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Finding the expression for the horizontal acceleration

  1. Sep 17, 2011 #1
    Hey folks its me again! The problem at hand gives me a potential energy function ( V(x) ) that defines the height of a particle in motion. Next to the problem is a picture of a curvy graph that is y vs x. It asks to find the expression for the horizontal acceleration. This is what I did..

    [PLAIN]http://img101.imageshack.us/img101/3631/prob2.gif [Broken]

    I started out with the conservation of Energy equation and solved for "v." Then I differentiated with respect to "t". I do not think this is correct however because "V" does not depend on "t" so, how could I differentiate "v" without getting zero? (Since no "t's" are in the equation and nothing depends on "t." Also, why would this be the horizontal acceleration? as far as I know, the "v" in the kinetic energy equation only gives speed, with no direction. So how could I get the horizontal direction?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 18, 2011 #2

    NascentOxygen

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    Staff: Mentor

    I don't understand the question about horizontal acceleration, but to answer your immediate question, could you use the chain rule? dV/dt = dV/dx * dx/dt
    and dx/dt = velocity
     
  4. Sep 18, 2011 #3
    Yeah I don't quite understand it either, all it says is a particle is moving on a friction-less wire with a function defining its height V(x). Then it says to find the horizontal acceleration.

    Hey I think I could! And dV/dx = F. Again I am not sure if this expression is correct though :\
     
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