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Who can differentiate this one?

  1. Nov 25, 2009 #1
    For a falling object, who can find the time in terms of the velocity for this journey( before reaching terminal velocity) with considering the resistive force?
     
  2. jcsd
  3. Nov 25, 2009 #2
    Force equals mass times acceleration
    Choose what resistive force to use, inverse velocity or inverse square velocity
    Write down the differential equation
    Solve it
     
  4. Nov 26, 2009 #3

    HallsofIvy

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    I can"! But are you clear on the terminology? Your title asked "how can differentiate this one" implying you have a function to differentiate but that doesn't appear to be the case. You are just asking about setting up the dynamic equation.

    As g edgar says, "Force equals mass times acceleration" so ma= m dv/dt= -g- f(v) where "f(v)" is the resistive force. That can be a very complicated function of the velocity depending on the situation. I do not agree with g edgar's "inverse" formulas. Typically, the faster something is going, the greater the drag, not the other way around. Normally, the drag is simplified to either -kv or -kv2 where k is the constant of proportionallity and v is the speed.
     
  5. Nov 26, 2009 #4
    When I tried to solve it, I started with this equation
    F-R=ma, where F= mg, snd R is the risitive force which equals bv (b is a constant)
    So, mg-bv=ma
    mg-bv=m dv/dt then we seperate the variables to get the time in terms of the velocity

    Am I correct with that?
     
  6. Nov 26, 2009 #5

    HallsofIvy

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    Yes, in that case, with resistive force proportional to v, you get a fairly simple equation:m dv/dt= mg- bv, a separable equation. But be careful about signs. If you are taking "upward" to be positive, then it is m dv/dt= -mg- bv. Since drag always acts opposite to velocity, the coefficient of v is always negative.
     
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