Starting from rest, a train traveIs a certain distance with a uniform acceleration `alpha`. Then it traveIs with a uniform retardation `beta` and finally comes to rest again. If the totaI time of motion is t,find (i) the maximum velocity attained and (ii) the totaI distance travelled by the train.

Explanation

(i) Let `t_(1)` be the time taken to travel a distance `s_(1)` with an acceleration a and `t_(2)` be the time taken to travel a farther distance `s_(2)` with retardation `beta`. Let the maximum velocity attained by the train be v.
Here t = `t_(1)+t_(2)`
`:.` For the motion of the train with acceleration `alpha`.
v = `at_(1) " ""or", t_(1) = (v)/(alpha) " " cdots(1)`
and `v^(2) = 2alphas_(1)" ""or", s_(1) = (v^(2))/(2alpha) " "cdots(2)`
Similarly, for the motion of the train with retardation `beta`,
`0 = v - betat_(2) " ""or",t_(2) = (v)/(beta) " "cdots(3)`
and `0^(2) = v^(2)-2betas_(2)" ""or",s_(2) = (v^(2))/(2beta) " "cdots(4)`
From (1) and (3) we get,
`t = t_(1)+t_(2) = v ((1)/(alpha)+(1)/(alpha))" ""or",v = (alphabetat)/(alpha+beta)`
(ii) From (2) and (4), the total distance travelled,
`s = s_(1)+s_(2)=(v^(2))/(2)((1)/(alpha)+(1)/(beta))`
=`((alphabetat)/(a+beta))^(2)xx(1)/(2)((alpha+beta)/(alphabeta)) = (alphabetat^(2))/(2(alpha+beta)).`