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 ?

Explanation

Initial speed of train A and train B, `v _(OA)= v_(OB) = 72` km/hr `=20 ms ^(-1)`
Suppose, distance between two train is x,
Now, for train `B, a _(B) =1 ms ^(-2) and t = 50` s
Distance covered by train B is `d _(B).`
`therefore d _(B) = v _(OB) t + 1/2 a _(B) t ^(2)`
`therefore d _(B) = 20 xx 50 + 1/2 xx 1 xx (50) ^(2)`
`= 100 0 + 1250`
`therefore d _(B)=2250 m`
Distance covered by train A is `d _(A)`
`d _(A) = v _(OA)-d_(A) = 2250- 1000 = 1250 m`
Second Method:
Both trains are moving with same speed initially Hence, `v _(AB)= v _(BA) =O`
Now for train `B, a _(B)=1 ms ^(-2) , v _(BA)= 0 and t = 50 s x (t) -x (o) =` ?
Now, `x (t) - x (o) = v_(BA) t + 1/2 a _(B ) t ^(2)`
`= O +1/2 xx 1 xx (50) ^(2) = 1250 m.`