A car accelerates from rest at a constant rate a for some time and then decelerates at a constant rate `b` to come to rest. The total time elapsed is `t`.
Total distance travelled by the car is .

Explanation

Let the car acceleratea for time acelerate for time `t_(1)` and decelerate for time `t_(2)` then
`t=t_(1)+t_(2) " "…..(i)`
and corresponding velocity-time graph will be as show in fig.

From the graph `alpha` = slop of line `OA=(v_("max"))//(t_(1)) or t_(2)=(v_("max"))/(alpha)" ".....(ii)`
and `beta` =- slop of line `AB=(V_("max"))/(t_(2)) or t_(2) =(V_("max"))/(beta) " "......(iii)`
From Eqs . (i) , (ii) and (iii) `(v_("max"))/(alpha)+(v_("max"))/(beta)=t or v_("max") ((alpha+beta)/(alpha beta))=t`
or ` v_("max")=(alpha beta t)/(alpha +beta)`
(b) Total distance =area under v-t graph `=(1)/(2)xxt xx v_("max") =(1)/(2)xx t xx(alpha betat)/(alpha+beta)`
Distance `=(1)/(2)((alpha beta t^(2))/(alpha+ beta))`
Note. This problem can also be solved by using equations of otion (v=u+at, etc.)