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A bus starts to move with an acceleration of 1m/s^2 .A man who is 48m behind the bus runs with constant velocity of10m/s to catch it.In hhow much time he will catch the bus? a)6sec b)8sec c)10sec d)12sec
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The answer is b) 8 seconds. You just have to try plugging the choices into the relative equations, and whatever works is the right answer. For b) 8 seconds, it is: a=1m/s^2 Bus displacement: x=0.5at^2 x=0.5t^2 x=0.5(64) x=32 (after acceleration). Plus the fact it's already 48m ahead of the man, so it's x location would be: 32+48= 80m Man displacement: x=vt x=10*8 x= 80m So they'll both be at the same place after 8 seconds.
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