We establish a coordinate system where the upward direction is positive. The motion of the bolt is analyzed from two reference frames: the elevator's frame and a fixed (ground) frame. Key variables are the elevator's acceleration $a_e = 1.22 \text{ m/s}^2$, its initial velocity $v_{e,0} = 2.44 \text{ m/s}$, the ceiling-to-floor distance $h = 2.74 \text{ m}$, and the acceleration due to gravity $g = 9.8 \text{ m/s}^2$.

[Q1] Calculate the time required for the bolt to fall from the ceiling to the floor.

To find the time of flight inside the elevator, we analyze the bolt's motion relative to the elevator. The initial velocity of the bolt relative to the elevator is zero, as they share the same initial velocity:

$v_{rel,0} = v_{bolt,0} - v_{elevator,0} = 0$.

The acceleration of the bolt relative to the elevator is the difference between their absolute accelerations:

$a_{rel} = a_{bolt} - a_{elevator} = (-g) - a_e = -(g+a_e)$.

The relative displacement of the bolt is from the ceiling to the floor, $\Delta y_{rel} = -h$. Using the kinematic equation for constant acceleration, $\Delta y = v_0 t + \frac{1}{2}at^2$:

$$-h = (0)t + \frac{1}{2}(-(g+a_e))t^2$$

Solving for the time, $t$:

$$t = \sqrt{\frac{2h}{g+a_e}}$$

Substituting the given values:

$$t = \sqrt{\frac{2(2.74 \text{ m})}{9.8 \text{ m/s}^2 + 1.22 \text{ m/s}^2}} = \sqrt{\frac{5.48 \text{ m}}{11.02 \text{ m/s}^2}} \approx 0.705 \text{ s}$$

[Q2] Determine the displacement and the total distance traveled by the bolt relative to a fixed post outside the elevator during this time.

This analysis is performed in the fixed (ground) reference frame. At $t=0$, the bolt has an initial upward velocity $v_{b,0} = v_{e,0} = 2.44 \text{ m/s}$ and is subject to gravitational acceleration, $a_b = -g$.

Displacement: The bolt's displacement, $\Delta y_b$, over the time interval $t$ is given by:

$$\Delta y_b = v_{b,0}t + \frac{1}{2}a_b t^2 = v_{e,0}t - \frac{1}{2}gt^2$$

Substituting the values, including the time $t$ calculated in Q1:

$$\Delta y_b = (2.44 \text{ m/s})(0.705 \text{ s}) - \frac{1}{2}(9.8 \text{ m/s}^2)(0.705 \text{ s})^2 \approx 1.72 \text{ m} - 2.44 \text{ m} \approx -0.716 \text{ m}$$

The negative sign indicates a net downward displacement of $0.716 \text{ m}$.

Total Distance Traveled: The bolt initially travels upward, reaches a maximum height, and then falls. The total distance is the sum of the distance traveled up and the distance traveled down.

1. Upward distance to the peak ($d_{up}$): Using $v_f^2 = v_i^2 + 2a\Delta y$, where $v_f=0$ at the peak: $$d_{up} = \frac{0^2 - v_{b,0}^2}{2a_b} = \frac{-v_{e,0}^2}{2(-g)} = \frac{v_{e,0}^2}{2g}$$ $$d_{up} = \frac{(2.44 \text{ m/s})^2}{2(9.8 \text{ m/s}^2)} \approx 0.304 \text{ m}$$
2. Downward distance ($d_{down}$): This is the magnitude of the displacement from the peak to the final position. The final position is $\Delta y_b$ below the start, and the peak is $d_{up}$ above the start. $$d_{down} = d_{up} - \Delta y_b$$
3. Total distance ($d_b$): $$d_b = d_{up} + d_{down} = d_{up} + (d_{up} - \Delta y_b) = 2d_{up} - \Delta y_b$$ Substituting the calculated values: $$d_b = 2(0.304 \text{ m}) - (-0.716 \text{ m}) \approx 0.608 \text{ m} + 0.716 \text{ m} = 1.324 \text{ m}$$