An automobile engine slows down from 4500 rpm to 1200 rpm in 2.5 s . Calculate \( (a) \) its angular accel assumed constant, and \( (b) \) the total number of revolutions the engine makes in this time. \( (-140,120) \)

To calculate the angular acceleration, we can use the formula: \[ \alpha = \frac{\Delta \omega}{\Delta t} \] where \(\Delta \omega = \omega_f - \omega_i\) (final angular velocity minus initial angular velocity) and \(\Delta t\) is the time duration. First, we need to convert the RPM values to radians per second (rad/s). \(\omega_i = 4500 \times \frac{2\pi}{60} \approx 471.24 \, \text{rad/s}\) \(\omega_f = 1200 \times \frac{2\pi}{60} \approx 125.66 \, \text{rad/s}\) So, \(\Delta \omega = 125.66 - 471.24 \approx -345.58 \, \text{rad/s}\) \(\Delta t = 2.5 \, \text{s}\) Now, substituting these values in: \[ \alpha = \frac{-345.58}{2.5} \approx -138.23 \, \text{rad/s}^2 \] For the total number of revolutions, we can use the kinematic equation: \[ \theta = \omega_i t + \frac{1}{2} \alpha t^2 \] Here, we need to convert time \(t\) into proper units: \[ \theta = 471.24 \times 2.5 + \frac{1}{2} \times (-138.23) \times (2.5)^2 \] Calculating: \[ \theta \approx 1178.1 - 429.06 \approx 749.04 \, \text{rad} \] Finally, converting radians to revolutions: \[ \text{Total revolutions} = \frac{749.04}{2\pi} \approx 119.05 \] Thus, the engine slows down with an angular acceleration of approximately \(-138.23 \, \text{rad/s}^2\) and completes about \(119\) revolutions in this time.