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InterviewSolution
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A man `16 m` behind the door of a train when it starts it starts moving with an acceleration `a=2 ms^(-1)`.The man runs with a msximum constant speed to get into the train. How for does he have to run and after what time does he get the train ? What is his maximum constant speed ? |
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Answer» Let (v) be the maximum constant speed ot theman. He get into the train after time (t), of the man. He will get into tnto the trainaftre time (t), when his velocity becomes equal to the velocity of train. :.`v=0 + 2 xx t =t ` ….(i) Distance moved by man in time (t), ltbRgt `S=ut =2t xx t=2 t^(2)` Distance moved by train in time (t) is `S_(1) =0 +1/2 xx at^(2) =1/2 xx 2 xx t^(2)=1/2 xx2 xx t^(2) =t^(2)` Mam will get into the train, when `S=S_91) +16` :. 2 t^(2) =t +16 or t^(2) =16` or `t=4s` Distance moved by by man `=2 xx (4)^(2) =32 m` Maximum constant speed of man, `v= 2 t= 2 xx 4 =8 ms ^(-1)`. |
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