Question

6. Two trains A and B of length 300 m each are moving on two parallel tracks with an
uniform speed of 54 kmph in the same direction, with the train A ahead of B. The driver
of train B decides to overtake A and accelerates by 2 ms. If after 25 s, the guard of
train B just brushes past the driver of A, what was the original distance between them.
(NCERT)​

Answer 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 = ut + (1/2)at2

= 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 

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