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A man is s=9m behind the door of a train when it starts moving with acceleration a = 2m/s^2 . the man runs at full speed. how far does he have to run and after what time does he get into the train? what is his full speed?
Dear Abhishek There is a huge conceptual error with the author of the question who thinks that the train can be boarded only when the speed of the body say v equals the speed of train, but blv me theres no such contraint practically; Anyhow let us honour the question then speed of body v= speed of train at the time of boarding say at time thence v = at or v=2talso the man travels a distance vt in time t which should be equal to the gap of 9m plus the distance moved by train in time t(which is 1/2at^2)i.e. vt = 9 + 1/2at^2replacing v=2t and a=2 we get2t^2 = 9 + t^2or t=3 Hence v = at = 6 m/sec
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