__Question 3.7__

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 ?

__Solution:__

*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 X 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) X 1 X (50)

^{2}= 2250 m

Length of both trains = 2 X 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.

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