How far up the hill will it coast before starting to roll back down?

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A car traveling at 22.0 m/s coasts up a 24.0-degree slope after running out of gas. The acceleration parallel to the slope is calculated as -3.986 m/s² using trigonometry. By applying the kinematic equation, the distance traveled up the slope is determined to be approximately 60.712 m, although the poster expresses concern that this seems too large. An alternative energy approach is suggested, equating kinetic energy at the bottom to potential energy at the top, which simplifies the calculation. The discussion highlights the effectiveness of using energy methods over basic kinematics for this problem.
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Homework Statement


A car traveling at 22.0 m/s runs out of gas while traveling up a 24.0deg slope. How far up the hill will it coast before starting to roll back down?


Homework Equations


(v_final)^2 = (v_initial)^2 + 2*(a_parallel)*(x_final - x_initial)
...and some trig

The Attempt at a Solution


I know the acceleration that is important is the acceleration parallel to the plane which is found by some trig manipulation:
(a_y)sin(24.0deg) = a_parallel, where (a_y) = -9.8 m/s^2.
Thus: a_parallel = -3.986 m/s^2.
Now would I use (v_initial) = 22.0 m/s and plug this with a_parallel, x_initial = 0, and v_final = 0 m/s into the above kinematic equation to get: 60.712 m ?
For some reason this just seems like too big of an answer. Any help appreciated. Thanks.
 
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That answer looks good to me!
It is interesting to check it with an energy approach. KE at the bottom equals PE at the top so
mgh = .5*mv²
Solve for h, then convert that to a distance along the slope using trigonometry.
 
Wow it is a lot easier with the energy approach. We haven't talked about that yet, we're still on basic kinematics. Thanks for your help!
 
Most welcome!
 
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