1m/s^2. A man, who is 48m behind the bus, starts witha uniform velocity of 10m/s. Then, what is the minimum time after which the man will catch the bus?
(1/2)*1*t^2 +48=10t,t^2-20t+96=0,t=12,8, therefore minimum time=8 seconds
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relative velocity of man with respect to bus is V=vman-v bus=10-0=10m/s
relative accleration of man ......................is A=aman-abus=0-1= -1m/s^2
relative displacement of man .................is 48 metres
now using eq of motion for this relative motion
s=ut+1/2at^2
we get t=8,12
so minimum time will bw 8 sec
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