A and B Two train length …

For train A:
Initial velocity, u = 72 km/h = 20 m/s
Time, t = 50 s
Acceleration, a_I = 0 (Since it is moving with a uniform velocity)
From second equation of motion, distance (S_I) covered by train A can be obtained as:
{tex}s _ { I } = u t + \frac { 1 } { 2 } a _ { I } t ^ { 2 }{/tex}
= 20 {tex}\times{/tex} 50 + 0 = 1000 m
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 (S_II) covered by train A can be obtained as:
{tex}s _ { II } = u t + \frac { 1 } { 2 } a t ^ { 2 }{/tex}
{tex}= 20 \times 50 + \frac { 1 } { 2 } \times 1 \times ( 50 ) ^ { 2 } = 2250 \mathrm { m }{/tex}
Length of both trains = 2 {tex}\times{/tex} 400 m = 800 m
Hence, the original distance between the driver of train A and the guard of train B is 2250 - 1000 - 800 = 450 m.