#### 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.