Rollback Distance Calculation for a Car on a Slope

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SUMMARY

The discussion focuses on calculating the rollback distance of a car traveling at 24.0 m/s up a 20.0-degree slope after running out of gas. The relevant equation used is Vf^2 = Vi^2 - 2a(g*sin(20)(Xf-X0), where Vf is the final velocity, Vi is the initial velocity, and g is the acceleration due to gravity. The user expresses uncertainty about the formula and the sine value of 20 degrees, which is approximately 0.34. The correct approach involves substituting the values into the equation to solve for Xf, the distance traveled up the slope before rolling back down.

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Homework Statement



A car traveling at 24.0m/s runs out of gas while traveling up a 20.0degree slope.
How far up the hill will it coast before starting to roll back down?

Homework Equations



Vf^2=Vi^2-2a(g*sin(20)(Xf-X0)

The Attempt at a Solution



I don't know if that is correct formula but I don't know how to solve for Xf
 
Physics news on Phys.org
Yes you may solve using this formula
 
when i enter in sin(20) my calculator is .34 is that on track? it doesn't seem right
 

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