2trains A and B of length 400m each are moving with the speed of 72 km/h . Train A is ahead of train B.If train B accelerate 1m/s^2 to overtake train A and after 50 sec guard of train B across guard of train A. Find the original distance between them.
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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 = ut + (1/2)a1t2
= 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 (sII) covered by train A can be obtained as:
sII = ut + (1/2)at2
= 20 X 50 + (1/2) × 1 × (50)2 = 2250 m
Length of both trains = 2 × 400 m = 800 m
Hence, the original distance between the driver of train A and the guard of train B is 2250 – 1000 – 800 = 450m.
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 = ut + (1/2)a1t2
= 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 (sII) covered by train A can be obtained as:
sII = ut + (1/2)at2
= 20 X 50 + (1/2) × 1 × (50)2 = 2250 m
Length of both trains = 2 × 400 m = 800 m
Hence, the original distance between the driver of train A and the guard of train B is 2250 – 1000 – 800 = 450m.
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