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

=>  S_B = u_Bt + (1/2)at²

And we are given that acceleration (a) = 1 m/s and time taken = 50 s

Upon substituting values => S_B = (20 x 50) + (1/2) x 1 x (50)²

S_B  = 2250 m

Similarly, distance covered by the train A can be determined =>  S_A = u_At + (1/2)at²

In this case the acceleration (a) = 0

Therefore, we are left with => S_A = u_At

= 20 x 50

S_A  = 1000 m

Therefore, the initial distance between the two trains is given by

S_B – S_A = 2250 – 1000

S_B – S_A  = 1250 m