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Proving a complex Force derivation with CalcConfused

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data
    If we know F(t) the force as a function of time for straight line motion, Newton's second law gives us a(t) the acceleration as a function of time. We can then integrate a(t) to find v(t) and x(t). However, suppose we know F(v) instead. a) The net force on a body moving along the x-axis equals -Cv[itex]^{2}[/itex]. Use Newton's second law written as ƩF = m [itex]\stackrel{dv}{dt}[/itex] and two integrations to show that x - x0 = (m/C) ln(v0/v). b) Show that Newton's second law can be written as ƩF = m*v*[itex]\stackrel{dv}{dx}[/itex]. Derive the same expression in part (a) using this form of the second law and one integration.

    2. Relevant equations
    F = ma
    F(v) = -Cv^2

    3. The attempt at a solution

    For the first part a), I had no idea so I started to differentiate the answer to try and get a method for simplifying it originally. The problem I encounter is that at X, If I take the 2nd derivative of the function I get a(t) = m/Cv^2, which seems impossible because you cant do that with -Cv^2 = m dv/dt. I solved that for dv/dt but i get nowhere really. I'm extremely stumped. Any pointers in the right direction would be extremely useful.

    Thanks in advance.
  2. jcsd
  3. Oct 11, 2011 #2
    For part a, you already know that

    F = ma
    F(v) = -Cv[itex]^{2}[/itex]

    We can then write the following:

    ma = -Cv[itex]^{2}[/itex]
    a = [itex]\frac{-Cv^2}{m}[/itex]

    [itex]\frac{dv}{dt}[/itex] = [itex]\frac{-Cv^2}{m}[/itex]

    Next I will express [itex]\frac{-C}{m}[/itex] as some other constant to simplify the algebra. So let [itex]\alpha[/itex] = [itex]\frac{-C}{m}[/itex].

    [itex]\frac{dv}{dt}[/itex] = [itex]\alpha v^2[/itex]

    We can then use the method of separation of variables from ordinary differential equations to write:

    [itex]\frac{dv}{v^2}[/itex] = [itex]\alpha dt[/itex]

    [itex]\int \frac{dv}{v^2}[/itex] = [itex]\alpha \int dt[/itex]

    You should be able to solve this integral for an expression for v. Then, if you isolate v you will be able to write v = dx/dt and repeat this procedure to find an expression for x.
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