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Velocity/Acceleration relation w/ constants

  1. Feb 16, 2015 #1
    1. The problem statement, all variables and given/known data

    The velocity of a particle is related to its position by: v2 = w2 (A2 - x2) where w and A are constants. Show that the acceleration is given by: a=-w2x



    2. Relevant equations


    3. The attempt at a solution

    a= v* dv/dt

    v=(A2w2-x2w2)1/2

    dv/dt= 1/2 (A2w2-x2w2)-1/2 * -2xw2

    v * dv/dt = 1/2*-2xw2 * (A2w2-x2w2)1/2/ (A2w2-x2w2)-1/2

    -w2x

    I think this is the proper way of showing the acceleration, but I'm not 100% with my chain rule and cancellations.
     
  2. jcsd
  3. Feb 16, 2015 #2

    haruspex

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    I think not.
     
  4. Feb 16, 2015 #3
    Sorry, I looked at v* dv/dx. Now definitely not sure how to go about it.
     
  5. Feb 16, 2015 #4

    haruspex

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    What do you get if you simply differentiate the given equation wrt time?
     
  6. Feb 16, 2015 #5
    Wouldn't we have to differentiate wrt x? As t doesn't appear in the equation.
     
  7. Feb 16, 2015 #6

    haruspex

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    Either will work, but you're right that differentiating wrt x is quicker here - so do it.
     
  8. Feb 16, 2015 #7
    After eliminating the ^2,

    v= w * √(a^2 - x^2)

    dv/dx= -xw/ √(a^2-x^2)
     
  9. Feb 16, 2015 #8

    haruspex

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    No need - just differentiate as is.
     
  10. Feb 16, 2015 #9
    I think I understand the derivation, but I do not understand the notation for v^2 and a

    v^2*dv/dx
    2v/dx? = 2w(a^2-x^2)+ w^2(2a -2x)

    but since 2w and 2a are constants,

    so 2v/dx?= w^2*-2x

    dividing by 2

    v/dx= -xw^2

    and, however notation allows, the v/dx? is equivalent to a.

    therefore a= -xw^2
     
  11. Feb 16, 2015 #10

    haruspex

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    No, you want ##\frac d{dx} v^2##. Apply the chain rule.
     
  12. Feb 16, 2015 #11
    2v * dv/dx= 0*(a^2-x^2)+ w^2*(0 -2x)= w^2*-2x

    2v dv/dx = -2x*w^2

    v dv/dx = -x*w^2

    where
    v dv/dx= a?
     
  13. Feb 16, 2015 #12

    haruspex

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    Yes.
     
  14. Feb 17, 2015 #13
    I appreciate the help, it's nice to be walked through the proper notation.
     
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