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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?

Answer:

For train A:

Initial velocity, u = 72 km/h = 20 m/s

Time, t = 50 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:

= 20 × 50 + 0 = 1000 m

For train B:

Initial velocity, u = 72 km/h = 20 m/s

Acceleration, a = 1 m/s

Acceleration, a = 1 m/s

^{2}
Time, t = 50 s

From second equation of
motion, distance (sII)covered by train A can be obtained

as:

as:

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

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