Answer: 2.3 m
The train started from rest. Initial velocity,
Acceleration of train
The distance travelled by train in t=3 s is:
Using the equation of motion:
Hence, distance travelled by train in 3 s is 2.7 m.
Initial velocity of man,
Acceleration of man,
Distance covered by man in 3s:
train door was already ahead of man by 5 m. Hence, train door is now at (5+2.7)m=7.7 m from the initial point of man.
Therefore, the distance between train's door and man after 3 s is=(7.7-5.4) m =2.3 m