This problem involves rotational kinematics with uniform angular acceleration. We will use the standard kinematic equations for constant acceleration, adapted for rotational motion.

First, define the given variables and convert units for consistency. Initial angular velocity: $\omega_0 = 900$ rev/min Final angular velocity: $\omega_f = 800$ rev/min Time interval: $t = 5$ s

Converting angular velocities from rev/min to rev/s:

$$\omega_0 = 900 \frac{\text{rev}}{\text{min}} \left( \frac{1 \text{ min}}{60 \text{ s}} \right) = 15 \text{ rev/s}$$ $$\omega_f = 800 \frac{\text{rev}}{\text{min}} \left( \frac{1 \text{ min}}{60 \text{ s}} \right) = \frac{40}{3} \text{ rev/s}$$

[Q1] Find the angular acceleration $\beta$. The relationship between initial and final angular velocities over a time interval $t$ with constant angular acceleration $\beta$ is given by:

$$\omega_f = \omega_0 + \beta t$$

Solving for the angular acceleration $\beta$:

$$\beta = \frac{\omega_f - \omega_0}{t}$$

Substituting the numerical values:

$$\beta = \frac{\frac{40}{3} \text{ rev/s} - 15 \text{ rev/s}}{5 \text{ s}} = \frac{\frac{40 - 45}{3}}{5} \frac{\text{rev}}{\text{s}^2} = -\frac{1}{3} \text{ rev/s}^2$$

[Q2] Find the total number of revolutions during these 5 seconds. The angular displacement, $\Delta \theta$, can be found using the average angular velocity for uniform acceleration:

$$\Delta \theta = \frac{1}{2}(\omega_0 + \omega_f)t$$

Substituting the numerical values:

$$\Delta \theta = \frac{1}{2} \left( 15 \text{ rev/s} + \frac{40}{3} \text{ rev/s} \right) (5 \text{ s})$$ $$\Delta \theta = \frac{1}{2} \left( \frac{45 + 40}{3} \right) (5) \text{ rev} = \frac{1}{2} \left( \frac{85}{3} \right) (5) \text{ rev} = \frac{425}{6} \text{ rev}$$

[Q3] Find how many more seconds it will take for the wheel to stop rotating. Let $t'$ be the additional time to stop. The initial angular velocity for this part is $\omega'_0 = \omega_f = 40/3$ rev/s, and the final angular velocity is $\omega'_f = 0$. The angular acceleration $\beta$ is constant. Using the same kinematic equation, $\omega'_f = \omega'_0 + \beta t'$: Solving for the time $t'$:

$$t' = \frac{\omega'_f - \omega'_0}{\beta}$$

Substituting the numerical values:

$$t' = \frac{0 \text{ rev/s} - \frac{40}{3} \text{ rev/s}}{-\frac{1}{3} \text{ rev/s}^2} = 40 \text{ s}$$