Question

An elevator in which a man is standing is moving upward with a constant acceleration of $2m{s^{ - 2}}.$ At some instant, when the speed of the elevator is $10m{s^{ - 1}}$ the man drops a coin from a height of $1.5m$ .Find the time taken by the coin to reach the floor.
(A) $\dfrac{1}{{\sqrt 3 }}\sec$
(B) $\dfrac{1}{2}\sec$
(C) $\dfrac{1}{{\sqrt 2 }}\sec$
(D) $1\sec$

Answer 1

In order to solve this question, we will use the concept of net acceleration on a body when affected by multiple acceleration and as we know as soon as a body is released in an elevator which is moving upward it will have same initial velocity as of the elevator so net velocity with respect to elevator is zero, here we will use the newton equation of motion to calculate the time taken by coin to reach the floor.
Formula used: If u , a and t are the initial velocity, acceleration and time taken by a body to cover a distance of S meter we have, $S = ut + \dfrac{1}{2}a{t^2}$
If ${a_1}{\kern 1pt} {\kern 1pt} and{\kern 1pt} {\kern 1pt} {a_2}$ are the acceleration of the body acting in same direction then, net acceleration on the body is ${a_{net}} = {a_1} + {a_2}$

Complete step-by-step solution:
According to the question we have given that acceleration of an elevator is in upward direction of magnitude $2m{s^{ - 2}}.$ so, as soon as coin starts to fall it will be affected by acceleration due to gravity in downward direction as well as the acceleration of $2m{s^{ - 2}}.$ will also act on coin in downward direction.
So, net acceleration on coin is given by ${a_{net}} = g + 2$ where $g = 10m{s^{ - 2}}$
${a_{net}} = 12m{s^{ - 2}}$
$u = 0$ initial velocity at the moment the coin is dropped will be zero because the coin will have the same velocity as the elevator and in the same direction, so net velocity with respect to the elevator will be zero.
$S = 1.5m$ Distance the coin will cover.
Let “t” be the time coin takes to fall, then using the formula
$S = ut + \dfrac{1}{2}a{t^2}$
$\Rightarrow 1.5 = 0 + \dfrac{1}{2}(12){t^2}$
$\Rightarrow {t^2} = \dfrac{3}{{12}}$
$\Rightarrow t = \dfrac{1}{2}\sec$
Hence, the correct option is (B) $\dfrac{1}{2}\sec$.

Note: It should be remembered that, here the initial velocity of the coin is the relative velocity of coin with respect to elevator where velocity of coin and elevator has same magnitude of $10m{s^{ - 1}}$ in the same direction and also, if the elevator was moving in downward direction then, net acceleration of coin will $(g - a)m{s^{ - 2}}$ acceleration on coin will be in direction opposite to that of elevator due to newton third law of motion which states that, every action has equal and opposite reaction.