3.7 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 ?
For train A:
Initial velocity, u = 72 km/h = 20 m/s
Time, t = 50 s
Acceleration, a = 0 (Since it is moving with a uniform velocity)
From second equation of motion, distance S1 covered by train A can be obtained as:
S1 = ut + (1/2) at2
= 20 × 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 S2 covered by train A can be obtained as:
S2= ut + (1/2) at2
= 20 50 + (1/2) × 1 × (50)2 = 2250 m
Hence, the original distance between the driver of train A and the guard of train B is 2250 - 1000 = 1250m.
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