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Both Quadratic and Linear Drag

  1. Jan 27, 2008 #1
    An object is coasting on the horizontal axis, in the positive direction and is subject to a drag force f = -bv - cv[tex]^{2}[/tex].
    Write down Newton's 2nd Law and solve for v using separation of variables.

    So first I wrote out Newton's law as:
    F= m(dv/dt) = -bv - cv[tex]^{2}[/tex]

    Solving the integral: dt = [tex]\frac{dv}{-bv-cv^{2}}[/tex], with boundaries from 0 to t and v(0) to v

    I got: t = [tex]\frac{-m}{b}[/tex] ln[tex]\frac{v}{1+\frac{c}{b}v^{2}}[/tex]

    Note: I haven't put in my boundaries on v yet. However, Once I put in my boundaries on v and try to rearrange to solve for v, I can't get anywhere. Any suggestions or help would be greatly appreciated.

    Thx, CB
     
  2. jcsd
  3. Jan 28, 2008 #2

    CompuChip

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    Homework Helper

    Are you sure about the ln? If I differentiate it, I first get
    [tex]\frac{ 1 + (c/b) v^2 }{ v } [/tex]
    times the derivative of that (chain rule!) which will be something like (... - ...) / v^2
    in total giving me something containing 1/v^3.

    If you want to do it analytically, you could try writing
    [tex]\frac{-dv}{v} \frac{1}{b + c v}[/tex]
    as
    [tex]\left( \frac{A}{v} + \frac{B}{b + c v}\right) dv[/tex]
    with A and B constants (which, in this case, works). Then split the integral and solve both parts separately.
     
  4. Jan 28, 2008 #3
    Well looks like I made an error in my calculations: the integral turns out to be :
    t =[tex]\frac{-m}{b} \frac{1}{b + c v}[/tex] ln[tex]\frac{v}{1+\frac{c}{b}v}[/tex]

    And yes, using partial fractions will also solve this integral, and yes a ln function will turn out to be there.

    Anyway, then the solution is solved for just rearranging for v in terms of t.
     
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