A man is s=9 m behind the door of a train when it starts moving with acceleration a=2ms^-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?
ans]6ms^-1
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Given:
The distance between man and train (s) = 9 meters
The acceleration of the train (a) = 2 m/s²
To Find:
The time and total distance covered by man in which he gets into the train, and the highest speed of man.
Solution:
The highest velocity of train (v) = v = u + at
⇒ v = 0 + 2 × t (∵ initially the train is at rest)
⇒ v = 2t
The total distance covered by train in time 't' = x = ut + at²/2
⇒ x = 0×t + 2t²/2 ⇒ x = t²
So, the total distance covered by man = (x+9) meters
And, the distance covered by man = (x +9) = ut
(where u =v = the highest speed of the train)
Putting the value of 'v' in the above equation
⇒ x+9 = (2t)t
⇒ x+9 = 2t²
And, t² is equal to x, the distance covered by the train
⇒ x+9 = 2x
⇒ x = 9 meters
Therefore, x = t²
⇒ x = 9 = t²
⇒ t = 3 seconds
And, the speed of man (u) = 2t = 6m/s
Hence, the man has to run 9 meters in 3 seconds. And his full speed is 6m/s.