Answer:
[tex]\alpha =1.5\ rad/s^2[/tex]
Explanation:
Given that,
Initial angular velocity, [tex]\omega_i=0[/tex]
Final angular speed, [tex]\omega_f=6\ rad/s[/tex]
Angular displacement, [tex]\theta=12\ rad[/tex]
We need to find the angular acceleration of the tire. We can find it using the third equation of rotational kinematics. So,
[tex]\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{\omega_f^2-\omega_i^2}{2\theta}\\\\\alpha =\dfrac{6^2-0^2}{2\times 12}\\\\\alpha =1.5\ rad/s^2[/tex]
So, the angular acceleration of the tire is equal to [tex]1.5\ rad/s^2[/tex].
