Question 7

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

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:

sI  =   ut  +  ½ aI t2

= 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:

sIl  =   ut  +  ½ a t2

= 20 x 50 x ½ x 1x (50)2 = 2250 m

Hence, the original distance between the driver of train A and the guard of train B

= 2250 –1000 = 1250 m.

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