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# A person in an elevator is accelerating upwards with an acceleration of $2m{s^{ - 2}}$, tosses a coin vertically upwards with a speed of $20m{s^{ - 1}}$. After how much time will the coin fall into his hands? (Take $g = 10m{s^{ - 2}}$)A) $\dfrac{5}{3}s$B) $\dfrac{3}{{10}}s$C) $\dfrac{{10}}{3}s$D) $\dfrac{3}{5}s$

Last updated date: 20th Sep 2024
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Hint: Imagine yourself in an elevator which is accelerating upwards. You feel heavy right and why does this happen. Try to recall what concept of mechanics this phenomenon is based on. This will help us solve the problem.

Let’s go to the example we provided in the hint. When in an elevator which is going upwards or if we say accelerating upwards, we feel heavy. As Newton has proposed, our weight is the product of our mass and the amount of our acceleration.
Well when we stand still, we experience acceleration due to gravity $(g)$. If we move vertically upwards or downwards this acceleration due to gravity changes. Our masses are constant so this change in our acceleration inside an elevator along with acceleration due to gravity makes us feel heavy. Similarly when we move vertically downwards in an elevator, we feel light.
Now this concept can be applied to our problem in hand.
When the elevator is moving vertically up, our net acceleration increases.
If the acceleration due to gravity is $g$ and the acceleration of the lift is $a$ upwards, the person inside the elevator will experience this $a$ acceleration downwards along the direction of $g$.
So the net acceleration will be,
${a_{net}} = g + a$
Substituting these values we get,
${a_{net}} = 10 + 2 = 12m{s^{ - 2}}$ ………. (1)
Now let’s consider the flipped coin. When it returns to the person’s hand its net displacement is zero inside the elevator.
For this scenario, applying the equation of motion, we get
$s = ut + \dfrac{1}{2}a{t^2}$
Where $s$ is the displacement of an object
$u$ is its initial velocity
$t$ is the time of motion
$a$ is the acceleration for the mentioned time frame
As the net displacement is zero,
$\Rightarrow 0 = ut - \dfrac{1}{2}a{t^2}$
The minus sign here implements that the net acceleration of the body is against the direction of the coin that is, the acceleration is opposite to the direction of motion of the coin.
The question mentions that the coin was tossed with the velocity of $20m{s^{ - 1}}$ and we have estimated the value of the net acceleration that is, $12m{s^{ - 2}}$.
Substituting these values we get,
$\Rightarrow 0 = 20t - \dfrac{1}{2} \times 12 \times {t^2}$
$\Rightarrow 3{t^2} - 10t = 0$
Solving this quadratic equation, we get the value of $t$
$\Rightarrow t = \dfrac{{10}}{3}s$
The time required by the coin to return to the hands of the person is $\dfrac{{10}}{3}s$

So the correct answer is option (C).

Note: The direction of motion is an important aspect of solving these problems. Note that however the elevator is accelerating upwards, we have to consider the acceleration of the person in the lift or the coin tossed inside it. Be careful around direction notations and system of units.