The driver of a train moving at a speed ` v_(1)` sights another train at a disane ` d`, ahead of him moving in the same direction with a slower speed ` v_(2)`. He applies the brakes and gives a constant teradation ` a` to his train. Show that here will be no collision if ` d gt (v_(1) -v_(2))^(2) //2 a`.

Explanation

For no collision the speed of the train must be reduced to `v_(2)` before the trains meet, i.e., the velocity of train relative to good train `v_(1) - v_(2)` should become zero before the trains meet.
Now if `s` is the displacement of train relative to goods trains when it comes to rest relative to goods train due to application of breakes, we have, from 3rd equation of motion, i.e.,
`v^(2) = u^(2) + 2as`,
`0^(2) = (v_(1) - v_(2))^(2) - 2as`
or `s = ((v_(1) - v_(2))^(2))/(2a)`
The train will not collide if
`d ge s, " i.e., " d ge ((v_(1) - v_(2))^(2))/(2a)`
Alternative solution: The collision will be avoided if the velocity of first trains `v_(1)` just before touching the other is reduced to `v_(2)` (the two trains being at same speed can never collide).Thus if `a` is the retardation, the time taken by first trains to reduce velocity from `v_(1) to v_(2)` will be given by
`v_(2) = v_(1) - at, " i.e., " t = (v_(1) - v_(2))//a`
In this time the train will move a distance `d_(1)` such that
`v_(2)^(2) = v_(1)^(2) - 2ad_(1), " i.e., " d_(1) = (v_(1)^(2) - v_(2)^(2))//2a`
During the same time, the other train will move a distance
`d_(2) = v_(2) t = v_(2) (v_(1) - v_(2))//a`
If the initial distance between the two trains is `d`, the collision will be avoided if
`d_(1) lt d_(2) + d`
i.e., `(v_(1)^(2) - v_(2)^(2))/(2a) le d + (v_(2) (v_(1) - v_(2)))/(a)`
which on solution gives `d ge (v_(1) - v_(2))^(2)//2a`.