Question

A lift performs the first part of its ascent with uniform acceleration, a, and the remaining with uniform retardation, 2a. If, t, is the time of ascent, the depth of the shaft is:
A. $\dfrac{{a{t^2}}}{4}$
B. $\dfrac{{a{t^2}}}{3}$
C. $\dfrac{{a{t^2}}}{2}$
D. $\dfrac{{a{t^2}}}{8}$

Answer 1

Hint: We know the equations of motion, and they should be applied in this question to find the depth of the shaft i.e. $S = ut + \dfrac{1}{2}a{t^2}$. Moreover, in these questions we need to let quantities such as time or displacement.

Complete step-by-step answer:

Formula used : $S = ut + \dfrac{1}{2}a{t^2}$

Given, time of ascent=t
Acceleration in first part =a
Acceleration in the second part = 2a
Let the time taken in 1st part = ${t_1}$
Let the time taken in 2nd part= ${t_2}$

Since, the acceleration in part2 is twice that of initial, so the time taken in second part will be half of that of time taken in 1st part.

${t_2} = \dfrac{{{t_1}}}{2}$
Total time, $t = {t_1} + {t_2}$
Therefore, ${t_1} = \dfrac{2}{3}t$
In the first part:
$S = ut + \dfrac{1}{2}a{t^2}$
Since u=0
${S_1} = \dfrac{1}{2}a{t^2}$……..(i)
In the second part:
$S = ut + \dfrac{1}{2}a{t^2}$
${S_2} = \dfrac{1}{2}(2a){t_2}^2$
${S_2} = \dfrac{1}{2}(2a){t_2}^2$……ii)
Let the total depth be S

Therefore, $S = {S_1} + {S_2}$
Using equation (i) and (ii),
$
  S = \dfrac{1}{2}at_1^2 + \dfrac{1}{2}(2a)t_2^2 \\
  S = \dfrac{1}{2}at_1^2 + \dfrac{1}{2}(2a){(\dfrac{{{t_1}}}{2})^2} \\
$
$S = \dfrac{3}{4}at_1^2$
$
  S = \dfrac{3}{4}a{(\dfrac{{2t}}{3})^2} \\
  S = \dfrac{{a{t^2}}}{3} \\
$

Hence, the answer to this question is $S = \dfrac{{a{t^2}}}{3}$.
Hence, the correct option is B.

Note: In this type of question we need to find the basic relationships between quantities such as time or displacement and then solve using the equations of motion.