A car starting from rest, travels with uniform acceleration x and then comes to rest with uniform retardation y. If the total time of travel is t sec, the maximum velocity of the car is

In accelerated motion,V=u+xt1=0+xt1∴ V=xt1....(1)When the car stops (after retardation)0=v−yt−t1where t =total time,t1 = time of acceleration, andt2 = time of retardation∴v=yt−t1....(2)From eqs. (1) and (2) xt1=yt−t1=yt−yt1Solving, we get t1=yt/(x+y)From eq,(1), v=xyt(x+y)