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Deriving constant acceleration formula

  1. Apr 27, 2012 #1

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Apr 27, 2012 #2

    I like Serena

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    Hey!

    Your first line is correct.
    The second line is not (how did you get it?).
    The third line is correct again and would indeed follow from the first line (although with v and v0 instead of vf and vi).
     
  4. Apr 27, 2012 #3

    Femme_physics

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    Well, I integrated from Vfinal to Vinitial as per the integration formula for "X2" case. Then I integrated acceleration from zero till t. No point in integrating zero because it's zero. Yes point in integrating acceleration and time. That's how I got my equation.

    Oh, I made a mistake, it should be AC^2/1+1 multiplied by T^2/1+1
     
  5. Apr 27, 2012 #4

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    You appear to have integrated the left hand side to ##{v^2 \over 1+1}##.
    If you take its derivative with respect to v, you should get what you were integrating, but you won't.


    What's the derivative of ##{v^2 \over 1+1}## with respect to v?


    Furthermore, what's the derivative of ##{a_c^2 \over 1+1} \cdot {t^2 \over 1+1}## with respect to t?
     
    Last edited: Apr 27, 2012
  6. Apr 27, 2012 #5

    Femme_physics

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    Don't I? Are you sure?..


    Here->

    http://img14.imageshack.us/img14/6018/yepyepm.jpg [Broken]



    [/QUOTE]

    http://img820.imageshack.us/img820/232/56382450.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  7. Apr 27, 2012 #6

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    Almost good.
    It should be just "v".
    Check your calculation.

    Anyway, your integral is ∫dv, which is the same as ∫1.dv.
    So you are taking the integral of just "1" with respect to v.
    Do you know an expression with "v", that has "1" as its derivative?



    You have the derivative of the t2 part right. :)

    However, "a" is a constant, meaning it does not change, nor does a2 change.



    Btw, the end result should match the integral that you are taking, which is just "a".
    It is not "a.t".
     
    Last edited by a moderator: May 5, 2017
  8. May 5, 2012 #7

    Femme_physics

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  9. May 5, 2012 #8

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    Cheater! ;)

    But you didn't do it right.
    You should have calculated ∫dv instead of ∫v dv.

    And even though wolframalpha did not understand the boundaries v0 and v, you should still have those in your formula.
     
  10. May 5, 2012 #9

    Femme_physics

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    Ohhh...you're right!

    http://img545.imageshack.us/img545/76/imacheater.jpg [Broken]

    Oh I'm such a cheater ;) I love this software!!

    If I weren't caught up with my non-calculus studies I'd have invested in figuring it out on my own. I just wanted to bring this topic to a closure for now and not leave it hanging.
    I'll get my calculus straight because I'll need it for a first degree anyway, but all in due time :) Thanks for showing me the software and the way to the solution.
     
    Last edited by a moderator: May 6, 2017
  11. May 5, 2012 #10

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    One thing left: the boundaries v0 and v.

    $$∫_{v_0}^v dv = [ v ]_{v_0}^v = v - v_0$$
     
  12. May 5, 2012 #11

    Femme_physics

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    ahh..wasn't sure how to add boundaries in wolfram-- thanks
     
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