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

The acceleration directed down the incline is $g~sin(\theta)$. Therefore; $d = \frac{v^2-v_0^2}{2a}$ $d = \frac{0-(30~m/s)^2}{(2)(-9.80~m/s^2)~sin(10^{\circ})}$ $d = 260~m$ The car will coast 260 meters up the hill before starting to roll back down.