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Calculating the velocity decrease going up an incline

  1. Nov 13, 2014 #1
    I know there are many factors that go into this. But in general how would one go about calculating the angular velocity for a car going up an incline. What I want to know in particular is if a car say is driving at a consistent speed of 60mph on an even plane and then hits an incline up wards at 45* going the speed the person is going at what rate is the car's acceleration decreased. I'd think it must be a gradient going slower and slower as the incline increases in steepness. If someone could explain this and then give a mathematical equation that would be greatly appreciated.

    Thanks
     
  2. jcsd
  3. Nov 13, 2014 #2
    Assuming its a circular incline going steeper and becomes vertical
    As the car covers angle x in the incline
    Conserve energy(neglecting friction)
    ##(1/2)(mv^2) =mgR(1-cosx)##
    From there you can find v(x)

    Can you make the question more clearer
     
  4. Nov 13, 2014 #3
    Well, if you discount the rolling resistance of the tires/drive train, at the fixed output power of the engine when travelling 60 MPH, the force vectors are all in the same horizontal plane: The force of air resistance is matched by the engine's output. For a given object, air resistance is proportional to velocity squared.

    On the inclined plane, you must break the velocity vector into the horizontal and vertical components. (At 45 degrees , the horizontal and vertical speed components are equal, and both equal the velocity divided by the square root of 2.)

    On an inclined plane, the fixed engine output must be sufficient to lift the car's mass against the force of gravity. Since overcoming the drag at 60 MPH would be on the order of 20 HP/15KW, it is entirely possible that this 'fixed' output would not be enough to lift the car at all, unless the speed dropped to near zero. (in real life, given most vehicles gear reduction range, it's probable that a car would not be able to go up this steep an incline at all.) At 45 degrees, the force vector due to gravity is split into the horizontal and vertical components by dividing mass by sqrt(2), also. And, the air resistance vector, which is parallel to the plane, would need to be included needs to be included in the solution as well.

    I really think you should take a look at the force diagrams and explanations at the lesson on inclined planes here. Diagrams are helpful in understanding the forces involved.
     
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