A car acceleration form 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 evaluate (a) the maximum velocity attained and (b) the total distance travelled.

Explanation

(a) Let the car accelerates for time `t_(1)` and decelerates for time `t_(2)`.
Then
`t = t_(1) + t_(2) ….. (1)`
From the corresponding velocity-time graph graph

`alpha` = slope of line OA ` = (V_(max))/(t_(1))`
or `t_(1) = (V_(max))/(alpha)` ....... (ii)
Also,
`beta` = - slope of line AB ` = (v_(max))/(t_(2))`
or `t_(2) = (v_(max))/(beta)" "...(iii)`
From equ, (i), (ii) and (iii) we get,
`(v_(max))/(alpha) + (v_(max))/(beta) = t rArr v_(max) ((alpha + beta)/(alphabeta)) = t`
`rArr v_(max) = (alphabetat)/(alpha + beta)`
(b) Total distance covered = displacement = Area under v -t graph
` = (1)/(2) xx t xxv_(max) = (1)/(2) xx t xx (alphabetat)/(alpha + beta)`
or Distance covered ` = (1)/(2) ((alphabetat^(2))/(alpha + beta))`