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Air Resistance of vehicle

  1. Aug 20, 2009 #1
    1. The problem statement, all variables and given/known data

    A vehicle of mass m experiences a constant frictional resistance ma and air resistance proportional to the square of its speed. It can exert a constant propelling force mb and attain a maximum speed V Show that, starting from rest, it can attain the speed V/2 in the time

    Vln3/2(b-a)​

    And that the friction and air resistance alone can then bring it to rest in a further time

    (V/(a(b-a))^1/2)tan-1(b-a/4a)^1/2​

    2. The attempt at a solution

    Using mc as the constant of proportionality for air resistance yields:

    x''=b-a-c(x')^2

    from there I'm just guessing as to the method but I've tried this:

    dx'/dt=b-a-c(x')^2

    int(dx')=int((b-a-c(x')^2)dt)
    =(b-a)int(dt)-c(int((x')^2dt)
    =(b-a)int(dt)-c(int(x'dx))

    but I'm not sure how to integrate x' w.r.t. x?

    Of course I could be going about it completely the wrong way. The motion is in one dimension so I could maybe import one of the equations of motion, like v^2=u^2+2as? I'm sure I'm missing a trick (or two) somewhere.
     
  2. jcsd
  3. Aug 20, 2009 #2

    kuruman

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    Use auxiliary variable v = dx/dt

    Then your equation is of the form

    [tex]\frac{dv}{dt}= A - B v^{2}[/tex]

    from which

    [tex]\frac{dv}{A-Bv^{2}}=dt[/tex]

    The variables are now separated and you can integrate.
     
  4. Aug 20, 2009 #3
    integrating:

    t=arctan(((-c^1/2)/((b-a)^1/2))v)/(b-a)^1/2(-c^1/2)

    rearranging:

    v=(((b-a)^1/2)/(-c^1/2))tan((b-a)^1/2)(-c^1/2)t

    sorry about the horrible presentation I don't know how to use LaTeX.

    The -c^1/2 makes me a little uneasy but I've used the tan form of the integral as it's easier to remember than tanh and judging from the question more likely to yield the result I'm looking for.

    OKay so now how do I actually answer the question?
     
  5. Aug 20, 2009 #4

    kuruman

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    I don't know how you got your answer, but I shoved the integral in Mathematica and got

    [tex]t = \frac{1}{\sqrt{AB}}ArcTanh(\frac{\sqrt{B}}{\sqrt{A}}v)[/tex]

    where A = a - b and B = c. Recheck your work.

    To complete the problem, you need to eliminate c in favor of V, i.e. find c in terms of V and the other constants. Hint: What does your diff. eq. look like when the object reaches terminal velocity?
     
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