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. $\frac{a{t}^2}{4}$
• B. $\frac{a{t}^2}{3}$
• C. $\frac{a{t}^2}{2}$
• D. $\frac{a{t}^2}{8}$

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 $a{t}^2=3h$

$\Rightarrow h=\frac{a{t}^2}{3}$