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Two trains A and B of length 300 m each ...

Two trains A and B of length 300 m each are moving on two parallel tracks with a uniform speed of `54" km h"^(–1)` in the same direction with A ahead of B. The driver of B decides to overtake A and accelerates by `2" ms"^(–2)`. If after 25s, the guard of B just crosses the driver of A, what was the original distance between them ?

Text Solution

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Step I: For train `A, u =54" km h"^(-1) =54xx(5)/(18)" ms"^(-1) =15" ms"^(-1)`
For t=25s, a=0 using `s=ut +(1)/(2) at^(2)`, we get `s_(1) =15xx25+0=375 m`
Thus, the distance traveled by trains A in 25s is 375m.
Step II: For train `B, u =54" km h"^(-1) =15" ms"^(-1)`
For t =25 s, using `s=ut +(1)/(2) at^(2)," we get "s_(2) =15xx25+(1)/(2) xx2xx2(25)^(2)=275+625=1000 m`.
Step III: Let `s_(o)` be the original separation between two trains. Then the distance traveled by train B in 25 sec, so that it just crosses the head of train A, is given by
`s_(2) =s_(o) +s_(1)` + length of train A + length of train B.
or `1000 = s_(0) + 375 + 300 + 300`
? `s_(0) = 25 m`
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