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Question: An elevator car whose floor to ceiling distance is equal to \[2.7{\text{ m}}\]starts ascending with ...

An elevator car whose floor to ceiling distance is equal to 2.7 m2.7{\text{ m}}starts ascending with constant acceleration1.2 m/s21.2{\text{ m/}}{{\text{s}}^2}, 2 s2{\text{ s}} after the start, a bolt begins falling from the ceiling of the car. The free fall time of the bolt is(g=9.8 m/s2)(g = 9.8{\text{ m/}}{{\text{s}}^2})?
(A) 2.79.8s\sqrt {\dfrac{{2.7}}{{9.8}}} {\text{s}}
(B) 5.49.8s\sqrt {\dfrac{{5.4}}{{9.8}}} {\text{s}}
(C) 5.48.6s\sqrt {\dfrac{{5.4}}{{8.6}}} {\text{s}}
(D) 5.411s\sqrt {\dfrac{{5.4}}{{11}}} {\text{s}}

Explanation

Solution

When an object is free falling, object is considered to have acceleration due to gravity. When there is relative motion of an object, with respect to the motion of the elevator, the net acceleration of the object changes accordingly. When two accelerations are acting in opposite directions the net acceleration of the object is the sum of the two accelerations. Otherwise, the net acceleration is the difference of the two accelerations.

Formula used:
The kinematic equation of displacement is written as,
y=vot+12at2y = {v_o}t + \dfrac{1}{2}a{t^2}
Here, yy is the displacement of the block, vo{v_o} is the initial velocity of the block, tt is the time and aa is the acceleration.

Complete step by step answer:
Understand that, bolt is falling in the opposite direction from the direction of the motion of the elevator. Therefore, the relative acceleration of a bolt is the sum of the acceleration due to gravity gg and acceleration of the elevator. And the bolt is at rest initially.
ab=g+ae{a_b} = g + {a_e}
Here, ab{a_b} is the acceleration of a bolt and ae{a_e} is the acceleration of an elevator.
Substitute 9.8 m/s29.8{\text{ m/}}{{\text{s}}^2} for gg and 1.2 m/s21.2{\text{ m/}}{{\text{s}}^2} for ae{a_e}

{a_b} = 9.8 + 1.2{\text{ m/}}{{\text{s}}^2} \\\ \Rightarrow{a_b}= 11{\text{ m/}}{{\text{s}}^2} \\\ \ $$ The kinematic equation of displacement is written as, $$y = {v_o}t + \dfrac{1}{2}a{t^2}$$ Here, $$y$$ is the displacement of the block, $${v_o}$$ is the initial velocity of the block, $$t$$ is the time and $$a$$ is the acceleration. Substitute $$0{\text{ m/s}}$$ for $${v_o}$$, $$11{\text{ m/}}{{\text{s}}^2}$$ for $$a$$ and $$2.7{\text{ m}}$$for $$y$$ $$\Rightarrow 2.7{\text{ m}} = (0{\text{ m/s}})t + \dfrac{1}{2}(11{\text{ m/}}{{\text{s}}^2}){t^2}$$ Rearrange for $$t$$ $$\therefore t = \sqrt {\dfrac{{5.4}}{{11}}} {\text{s}}$$ **Therefore, Option D is the correct choice.** **Note:** The relative acceleration of the bolt is calculated with respect to the motion of the ascending elevator using the expression for the relative acceleration. Then, the free fall time of the bolt is calculated using the kinematics equation of displacement. Bolt is considered at rest initially.