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Integration Newtons Second Law

  1. Feb 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi

    In a book they state
    [tex]
    \frac{dv}{dt} = a \quad \Rightarrow \quad v(z) = \sqrt{2a(z-z_{max})}
    [/tex]
    I am trying to reproduce this. Here is what I have so far:

    dv/dt = (dv/dz)(dz/dt) = v(dv/dz) = a

    Since a is constant (I assume?), I get
    [tex]
    \int vdv = \int adz \quad \Rightarrow\quad \frac{1}{2}v^2 + C = az
    [/tex]
    Here C denotes a constant. What should I do from here?

    Best,
    Niles.
     
  2. jcsd
  3. Feb 17, 2012 #2
    what is z?

    if [itex]\frac{dv}{dt} = a[/itex] then [itex]\int dv = \int a\ dt \ \rightarrow v = \int a\ dt[/itex]
     
  4. Feb 17, 2012 #3

    tiny-tim

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    Hi Niles! :smile:
    erm :wink:

    you're there! :smile:
     
  5. Feb 17, 2012 #4
    Thanks. But how is it seen that C=2azmax?
     
  6. Feb 17, 2012 #5

    Pengwuino

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    Gold Member

    You really might want to give us a lot more context on what you're doing. What is z? What is [itex]z_{max}[/itex]? If you tell us what the problem is, we can even tell you if the acceleration is constant.
     
  7. Feb 17, 2012 #6
    It is from a book on how to slow atoms with a Zeeman slower. Here [itex]z[/itex] is the coordinate (longitudinal, along the magnetic field), and [itex]z_{max}[/itex] I believe is the longitudinal coordinate, where the velocity is zero. Sorry, I should have stated that at first.
     
  8. Feb 17, 2012 #7

    tiny-tim

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    Hi Niles! :smile:
    v= 0 when z = zmax, so C = azmax :wink:

    (the "2" would be for a different C)
     
  9. Feb 18, 2012 #8
    Thanks. It is very kind of all of you to help me.

    Best,
    Niles.
     
  10. Feb 18, 2012 #9
    Is it correct that there are other ways to define C? E.g. as the initial velocity? In that case I would say that when z=0, then C=-0.5v2initial, so I get
    [tex]
    2az = v^2 - v^2_{initial}
    [/tex]
     
  11. Feb 18, 2012 #10

    tiny-tim

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    Yes, C can be anything, and can be re-named later.

    We often get C in an integration, then we tidy it up, and we find we have something awkward like 2π/C …

    so we call that C (or vo or zo) instead! :wink:

    (in this case, I think it was C/2)
     
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