Question

If a lift started from floor 1 ascents with uniform acceleration $a$ and retards with uniform retardation 2$a$ to stop at floor 2, in a time interval of $t$. Then the height between floors will be

• 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

The correct option is B $\frac{a{t}^2}{3}$

From the question, the a - t graph can be

Initial velocity = 0 and final velocity = 0
Area of a - t graph = 0
$a{t}_1-2a(t-t_1)=0$
$t_1-2t+2t_1=0$
$t_1=\frac{2t}{3}$

And Max velocity is at $t_1$, and it is $\frac{2at}{3}$ as the lift started from rest and moved with uniform acceleration $a$.

Area of v - t graph is the displacement

Displacement $=$ Area
$=\frac{1}{2}\times t\times \frac{2at}{3}$
$=\frac{1}{3}a{t}^2$