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