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_(max)` is the maximum velocity attained and `t_(1)` is the time at which maximum velocity will occur, the velocity vs time graph can be The slope of line `OP`.
.
`alpha =(v_(max))/(t_(1))`
The slope of line `PQ`,
`beta=(v_(max))/(t-t_(1))`
From Eqs. (i) and (ii) we get (ii)
which gives `t_(1)=(betat)/(alpha+beta)` (iii)
Substituting value of `t_(1)` in Eq. (i), we get
a. `v_(max)=(alpha beta t)/(alpha+beta)`
b. Total displacement,`vec s` =Area of `vec t` v-t graph
`=(1)/(2) xxv_(max)xxt`
`=(1)/(2) xxd(alpha beta t)/(alpha+beta)xxt=(apha beta t^(2))/(2(alpha+beta))`.