
An elevator is accelerating upward at a rate of $$6\,ft\,{\sec ^{ - 2}}$$ when a bolt from its ceiling falls to the floor of the lift (Distance $ = 9.5\,feet$). The time (in seconds) taken by the falling bolt to hit the floor is (take $g = 32\,ft\,{\sec ^{ - 2}}$)
A. $\sqrt 2 $
B. $\dfrac{1}{{\sqrt 2 }}$
C. $2\sqrt 2 $
D. $\dfrac{1}{{2\sqrt 2 }}$
Answer
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Hint: We will use the distance formula to calculate the time taken by the falling bolt to hit the floor. At first, we will calculate the acceleration due to gravity acting on the object by subtracting the value of acceleration from the given value of acceleration due to gravity. Also, here the initial velocity of the bolt is zero.
Formula used:
The formula used for calculating the time taken by the bolt to hit the floor is given below
$s = ut + \dfrac{1}{2}a{t^2}$
Here, $s$ is the distance covered by the object, $u$ is the initial velocity of the object, $t$ is the time taken by the object to travel a certain distance and $a$ is the acceleration due to gravity acting on the object.
Complete step by step answer:
The following are the terms that are given in the question.Distance covered by the bolt from the top to the floor of the lift is, $s = 9.5\,feet$. Now, the acceleration due to gravity acting on the bolt is,
$a = 32 - 6$
$ \Rightarrow \,a = 26\,ft\,{\sec ^{ - 2}}$
Now, for calculating the time taken by the bolt to fall on the floor of the lift, we will use the distance formula which is given below,
$s = ut + \dfrac{1}{2}a{t^2}$
Now, for initial velocity $u = 0$, the above equation becomes
$ \Rightarrow \,9.6 = \dfrac{1}{2} \times 26 \times {t^2}$
$ \Rightarrow \,\dfrac{{9.5 \times 2}}{{26}} = {t^2}$
$ \Rightarrow \,{t^2} = \dfrac{{19}}{{26}}$
$ \Rightarrow \,t = \sqrt {\dfrac{{19}}{{26}}} $
$ \Rightarrow \,t = \sqrt {0.5} $
$ \therefore \,t = \dfrac{1}{{\sqrt 2 }}\,\sec $
Therefore, the time taken by the bolt to fall on the floor of the lift is $\dfrac{1}{{\sqrt 2 }}\,\sec $.
Hence, option B is the correct option.
Note:As we know that the time is frame-independent quantity, therefore, the time of free fall of bolt from any frame will remain the same. Also, note that values of all the three parameters i.e. initial velocity, distance and acceleration are with respect to the time frame.Here, it is not necessary to convert the units of parameters given in the question.
Formula used:
The formula used for calculating the time taken by the bolt to hit the floor is given below
$s = ut + \dfrac{1}{2}a{t^2}$
Here, $s$ is the distance covered by the object, $u$ is the initial velocity of the object, $t$ is the time taken by the object to travel a certain distance and $a$ is the acceleration due to gravity acting on the object.
Complete step by step answer:
The following are the terms that are given in the question.Distance covered by the bolt from the top to the floor of the lift is, $s = 9.5\,feet$. Now, the acceleration due to gravity acting on the bolt is,
$a = 32 - 6$
$ \Rightarrow \,a = 26\,ft\,{\sec ^{ - 2}}$
Now, for calculating the time taken by the bolt to fall on the floor of the lift, we will use the distance formula which is given below,
$s = ut + \dfrac{1}{2}a{t^2}$
Now, for initial velocity $u = 0$, the above equation becomes
$ \Rightarrow \,9.6 = \dfrac{1}{2} \times 26 \times {t^2}$
$ \Rightarrow \,\dfrac{{9.5 \times 2}}{{26}} = {t^2}$
$ \Rightarrow \,{t^2} = \dfrac{{19}}{{26}}$
$ \Rightarrow \,t = \sqrt {\dfrac{{19}}{{26}}} $
$ \Rightarrow \,t = \sqrt {0.5} $
$ \therefore \,t = \dfrac{1}{{\sqrt 2 }}\,\sec $
Therefore, the time taken by the bolt to fall on the floor of the lift is $\dfrac{1}{{\sqrt 2 }}\,\sec $.
Hence, option B is the correct option.
Note:As we know that the time is frame-independent quantity, therefore, the time of free fall of bolt from any frame will remain the same. Also, note that values of all the three parameters i.e. initial velocity, distance and acceleration are with respect to the time frame.Here, it is not necessary to convert the units of parameters given in the question.
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