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/s2. 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/s2
Acceleration, a = 1 m/s2
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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