A car accelerates from rest at a constant rate ` alpha ` for some time after which it decelerates at a constant rate ` beta` to come to rest. If the total time lapse is ` t` seconds , evauate.
(i) maximum velocity reached , and
(ii) the total distance travelled .

Explanation

Let the maximum velocity be v.
`A` to `C`: (constant acceleration)
`v=0+alphat_(1) …(i)`
`v^(2)=0+2alphas_(1) …(ii)`
`C` to `B`: (constant retardation)
`0=v-betat_(2) …(iii)`
`0=v^(2)-2beta s_(2) …(iv)`
From (i), `t_(1)=v/alpha`, from (ii), `t_(2)=v/beta`
`t_(1)+t_(2)=t(given)`
`v/alpha+v/beta=t`
`v=(alphabetat)/(alpha+beta)`
from (i), `s_(1) =v^(2)/(2d)`, from (ii), `s_(2)=v^(2)/(2beta)`
Total distance `s=s_(1)+s_(2)=v^(2)/2(1/alpha+1/beta)`
`=1/2((alphabetat)/(alpha+beta))^(2) ((alpha+beta)/(alpha beta))`
`=(alphabetat^(2))/(2(alpha+beta))`