A bus is begins to move with an acceleration of 1 m/s2. A boy who is 42m behind the bus starts running at 10 m/s towards the bus. After what time bus will cross the boy?

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Answer 1 · 2018-05-05T06:39:16

#14# seconds.

Assume that bus crosses the boy after #t# seconds.

As the boy is running at #10m/s#, in #t# seconds he covers #10t# meters.

The bus starts with an initial velocity of #0m/s# and acceleration of #1m/s^2#, hence using formula #S=ut+1/2at^2#

or here #S=1/2*1*t^2=t^2/2# meters

As boy must cover #42m# additional, we should have

#10t=1/2t^2+42# or #t^2-20t+84=0#

or #(t-6)(t-14)=0#

Hence, bus will cross the boy after #6# seconds as well as #14# seconds.

How is that? This is so, as initially boy and bus are together after #6# seconds, when boy is surging ahead of the bus

and after #14# seconds as bus accelerates it crosses the boy.

Hence, answer is #14# seconds.