Ans:
According to the question, length of the train A and B is 400 m and the speed of both the trains is given
= 72 km/h = 72 x (5/18)
= 20m/s
Using the following expression,
s = ut + (1/2)at2
We can write an expression for distance covered by the train B as follows
=> SB = uBt + (1/2)at2
And we are given that acceleration (a) = 1 m/s and time taken = 50 s
Upon substituting values => SB = (20 x 50) + (1/2) x 1 x (50)2
SB = 2250 m
Similarly, distance covered by the train A can be determined => SA = uAt + (1/2)at2
In this case the acceleration (a) = 0
Therefore, we are left with => SA = uAt
= 20 x 50
SA = 1000 m
Therefore, the initial distance between the two trains is given by
SB – SA = 2250 – 1000
SB – SA = 1250 m