Answer
(a) The length of the incline is 63 meters.
(b) It takes 7.1 seconds to reach the bottom of the incline.
Work Step by Step
(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.
