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 elapsed is t seconds. Then evalute (a) the maximum velocity reached and (b) the total distance travelled.

Explanation

Let` v_(m)` be the maximum velocity and let `t_(1)` be time taken to attain it . Then using
`v=u+at` we get `v_(m)=alpha t_(1)`……(1)
Let `t_(2)` be the time taken by the car to stop under retardation `beta`. Then
`0=v_(m) beta t_(2)"or"v_(m)beta t_(2)`....(2)
Eqns.(1)and(2) given
`t_(2)/t_(1)=alpha/beta "or" t_(2)/t_(1)+1-alpha/beta+1 "or" t/t_(1)=(alpha+beta)/beta "or" t_(1)=beta/(alpha+beta)t`
Substituting this value of `t_(1)`in eqn.(1),
`v_(m)=(alphabeta)/(alpha+beta)t`t
Now , let `s_(1)`be the distance travelled during acceleration and let `s_(2)`be the distance travelled during retardation . Then using the equation
`v^(2)=u^(2)+2as`,we get `v_(m)^(2)=2alphas_(1)` ........(3)
and `0=v_(m)^(2)-2betas_(2)`.......(4)
`:. s_(1)=v_(m)^(2)/(2alpha)=(alphabeta^(2))/(2(alpha+beta)^(2)t^(2)"and"s_(2)`
`=(alpha^(2)beta)/(2(alpha+beta))^(2)t^(2)`
`:. `Total distance `s=s_(1)+s_(2)`
`=(alphabetat^(2))/(2(alpha+beta))^(2)(alpha+beta)=(alphabetat^(2))/(2(alpha+beta))`