The distance between two stations is 40 km. A train takes 1 hour to travel this distance. The train, after starting from first station, moves with constant acceleration for 5 km, then it moves with constant velocity for 20 km and finally its velocity keeps on decreasing continuously for 15 km and it stops at the other station. Find the maximum velocity of the train.

Explanation

As the train starts from A station, its velocity at A is 0 & B, it is `v_(x) = v .`
`therefore` The velocity at `A = v _(0) = 0` and velocity at
`B = v_(x) = v`
`therefore` For distance AB,
Average velocity` = (v _(0) +v )/( 2) = (v + v _(x))/(2) = (v_(x))/(2) `
`therefore` Time for covering distance `AB = t _(1) = (10)/(v_(x))" "...(1)`
Now, velocity `v _(x)` from B to C is constnat.
`therefore` Time for covering distance BC.
`t _(2) = (BC)/("constant velocity") = (20)/(v _(x)) " "...(ii)`
Now velocity at `C = v _(0) = v _(x) and ` velocity at `D = v =0`
`therefore` For distance CD,
Average velocity `= (v _(0) + v )/(2) = (v_(x) + 0)/(2) = (v _(x))/(2)`
Now time for covering the distance CD
`t _(3) = (CD)/("Average velocity") = (CD)/((v _(x))/(2))= (15 xx 2 )/(v_(x)) = (30)/(v_(x))`
Now if t = time in covering the distance A to D.
i.e. 40 km and t =1 hr. then from,
`t = t _(1) + t_(2) +t_(3) = (10)/(v _(x)) + (20)/(v_(x)) + (30)/(v _(x))`
`therefore v _(x) = 60 kmh ^(-1)=` maximum velocity.