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

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 decleratioin are constants the velocity time graph is a straight line as shown in the figure with +ve slope for acceleration and -ve slope for decleration. From the graph, the slope of the line OA gives the acceleration `alpha`.
`alpha ="slope of the line "OA=(v_("max"))/(t_(1)) rArr t_(1)=(V_("max"))/(alpha)`
the slope of AB gives the decleration `beta`
`beta="slope of AB"=(v_("max"))/(t_(2)) rArr t_(2)=(v_("max"))/(beta)`
`t=t_(1)+t_(2)=(v_("max"))/(alpha)+(v_("max")).(beta)`
`t=v_(max) ((alpha+beta)/(alpha beta))`

`v_("max")=((alpha beta)/(alpha+beta))t`
(b) Displacement =area under the v-t graph =area of `triangle OAB`
`=1/2 ("base") ("height") =1/2 t v_("max") =1/2 t ((alpha beta t)/(alpha +beta))=1/2 ((alpha beta t^(2))/(alpha +beta))`