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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?

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?

Grade:11

1 Answers

Arun
25750 Points
5 years ago
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 t
hence v = at   or v=2t
also 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^2
replacing v=2t and a=2 we get
2t^2 = 9 + t^2
or t=3
 
Hence
 
v = at = 6 m/sec

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