For train A:
Initial velocity, u = 72 km/h = 20 m/s
Time, t = 50 s
Acceleration, aI = 0 (Since it is moving with a uniform velocity)
From second equation of motion, distance (sI) covered by train A can be obtained as:

For train B:
Initial velocity, u = 72 km/h = 20 m/s
Acceleration, a = 1 m/s2
Time, t = 50 s
From second equation of motion, distance (sII) covered by train A can be obtained as:

Hence, the original distance between the driver of train A and the guard of train B = 2250 –1000 = 1250 m.