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

  • Thread starter spartan55
  • Start date
  • #1
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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.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Delphi51
Homework Helper
3,407
10
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.
 
  • #3
4
0
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!
 
  • #4
Delphi51
Homework Helper
3,407
10
Most welcome!
 

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