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Drag forces

  1. Nov 6, 2004 #1
    Could someone please check my work here?
  2. jcsd
  3. Nov 6, 2004 #2


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    Urban, what was the problem with the solution in the other thread?
  4. Nov 6, 2004 #3
    I couldn't really follow what was going on. I didn't know there was a solution. In physics, I was given 2 formulas:

    dv/dt=-g+bv/m <- this being the differentiable equatioin
    and also:
    v=mg/b(1-e(-bt/m)) <- which showed the velocity at any time

    My homework asks to "solve" for the equations but I'm in high school so it only expects me to jump from


    without actaully solving it
    From that, they ask questions like...what if the drag force = -bv^2 and they change it. Then they tell me to find the differential equation for the object's motion. I THINK that all I have to do is sub in the numbers but I'm not quite sure. Cyclovenom, you did a great job but I'm only on the basics :rolleyes:
  5. Nov 6, 2004 #4


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    Urban if you want to check your solution then derivate it with respect to time, and it should give the right side of the equation.
  6. Nov 6, 2004 #5
    what does that mean? "give the right side of the equation"? Do I have the wrong equations?
  7. Nov 6, 2004 #6


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    For example:

    For this differential equation

    [tex] \frac{dy}{dt} = ay - b [/tex]

    its solution is: [itex] (y_{o} = 0) [/itex]

    [tex] y = \frac{b}{a} - \frac{b}{a}e^{at} [/tex]

    so if we substitute in the differential equation we should get both sides the same.

    [tex] \frac{d(\frac{b}{a} - \frac{b}{a}e^{at})}{dt} = a(\frac{b}{a} - \frac{b}{a}e^{at}) - b [/tex]

    we get:

    [tex] be^{at} = be^{at} [/tex]

    so that's the solution for our differential equation.
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