## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) The acceleration down the incline is $g~sin(\theta)$. Therefore; $d = \frac{v^2-v_0^2}{2a}$ $d = \frac{(15~m/s)^2-(3.0~m/s)^2}{(2)(9.80~m/s^2)~sin(10^{\circ})}$ $d = 63~m$ The length of the incline is thus 63 meters. (b) $t = \frac{v-v_0}{a}$ $t = \frac{(15~m/s)-(3.0~m/s)}{9.8~m/s^2~sin(10^{\circ})}$ $t = 7.1~s$ It takes 7.1 seconds to reach the bottom of the incline.