In case of train A:
Initial velocity, u = 72 km/h = 20 m/s
Time, t = 50 s
Since the train is moving with uniform velocity,
Acceleration, aI = 0
Now, using second equation of motion,
Distance covered by train A is given by,
For train B,
Initial velocity, uB = 72 km/h = 20 m/s
Acceleration, a = 1 m/s2
Time, t = 50 s
Now, using second equation of motion,
Distance travelled by second train B is,
Length of both trains = 2
Therefore, the original distance between the driver of train A and the guard of train B = 2250 - 1000 - 800
= 450m.