Complete question is: You pull your car into your driveway and stop. The drive shaft of your car engine, initially rotating at 2400 rpm,slows with a constant rotational acceleration of magnitude 30 rad/s². How long does it take for the drive shaft to stop turning?
Answer:
8.37 s
Step-by-step explanation:
Initial rotational velocity, ω₀ = 2400 rpm = 2400 × 2π/60 = 251.2 rad/s
Final velocity, ω = 0
rotational deceleration, α =- 30 rad/s²
Use the first equation of rotational motion:
![t = (\omega - \omega_o)/(\alpha)](https://img.qammunity.org/2021/formulas/physics/high-school/zy6n8pxj117qgeb8ayg0c12b61uchs5u27.png)
Substitute the values:
![t = (0 - 251.2)/(-30)=8.37 s](https://img.qammunity.org/2021/formulas/physics/high-school/kbssdpvzkvntfyo7qtleveyador9a4erh2.png)