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

Answer

B

Explanation

From `A` to `B` applying `sut +(1)/(2) at^(2)`, we get
`S_(1) =0 xx t_(1) +(1)/(2) alph t_(1)^(2) =(1)/(2) (alpha t_(1)) t_(1) =(1)/(2) v_(0)t_(1)` (1)
Sinmilarly from `B` to `C`:
`S_(2) =v_(0) t_(2) +(1)/(2) (-beta) t_(2)^(2) =v_(0)t_(2) -(1)/(2) (betat_(2)) t_(2) =(1)/(2) v_(0) t_(2) (2)
Now total distance travelled is
`S_(1) +S_(2) =(1)/(2) v_(0)t_(1) +(1)/(2) v_(0)t_(2) =(1)/(2) v_(0) (t_(1) _t_(2))`
`=(1)/(2) (alph beta t)/(2 alpha+beta) t =(alpha beta t^(2))/(2 (alpha+beta) t =(alpha beta t^(2))/(2(alpha +beta)`
Also, we can find (i) and (ii) by using the formula, `s=(u+v)/(2) t`.