lift performs the first part of ascent with uniform acceleration a and the remainder with uniform retardation 2a. The lift starts from rest and finally comes to rest. If t is the time of ascent. Find the height ascended by lift.

Let OAB be the velocity-time graph of the lift. The ordinate at A (i.e., AM) represents maximum velocity.Total distance travelled = Area of the  ΔOAB = 12×OB×AMAM = v, OM = t1, t1+t2= OB = t, MB = t2 ∴    ΔOAB = 12×tv = hOr vt = 2h   ..(i)Now  vt1 = a  or  t1 = va   .....(ii)and  vt2 = 2a  or  t2 = v2a   .....(iii)Adding (ii) and (iii)t = t1+t2 = va+v2a = 3v2a = 32a×2htor  at2 = 3h   ⇒ h = at23