### Q.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 s^{-2}. 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, *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 = ut + (1/2)a_{1}t^{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*_{II} = ut + (1/2)at^{2}

= 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.