A 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.

Options

A. `(at^2)/(4)`
B. `(at^2)/(3)`
C. `(at^2)/(2)`
D. `(at^2)/(8)`

Answer

B

Explanation

Let `OAB` be the velocity-time graph of the lift. Time ordinate at `A` (i.e., `AM`) represents maximum velocity.
Total distance travelled
=area of the `Delta OAB = (1)/(2) xx OB xx AM`
`AM = v, OM = t_1, t_1 + t_2 = OB = t, MB = t_2`
`:. Delta OAB = (1)/(2) xx tv = h`
or `vt = 2h` ....(i)
Now `(v)/(t_1) = a` or `t_1 = (v)/(a)`...(ii)
and `(v)/(t_2) = 2 a` or `t_2 = (v)/(2 a)` ...(iii)
Adding (ii) and (iii)
`t = t_1 + t_2 = (v)/(a) +(v)/(2a) = (3v)/(2a) = (3)/(2a) xx (2h)/(t)`
or `at^2 = 3h rArr h = (at^2)/(3)`.