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Help with integrating the kinematic equations

  1. Mar 13, 2010 #1

    Just wondering if someone can help me make sense of something. I realize it's probably a simple problem, but my math skills aren't the best and I can't see through it.

    I'm trying to end up with this kinematic equation:

    X= Xo + Vo(t-to) + 1/2(a)(t-to)^2

    And to do this the book tells me to insert this equation:
    V=Vo + a(t-to)

    Into this one:

    X=Xo + ∫ vdt (integrated from to -> t)

    And then solve this:

    X=Xo + Vo∫ dt (from to->t) + a ∫ (t - to)dt (from to -> t)

    I am stuck on the second integration.

    How do I integrate (t-to)dt from to --> t ??

    I realize that there are a few methods on how to get these equations, it seems like everybook I look at does it in a different way, but I want to understand this method.

    (also, as you can probably tell, I'm new here and I'm not sure the best method of writing math symbols on a computer, can anyone sugest how to do it?)

  2. jcsd
  3. Mar 13, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    How would you solve this?: ∫x dx
  4. Mar 13, 2010 #3
    This would be 1/2[ X^2 - Xo^2]

    And I know that's what I need to get for the kinematic equation but I don't see how.

    If I have:

    a∫(t-to)dt = a[ ∫ t dt -∫ to dt] <--- here I assume to is a constant

    So integrating from to-->t give me:

    a[ (t^2- to^2)/2 - to(t-to) ]

    And this seems to lead to no where. arg.

    Thanks for your quick reply
  5. Mar 13, 2010 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Expand that out and simplify. (Hint: Factor the resulting expression.)
  6. Mar 13, 2010 #5
    ah, ok. Makes sense now. I was just making a mistake with the algebra and getting confused.

    Thanks so much your speedy help, and your patience with me.
    This is a really great website.

    But does anyone have any ideas on how to type math symbols a little less akwardly than I'm doing???
  7. Mar 13, 2010 #6
    Indeed! :)

    You can use the LaTeX typesetting available on these forums. Check out this post: https://www.physicsforums.com/showthread.php?t=8997 for more details
  8. Mar 13, 2010 #7
    Cool, I'll check that out for sure.

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