A car accelerates from rest at a constant rate  α for some time after which it decelerates at a constant rate  β to come to rest. If the total time elapsed is t seconds, the total distance travelled is:

If we represent the given car motion in V-t graph then we obtain as followsArea under graph gives displacement   =12base heights=12(t)(h)=12tV0 here, wew can sayV0=αt1 and   V0=βt2    and     t1+t2=t⇒t=V0α+V0β=V01α+1β ∴V0=t1α+1β∴S=12tt1α+1β ⇒S=αβ2(α+β)t2