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 the 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)-betat_(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)(alpha)/(alpha+beta)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)
`therefore S_(t)=(v_(m)^(2))/(2alpha)=(alphabeta^(2))/(2(alpha+beta)^(2))t^(2)` and `s_(2)`
`=(alpha^(2)beta)/(2(alpha+beta)^(2))t^(2)`
`therefore` Total distance s=`s_(1)+s_(2)`
`=(alphabetat^(2))/(2(alpha+beta)^(2))(alpha+beta)=(alphabetat^(2))/(2(alpha+beta))`