Question

A transparent lift A is going upwards with velocity $20m{s^{ - 1}}$ and retarding at the rate of $8m{s^{ - 2}}$ . Second transparent lift B is located in front of it and is going down at $10m{s^{ - 1}}$ with retardation of $2m{s^{ - 2}}$ . At the same instant a bolt from the ceiling of lift A drops inside lift A. Height of car of lift A is $16m$ . Find the time taken by bolt to hit the floor of A.
(A) $2s$
(B) $3s$
(C) $4s$
(D) $6s$

Answer 1

To solve this question, we need to use the second kinematic equation of motion. We have to find out the values of the initial velocity, the displacement, and the acceleration of the bolt with respect to the frame of reference of the lift A. Substituting these in the second equation of motion, we will get the required value of the time taken.

Formula used: The formula used to solve this equation is given by
$s = ut + \dfrac{1}{2}a{t^2}$ , here $s$ is the displacement, $u$ is the initial velocity, $a$ is the acceleration, and $t$ is the time.

Complete step-by-step solution
We consider the situation from the frame of reference of the lift A.
Let us take the vertically downward direction as a positive direction.
Since the bolt drops from the ceiling of lift A, so before dropping it will have the same velocity as that of the lift. Therefore in the frame of reference of the lift, the initial velocity of the bolt should be equal to zero, that is,
$u = 0$ .................(1)
SInce the bolt drops from the ceiling of the lift and drops onto the floor of the lift. So with respect to the lift A, the total displacement of the bolt must be equal to the height of the car of the lift, that is,
$s = h$ .................(2)
According to the question, the height of the car of the lift is given to be equal to $16m$ , that is,
$h = 16m$
So from (2) we have
$s = 16m$ .................(3)
The velocity of the lift A is given in the upward direction. So the retardation of the lift must be in the downward direction. According to the question, the acceleration of the lift is
$A = + 8m{s^{ - 2}}$ .................(4)
We know that an object falls under the acceleration due to gravity, which is in the vertically downward direction. So the acceleration due to gravity of the bolt is given by
$g = + 10m{s^{ - 2}}$ .................(5)
Therefore, the acceleration of the bolt with respect to the lift A is given by
$a = g - A$
Putting (4) and (5) in the above equation, we have
$a = 10 - 8$
$\Rightarrow a = 2m{s^{ - 2}}$ .................(6)
From the second kinematic equation of motion, we have
$s = ut + \dfrac{1}{2}a{t^2}$
Substituting (1), (3), and (6) in the above equation we get
$16 = 0t + \dfrac{1}{2} \times 2 \times {t^2}$
$\Rightarrow {t^2} = 16$
Taking square root both the sides, we finally get
$t = 4s$
Thus, the time taken by bolt to hit the floor of A is equal to $4s$ .
Hence, the correct answer is option C.

Note
The information regarding the lift B which is given in the question is totally irrelevant to the solution. It is just the extra information which is given to confuse us. Also, the value of the time could also be obtained by considering the values of the velocities and the displacement of the bolt in the ground frame of reference. But solving this question in the frame of reference of the lift is much easier.