Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h–1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m s–2. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them ?

Solution

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, a_I = 0
Now, using second equation of motion,
Distance covered by train A is given by,
$straight s space equals space ut space plus space left parenthesis 1 divided by 2 right parenthesis straight a subscript 1 straight t squared space space space space equals space 20 cross times 50 plus space 0 space equals space 1000 space straight m$
For train B,
Initial velocity, u_B = 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,
$straight s subscript II space equals space ut space plus space 1 half at squared space space space space space space equals space 20 space straight X space 50 space plus space left parenthesis 1 divided by 2 right parenthesis space cross times space 1 space cross times space left parenthesis 50 right parenthesis squared space space space space space space equals space 2250 space straight m space$
Length of both trains = 2 $cross times 400 space straight m space equals space 800 space straight m$
Therefore, the original distance between the driver of train A and the guard of train B = 2250 - 1000 - 800
= 450m.