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Question
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?
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Solution 1
For train A:
Initial velocity, u = 72 km/h = 20 m/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:
`s_1 = ut + 1/2 a_1t^2`
= 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 the second equation of motion, the distance (sII)covered by train A can be obtained as:
`s_u = ut + 1/2 at^2`
`=20 xx 50 + 1/2 xx 1 xx (50)^2 m`
= 1000 + 1250
= 2250 m
Hence, the original distance between the driver of train A and the guard of train B is 2250 – 1000 = 1250 m
Solution 2
Here length of train A = length of train B = l = 400 m. As speed of both trains u = 72 km h-1 = 20 ms-1 in same direction, hence their relative velocity uBA = 0.
Let initial distance between the two trains be ‘S’ then train B covers the distance (S + 11) = (S + 800) m in time t = 50 s when accelerated with a uniform acceleration a = 1 m/s2.
`:. (s+800) = u_(AB) xx t + 1/2 at^2`
`= 0 + 1/2 xx1xx(50)^2 = 1250 m`
=> S = 1250 - 800 = 450 m
and initial distance between guard of train B from driver of train A = 450 + 800 = 1250 m.
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