Question

An elevator is accelerating upward at a rate of $6ft/{\sec ^2}$ when a bolt from its ceiling falls to the floor of the lift (Distance $= 9.5feet$). The time (in seconds) taken by the falling bolt to hit the floor is: (take $g = 32ft/{\sec ^2}$).

Answer 1

To solve this question, we need to find the total acceleration of the elevator considering that it is accelerating in the upward direction. We will consider this total acceleration as the acceleration of the falling bolt. After that we will use one of the equations from the laws of motion to find the required answer.

Formula used:
$s = u + \dfrac{1}{2}a{t^2}$
where $s$ is the displacement, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time.

Complete step by step answer:
We are given that an elevator is accelerating upward at a rate of $6ft/{\sec ^2}$. Therefore, the total acceleration by which the bolt will fall on the floor of the lift is : $6 + 32 = 38ft/{\sec ^2}$
The initial velocity of the bolt is zero.
So we have: $s = 9.5feet$, $u = 0ft/\sec$, $a = 38ft/{\sec ^2}$, $t = ?$
s = u + \dfrac{1}{2}a{t^2} \\\ \Rightarrow {t^2} = \dfrac{{2\left( {s - u} \right)}}{a} \\\ \Rightarrow {t^2} = \dfrac{{2\left( {9.5 - 0} \right)}}{{38}} \\\ \Rightarrow {t^2} = \dfrac{{9.5}}{{19}} \\\ \Rightarrow {t^2} = 0.5 \\\ \Rightarrow {t^2} = \dfrac{1}{2} \\\ \therefore t = \dfrac{1}{{\sqrt 2 }}\sec \\\
Thus, the time taken by the falling bolt to hit the floor is $\dfrac{1}{{\sqrt 2 }}$ seconds.

Note: Here, we have seen that when the elevator is accelerating in upward direction, the acceleration by which the bolt is falling is the total acceleration of the acceleration and the gravitational acceleration. But, if the accelerator is moving in a downward direction, then the acceleration of the falling bolt is taken as the difference between the gravitational acceleration and acceleration of the elevator. Thus, it is important to consider the direction of the elevator to solve this question.