Physics - Kinematics Question with Solution | TestHub

When the elevator is stationary, the coin falls under gravity. When the elevator moves with constant velocity, it is an inertial frame of reference. The relative acceleration of the coin with respect to the elevator floor remains $g$. Thus, the time taken for the coin to reach the floor is the same in both cases.

Explanation:

Let $h$ be the height from which the coin is dropped to the elevator floor.

When the elevator is stationary, the acceleration of the coin is $g$.

Using the kinematic equation $s=ut+\frac{1}{2}a{t}^2$, with $u=0$, $s=h$, and $a=g$:

$$h=0\cdot t_1+\frac{1}{2}g{t}_1^2$$

$$h=\frac{1}{2}g{t}_1^2$$

$$t_1=\sqrt{\frac{2h}{g}}$$

When the elevator is moving with a constant velocity (either upwards or downwards), the elevator is an inertial frame of reference. This means that Newton's laws of motion apply exactly as they would in a stationary frame. The relative acceleration of the coin with respect to the elevator floor is still $g$.

From the perspective of an observer inside the elevator, the coin is dropped from a height $h$ with an initial relative velocity of zero, and it accelerates downwards at $g$ relative to the elevator floor.

So, the time $t_2$ taken for the coin to reach the floor is given by the same equation:

$$h=\frac{1}{2}g{t}_2^2$$

$$t_2=\sqrt{\frac{2h}{g}}$$

Comparing $t_1$ and $t_2$, we find that $t_1=t_2$.

The constant velocity of the elevator does not affect the relative acceleration between the coin and the elevator floor, nor does it affect the time it takes for the coin to cover the vertical distance $h$ relative to the elevator. This is a fundamental principle of inertial frames of reference.