Answer
(a) $35.7\text{ mi/hr}$
Work Step by Step
(a)
Let $\omega$ be the angular speed and $v_t$ be the linear speed.
$r=\frac{25~in}{2}=12.5~in\times\frac{1~ft}{12~in}\times\frac{1~mi}{5,280~ft}=1.973\times10^{-4}\text{ mi}$
$\omega=480~\frac{rev}{min}\times\frac{2\pi~rad}{1~rev}\times\frac{60~min}{1~hr}=180,956.2\text{ rad/hr}$
$v_t=r\omega=1.973\times10^{-4}(180,956.2)=35.7\text{ mi/hr}$
