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Find position function of freefall with air resistence ODE

  1. Sep 20, 2013 #1

    1s1

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    1. The problem statement, all variables and given/known data
    An object of mass ##5##kg is released from rest ##1000##m above the ground and allowed to fall under the influence of gravity. Assuming the force due to air resistance is proportional to the velocity of the object with proportionality constant ##b=50##N-sec/m, determine the equation of motion (for the position ##x(t)##) of the object.


    2. Relevant equations
    $$m\frac{dv}{dt} = mg-kv$$


    3. The attempt at a solution
    The given differential equation describes the velocity with respect to time. How do I turn it into a differential equation that describes the position with respect to time?
     
  2. jcsd
  3. Sep 20, 2013 #2

    Zondrina

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    Your relevant equation is a bit off, shouldn't it read ##m\frac{dv}{dt} = (-1)(mg+kv)##?

    If that's the case then solving it with the constraint that ##v(0) = 0## should simplify things a bit.
     
  4. Sep 20, 2013 #3

    1s1

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    You may be right about the relevent equation being negative on the RHS. My book has it written as I did, but my prof mentioned that physicists sometimes make sign changes to the constant.

    But I am still confused about how to make the function into a function of position? I assume you use the relationship ##d=vt##? Or maybe there is an entirely different formula derived for the position rather than the velocity?
     
  5. Sep 20, 2013 #4

    Zondrina

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    Notice this was part of your question:

    This means at ##t=0##, the velocity of the object is zero. That's why I mentioned solving the equation with the constraint ##v(0) = 0##. Right now you have an acceleration equation since it involves derivatives of velocity, but if you integrate acceleration, you get velocity;if you integrate vel...
     
  6. Sep 20, 2013 #5

    vanhees71

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    the sign is given by the direction of the [itex]x[/itex] axis. If you define it to point "down", i.e., in direction of the constant gravitation force, the positive sign [itex]+mg[/itex] is correct. The sign in [itex]-k v[/itex] (assuming [itex]k>0[/itex]) is the same for both directions of the [itex]x[/itex] axis, because the friction force must always be such that it tends to decrease [itex]v[/itex].

    Further note, that you have a nice inhomogeneous linear differential equation of first order with constant coefficients, which is pretty easy to solve by separation of variables. The gives you [itex]v(t)[/itex].

    Then you have to solve another very simple equation
    [tex]v=\frac{\mathrm{d} x}{\mathrm{d} t}.[/tex]
    Given [itex]v[/itex], it's very clear, how to get [itex]x(t)[/itex]!
     
  7. Sep 20, 2013 #6
    Why? Air RESISTANCE implies that it will go against the direction of acceleration. This is a physics problem, primarily.

    We're supposed to solve the equation and make a position function, right? I will denote ##\frac{dv}{dt}## as ##v'##, for clarity and ease of notation (id est, laziness. It's a friday. Give me a break :tongue:).

    I'm assuming we have ##k=b##, right? So, we have $$mv'=mg-bv \\ v' +\frac{b}{m}v=g.$$ In this form, we can introduce an integrating factor. Can you take it from there?

    Edit: vanhees71 got there first by noting the inhomogeneous equation.

    Indeed. Finding ##\Gamma(x(t)+1)## is easy once we have ##v##. :biggrin:
     
  8. Sep 20, 2013 #7

    1s1

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    This is very helpful, thanks!
     
  9. Sep 20, 2013 #8

    Zondrina

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    I suppose it depends on how you decide to define up and down.

    For me that's dependent on the day of the week :P.
     
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