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.

Options

A. `(alph beta)/(2(alpha+beta))t`
B. (alpha beta)/(alpha+beta) t`
C. `(2alph beta)/(alph +beta)t`
D. `(4 alpha beta)/(alpha+beta) t`

Answer

B

Explanation

.
From `A` to `B` applying `v =u +at`, we get
`v_(0)=0 +at_(1) rArr t_(1) t_(1) -v_(0)//a`
From `B` to `C`, again applying `v=u +at`, we get `0=v_(0) bt_(2) rArr t_(2) =v_(0)//b`
Given `t_(1) +t_(2) =t rArr (v_(0))/(alpha) + (v_(0))/(beta) =t rArr v_(0) =(alpha beta t)/(alpha +beta)`
`v_(0)` is the maximum velocity attained.