A train starts from station `A` with uniform acceleration `a_(1)`. For some distance and then goes with uniform retardation `a_(2)` for some more distance to come to rest at station `B`. The distance between stations `A` and `B` is `4 km` and the train takes `1//5 h` to compete this journey. If accelerations are in km per minute unit, then show that `(1)/(a_(1)) +(1)/(a_(2)) =x`. Find the value of `x`.

Explanation

`A` to `C`: `v=0+a_(1)t_(1) …(i)`
`v^(2)=0+2a_(1)s_(1) …(ii)`
`C` to `B`: `0=v-a_(2)t_(2) …(iii)`
`0=v^(2)-2a_(2)s_(2) …(iv)`
Given
`t_(1)+t_(2)=4 …(v)`
`s_(1)+s_(2)=4 …(vi)`
From (i), (ii) and (v)
`v/a_(1)+v/a_(2)=4`
`v=4/(1/a_(1)+1/a_(2))`
From (ii), (iv) and (vi)
`v^(2)/(2a_(1))+v^(2)/(2a_(2))=4`
`v^(2)/2(1/a_(1)+1/a_(a))=4`
`(1)/(2).(4^(2))/((1/a_(1)+1/a_(2))^(2)).(1/a_(1)+1/a_(2))=4`
`1/a_(1)+1/a_(2)=2`
or `s_(1)+s_(2)=1/2 a_(1)t_(1)^(2)+1/2a_(2)t_(2)^(2)=1/2a_(1)t_(1)t_(1)+1/2a_(2)t_(2)t_(2)`
`=v/2 (t_(1)+t_(2))`
`4=v/2(4)impliesv=2`
`t_(1)+t_(2)=v/a_(1)+v/a_(2)=v(1/a_(1)+1/a_(2))`
`4=2(1/a_(1)+1/a_(2))implies1/a_(1)+1/a_(2)=2`