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 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 constants 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` = slope of the line `OA=v_"max"/t_1 rArr = v_"max"/alpha`
The slope of AB gives the deceleration `beta`
`therefore 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)/(alphabeta))`
`therefore v_"max"=((alphabeta)/(alpha+beta))t`

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