Question

1. determine the angular acceleration of a tire that accelerates from 750 rpm to 350 rpm in 6 seconds. \\( 13.96 \frac{\text{rad}}{\text{s}^2} \\) \\( 3.49 \frac{\text{rad}}{\text{s}^2} \\) \\( -6.98 \frac{\text{rad}}{\text{s}^2} \\) \\( 6.98 \frac{\text{rad}}{\text{s}^2} \\)

Explanation:

Step1: Convert initial RPM to rad/s

$\omega_0 = 750 \times \frac{2\pi}{60} = 25\pi \ \text{rad/s}$

Step2: Convert final RPM to rad/s

$\omega = 350 \times \frac{2\pi}{60} = \frac{35\pi}{3} \ \text{rad/s}$

Step3: Apply angular acceleration formula

$\alpha = \frac{\omega - \omega_0}{t} = \frac{\frac{35\pi}{3} - 25\pi}{6}$

Step4: Calculate the result

$\alpha = \frac{\frac{35\pi - 75\pi}{3}}{6} = \frac{-40\pi}{18} \approx -6.98 \frac{\text{rad}}{\text{s}^2}$

Answer:

C. $-6.98 \frac{\text{rad}}{\text{s}^2}$