Correct Answer - C
For train A,
`u_A = 72 km h^(-1) = 72 xx 5/18 m s^(-1) = 20 m s^(-1) , a_A = 0`, t = 50 s,
`therefore S_A = u_(A)t - (20)(50) = 1000 m`
For train B,
`U_b = 72 km h^(-1) = 72 xx 5/18 m s^(-1) = 20 m s^(-1)`
`a_B = 1 m s^(-2)` , t = 50 s
`therefore S_B = u_(B)t + 1/2 a_(B)t^(2)`
`= (20 xx 50) + 1/2 xx 1 xx (50)^(2) = 2250` m
Original distance between A and `B = S_B - S_A = 2250 m - 1000 m = 1250 m`