Question

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.

• A. $\frac{at^2}{4}$
• B. $\frac{at^2}{3}$
• C. $\frac{at^2}{2}$
• D. $\frac{at^2}{8}$

Answer 1

Answer: (B)

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 $\Delta OAB = \frac{1}{2} \times OB \times AM$
$AM = v, OM = t_1, t_1 + t_2 = OB = t, MB = t_2$
$\therefore \Delta OAB = \frac{1}{2} \times tv = h$
or $vt = 2h \,\,...(i)$
Now $\frac{v}{t_1} = a$ or $ t_1 = \frac{v}{a} \,\,...(ii)$
and $\frac{v}{t_2} = 2a $ or $t_2 = \frac{v}{2a} \,\,...(iii)$
Adding (ii) and (iii)
$t = t_1 + t_2 = \frac{v}{a} + \frac{v}{2a}$
$ = \frac{3v}{2a} = \frac{3}{2a} \times \frac{2h}{t}$
or $at^2 = 3h $
$\Rightarrow h = \frac{at^2}{3}$