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F(x,v) = kvx

  1. Sep 11, 2005 #1
    I was given the following problem for a homework assignment:

    Given F(x,v) = kvx where k is positive and the velocity is v-nought at x = 0, t = 0, what is x as a function of time?

    First I derived v as a function of position:

    v = v-nought + kx^2/2m

    Then, I derived x as a function of time:

    x = Sqrt[2mv-nought/k]tan(t*Sqrt[kv-nought/2m])

    My problem with this solution is that x is then undefined for various points in time, plus the fact that it jumps from inifnity to negative infinity when it is undefined.

    Now, if this is correct, it can be used as proof that the function as given cannot conform to anything in reality. However, it just may be plain out wrong. If it is wrong, I was wondering if someone might supply me with the correct answer as well as how to go about getting it.

    Thank you very much for your time.
    Last edited: Sep 11, 2005
  2. jcsd
  3. Sep 12, 2005 #2

    James R

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    F(x,v) doesn't seem to tell us anything useful. It's just a random function, as far as I can see. Therefore, v is unconstrained, and x could be anything as a function of time.

    Edit to add:

    Ah...hang on. Is F a force?
  4. Sep 12, 2005 #3

    James R

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    If F = kvx is the force on the particle, then:

    ma = kvx,

    which means

    [tex]m \frac{d^2 x}{dt^2} = kx \frac{dx}{dt}[/tex]

    Can you solve that differential equation?
  5. Sep 12, 2005 #4
    Yes, it is a force, and I wrote kvx = ma = mv(dv/dx).

    So, kx/m dx = dv, and

    kx^2/2m = v - v-nought.

    Therefore, v = v-nought + kx^2/2m = dx/dt.

    I then seperated variables again and solved for x as a function of time, which resulted in my result written up in my first post, aloing with the problems it carries.
  6. Sep 12, 2005 #5

    James R

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    Ah, yes. I forgot that a = v(dv/dx).

    I haven't checked your final answer, but it's probably correct. Your conclusion that the force is unrealistic seems sensible to me.
  7. Sep 12, 2005 #6
    Thank you.
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