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Equations of motion problem

  1. Mar 24, 2012 #1
    1. The problem statement, all variables and given/known data

    So my main issue is with regards to when you integrate Newton's second law twice to get the position of a particle with respect to time. Why does everyone say that the first constant of your integration is initial velocity and second constant is initial position. Is there any logic behind that or is it just arbitary?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 24, 2012 #2
    Because the second derivative of position is acceleration.
     
  4. Mar 24, 2012 #3

    BruceW

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    Homework Helper

    It actually depends on what kind of function the acceleration is. But If the acceleration is a polynomial of time, then the constant of integration does equal the velocity at t=0.

    You can realise this logically. If the acceleration is a polynomial, which you then integrate, then what must be the value of the polynomial at t=0?

    P.S. welcome to physicsforums :)

    Edit: I mean 'what is the value of the integrated polynomial, without the constant of integration, at t=0' Hmm, maybe I asked for too many steps at once. First, start off with a polynomial, then integrate it.
     
    Last edited: Mar 24, 2012
  5. Mar 24, 2012 #4
    For a constant acceleration the position of the object is

    [itex]S=S{_0}+U{_0}t-\frac{1}{2}at^2[/itex]

    [itex]\frac {ds}{dt}=U{_0}-at[/itex]

    [itex]\frac {dv}{dt}=-a[/itex]
     
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