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Kinematics motion-setting up instantaneous velocity function

  1. Dec 20, 2013 #1
    1. The problem statement, all variables and given/known data

    An object starts moving in a straight line from position xi, at time t =0, with velocity vi. Its acceleration is given by a = ai + bt, where ai and b are constant. Find expressions for
    a)instantaneous velocity
    b) position as functions of time.



    2. Relevant equations

    none

    3. The attempt at a solution
    Capture.JPG
     
  2. jcsd
  3. Dec 20, 2013 #2
    The equation you have there (xf=xi+vi t+0.5 ai t^2) is only true for constant acceleration. You should workout the instantaneous velocity by integrating the acceleration with respect to time. THen do the same thing to get the position.
     
  4. Dec 20, 2013 #3
    But the acceleration that is given, a = ai + bt, is already a function of time.
     
  5. Dec 20, 2013 #4
    How is the derivative of the velocity with respect to time related to the acceleration?
     
  6. Dec 20, 2013 #5
    I'm lost. Could someone provide an exposition?
     
  7. Dec 20, 2013 #6
    Differentiating displacement with respect to time gives me instantaneous velocity, no?
     
  8. Dec 20, 2013 #7
    [tex]\frac{dv}{dt}=a=a_i+bt[/tex]
    Do you know how to integrate this differential equation?
     
  9. Dec 20, 2013 #8
    I do.

    Capture.JPG

    but why can't I set up a displacement equation as a function of time as like in my first post and differentiate it?
    What's the rational?
     
    Last edited: Dec 20, 2013
  10. Dec 20, 2013 #9
    The rationale is that the equation in your first post is incorrect if the acceleration is a function of time. So it's of no use in this problem. You need to start with the acceleration and work backwards by getting the velocity first and then the displacement. Compare what you get doing it this way your the equation in your first post.
     
  11. Dec 20, 2013 #10
    I have inserted the antiderivative in the above post after you quote.
     
  12. Dec 20, 2013 #11
    Yes. That's the increase in velocity. If you want the velocity itself, you need to add in the initial velocity. Then, integrate the resulting equation to get the location x.
     
  13. Dec 20, 2013 #12


    Capture.JPG
     
  14. Dec 20, 2013 #13
    This was done correctly, but the problem statement asked for the position as a function of time.
     
  15. Dec 20, 2013 #14
    Alright. This is delta x. To get xf, xf= xi+ delta x.
     
  16. Dec 20, 2013 #15
    Yeah!!! Excellent!!!
     
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