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Differential equations particle motion

  1. Sep 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Hey all. I have F(v) and I'm looking for x(t).

    F(v) = -bV2

    I'm a physics undergrad student who hasn't done diff. equations in a while, so I'm very unsure of my work so far... but here it is.

    3. The attempt at a solution

    I write F=ma as F(v)=mdv/dt:

    -bV2=mdv/dt

    -m∫ dv/v2 = -b ∫dt

    -m/v = -bt
    m/v = bt

    v = m/bt = dx/dt

    ∫ dx = m/b ∫ dt/t

    Finally x(t) = m/b*ln(t)

    Does this look good?
     
  2. jcsd
  3. Sep 18, 2009 #2

    rock.freak667

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    don't forget the constant.

    Otherwise it looks alright. Also remember that ma can be written as mv dv/dx as well
     
  4. Sep 18, 2009 #3
    I forgot to add that the initial position and initial velocity are both zero so I don't need to worry about constants. Thanks!
     
  5. Sep 18, 2009 #4

    Redbelly98

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    Are you sure the initial velocity is zero?

    Differentiate that expression. Do you get dx/dt=0 at t=0???
     
  6. Sep 19, 2009 #5
    I misinformed you all, the particle has an initial velocity Vo. My bad. Even still I see that if I put 0 for t in my equation, it is undefined, when it should be zero.

    I don't really know how to take it from here, any hints? How do I take into account the constant?
     
  7. Sep 19, 2009 #6

    Redbelly98

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    This is where things go awry, since the equation is contradictory to v=v0 when t=0.

    Yes, exactly. When you integrate, add a constant. What constant will give v=v0 when t=0?
     
  8. Sep 19, 2009 #7
    How does this seem?

    x(t) = m/b*ln(t) + Vo*t
     
  9. Sep 19, 2009 #8

    Redbelly98

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    Unfortunately, that does not give x(0)=0, due to the ln(t) term.

    Also, the initial velocity appears to be infinite, again due to the ln(t) term.

    If you show your work, we can help figure out what went wrong.
     
  10. Sep 19, 2009 #9
    Here's my work:

    -bV2=mdv/dt

    -m∫ dv/v2 = -b ∫dt

    (note: so the integral of 1/v^2 is -1/v, and this integral is from initial v to final v) so:

    bt = m(1/v - 1/vo) where vo is initial velocity

    bt/m = 1/v - 1/vo

    m/bt + vo = v

    m/bt + vo = dx/dt

    ∫(m/bt + vo)dt = ∫dx

    m/b*ln(t) + vo*t = x(t)
     
  11. Sep 19, 2009 #10

    Redbelly98

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    Okay so far.

    This step does not follow from the previous one. (We can tell because t=0 implies v=∞, instead of v=vo, in this equation.)

    Instead, this step could have been

    (bt/m) + (1/vo) = 1/v​
    See if you can take if from there.
     
  12. Sep 19, 2009 #11
    (bt/m) + (1/vo) = 1/v

    doesn't this mean v = m/bt + Vo?
     
  13. Sep 19, 2009 #12

    Redbelly98

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    No, what it means is that v is the reciprocal of

    (bt/m) + (1/vo)​

    which is

    [tex]\frac{1}{(bt/m) + (1/v_o)}[/tex]
     
  14. Sep 20, 2009 #13
    Please tell me I've done this right:

    from there:

    ∫dx = ∫https://www.physicsforums.com/latex_images/23/2356837-0.png [Broken]

    x = m/b*ln(bt/m + 1/vo) - m/b*ln(1/vo)

    for t=0, x=0 and when differentiated, at t=0,v=vo
     
    Last edited by a moderator: May 4, 2017
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