A train begins its journey from station A and stops at station B after 45 min. C is a certain point between A and B where the train attains its maximum velocity of 50 km.`"h"^(-1)` . If the train travels from A to C with a uniform acceleration and from C to B with a uniform retardation, calculate the distance between A and B.

Explanation

Let the train start from A with a uniform acceleration `a_(1)` and reach C in time `t_(1)`. From here it travels to B in time `t_(2)` with a uniform retardation `a_(2)`. The maximum velocity of the train is v at the point C. Let AC = `s_(1)` and CB = `s_(2)` So for the motion from A to C, `v = a_(1)t_(1) "and " s_(1)=(1)/(2)a_(1)t_(1)^(2)` and for the motion from C to B, `0 = v-a_(2)t_(2)" ""or"v = a_(2)t_(2)`
and `s_(2) = vt_(2)-(1)/(2)a_(2)t_(2)^(2) = a_(2)t_(2)^(2)-(1)/(2)a_(2)t_(2)^(2) = (1)/(2) a_(2) t_(2)^(2)`
So the distance between A and B ,
s = AB = AC + CB = `s_(1)+s_(2)`
`=(1)/(2)a_(1)t_(1)^(2)+(1)/(2)a_(2)t_(2)^(2)`
`= (1)/(2)vt_(1)+(1)/(2)vt_(2)" "[because v = a_(1)t_(1) = a_(2)t_(2)]`
`= (1)/(2)v(t_(1)+t_(2)) = (1)/(2)xx50xx(3)/(4)`
`[because v = 50 "km.h"^(-1),t_(1)+t_(2) = 45 "min" = (3)/(4)`h]
= 18.75 km