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/s^{2}. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?

#### Solution 1

*For train A:*

Initial velocity, *u* = 72 km/h = 20 m/s

Time, *t* = 50 s

Acceleration, *a*_{I} = 0 (Since it is moving with a uniform velocity)

From second equation of motion, distance (*s*_{I})covered by train A can be obtained as:

`s_1 = ut + 1/2 a_1t^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*_{II})covered by train A can be obtained as:

`s_u = ut + 1/2 at^2`

`=20xx50+1/2xx1xx(50)^2250 m`

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

#### Solution 2

Here length of train A = length of train B = l = 400 m. As speed of both trains u = 72 km h^{-1} = 20 ms^{-1} in same direction, hence their relative velocity u_{BA} = 0.

Let initial distance between the two trains be ‘S’ then train B covers the distance (S + 11) = (S + 800) m in time t = 50 s when accelerated with a uniform acceleration a = 1 m/s^{2}.

`:. (s+800) = u_(AB) xx t + 1/2 at^2`

`= 0 + 1/2 xx1xx(50)^2 = 1250 m`

=> S = 1250 - 800 = 450 m

and initial distance between guard of train B from driver of train A = 450 + 800 = 1250 m.