Question

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

Answer 1

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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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Answer 2

Explanation:

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.