Question

two trains A and B of length 400m each are moving on two parallel tracks with a uniform speed of 72 km/h in the same direction,with A ahead of B.The driver of B decides to overtake A andaccelerates by 1m/s^2.If after 50s ,the guard of B just brushes past the driver of A,what was the original distance between them?

Answer 1

Relativive velocity =zero
Relative acceleration = 1
Relative distance covered
S =1/2*1*50*50=1250m
S = distance between them + 500
D =1250 - 500=750m

Answer 2

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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 (s') covered by train A can be obtained as:

S = ut + 1/2 at^2

= 20 × 50 + 0 = 1000 m

For train B:

Initial velocity,

u = 72 km/h = 20 m/s

Acceleration, a = 1 m/s^2

Time, t = 50 s

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

S = ut + 1/2 at^2

= 20× 50 + 1 × (50)^2 = 2250 m

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

I hope, this will help you

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