A bus 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, evaluate.
(a)the maximum velocity achieved and (b)the total distance travelled graphically.

Explanation

(a) Let `t_(1)` be the time of acceleration and `t_(2)` that of deceleration of the bus.
The total time is `t=t_(1)+t_(2)`.
Let `v_(max)` be the maximum velocity.
As the acceleration and deceleration are constant the velocity time graph is a straight line as shown in the figure with `+ve` slope for acceleration and `-ve` slope for deceleration.
From the graph,
the slope of the line OA gives the acceleration `alpha`.
`therefore alpha` = slpe of the line `OA=(v_(max))/(t_(1))impliest_(1)=(v_(max))/(alpha)` the slope of AB gives the deceleration `beta`
`therefore beta` = slope of `AB=(v_(max))/(t_(2))impliest_(2)=(v_(max))/(beta)`
`t=t_(1)+t_(2)=(v_(max))/(alpha)+(v_(max))/(beta)`
`t=v_(max)((alpha+beta)/(alphabeta))`
`therefore v_(max)=((alphabeta)/(alpha+beta))t`
(b) Displacement = area under the v-t graph
= area of `DeltaOAB`
`=(1)/(2)("base")("height")=(1)/(2)tv_("max")`
`=(1)/(2)t((alphabetat)/(alpha+beta))=(1)/(2)((alphabetat^(2))/(alpha+beta))`