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Homework Help: Algebra in velocity dependent forces

  1. Jan 28, 2008 #1
    I'm having an issue getting two equations to look alike.

    How can you get:

    k*v + g = e^(-t*k + k*v0)

    to look like:

    v = -g/k + [(k*v0 + g) / k] * e^(-kt) ???


    I know I can split the exponent up, and so I guess the real question is, how do I get:

    e^(k*v0) / k

    to look like:

    (k*v0 + g) / k ?
  2. jcsd
  3. Jan 28, 2008 #2


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    The answer is: you don't, with the information given.
    The only thing I can think of is expanding the exponential into
    [tex]e^{k v_0} = k v_0 + 1 + \tfrac12 k^2 v_0^2 + \cdots[/tex]
    and defining g to be that expression except the first term (or letting g = 1 and neglecting terms of order k).

    Are you sure you gave us all the information?
  4. Jan 28, 2008 #3
  5. Jan 28, 2008 #4


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    OK, it looks like you got the physics right but you're just having trouble with the algebra. Therefore, I'm going to give you my answer; please go through the steps and tell me which one you don't understand / did differently.

    1. You have the equation of motion,
      [tex]\frac{dv}{dt} = - g - k v [/tex]
    2. Rewrite into
      [tex]\frac{dv}{g + k v} = - dt[/tex]
    3. Integrate both sides (I'm doing a definite integral here, you can do an indefinite one and later apply the boundary conditions, if you like) - calling initial time t = 0:
      [tex]\int_{v_0}^v \frac{dv'}{g + k v'} = \int_0^t -dt'[/tex]
    4. Solving both integrals
      [tex]\frac1k \ln( g + k v) - \frac1k \ln(g + k v_0 ) = - t[/tex]
      (I think you got this far, judging by your earlier post)
    5. Now the rest is algebra: multiply both sides by k and bring the second logarithm to the other side
      [tex]\ln( g + k v) = \ln(g + k v_0 ) - k t[/tex]
    6. Exponentiate everything (note: some steps are missing here, work them out yourself!)
      [tex] g + k v = ( g + k v_0 ) e^{- k t} [/tex]
    7. Now solve for v.
    I think your problem is somewhere in the last two steps, so try to work them out exactly and find out how I only left (- k t) in the exponent and not the k v_0 stuff.
  6. Jan 28, 2008 #5
    I'm a little confused about step 3. How did you know you were supposed to integrate with repsect to time?

    Here's what I understood:
    We obtain the equation representing velocity, integrate and obtain position. Where at during this process do you think of taking an indefinite intergral? (I'm having problems connecting calculus(which is 2 years ago for me now) concepts to physics ones.)

    For example, if I recall correctly, in calc when they ask you to solve the area under a curve.. well it makes sense that you would travel from end point to end point. I don't get the relationship = \. I completely understand everything now, except the logical step from 2 -> 3.

    Thanks a ton for the help, I appreciate it!
  7. Jan 28, 2008 #6


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    Actually it is a "physicists shortcut". The point is that on both sides of the equation you have a d(something): dv on one side and dt on the other. These are "infinitesimal" quantities, that don't make sense except in "fractions" like dv/dt and under an integral sign. To make sense of an equation like
    [tex]f(v) dv = g(t) dt[/tex]
    you could either write
    [tex]\frac{dv}{dt} = \frac{ g(t) }{ f(v) }[/tex]
    (treating them like ordinary fractions) or you can integrate both sides, getting
    [tex]\int f(v) dv = \int g(t) dt[/tex].
    You will have to do it on both sides though (like in any equation you solve), something like
    [tex]\int f(v) dv = g(t) dt[/tex]
    is nonsense. If you want to do a definite integration, consider the equation f(v) dv = g(t) dt as relating the change in velocity dv to the change in time dt. To get the change in velocity from the start of the motion to the end of the motion, you want to add up all these infinitesimal quantities during the motion. The time runs from time 0 to time t, while the velocity runs from the initial velocity [itex]v_0[/itex] to the final velocity v. The other way to solve it would be
    [tex]\int \frac{dv'}{g + k v'} = \int -dt'[/tex]
    without boundaries, which would give you
    [tex]\frac1k \ln( g + k v) = - t + C[/tex]
    where [itex]C[/itex] is an integration constant (technically, there should be one on the other side as well, but subtracting it from both sides just gives a new constant on one side only). Now you can do the same manipulations as before, and get
    [tex]v = \tilde C \frac{g}{k} e^{- t k } [/tex]
    where [itex]\tilde C[/itex] is still unknown (related to C through [itex]\tilde C = e^{- k C}[/itex], as you can check). Now you can plug in [itex]t = 0, v = v_0[/itex] and this will give you an equation for [itex]\tilde C[/itex], of which the solution should be [itex]\tilde C = 1 + \frac{k v_0}{g}[/itex] (I think) such that you get the same result as with the definite integration).

    It probably sounds all elaborate and complicated, but I recommend you go through the steps to check that it indeed works out (and hopefully, to find out it's not as bad as it seems).

    Finally, if you want to do it all rigorously (the "mathematicians way") you would start from
    [tex]\frac{dv}{dt} = - g - k v [/tex]
    and integrate this equation over t and using the fundamental theorem of calculus:
    [tex]\int_0^T \frac{dv}{dt} dt = v(t = T) - v(t = 0)[/tex]
    which will theoretically give you the same result as well (but requires more care, which is why it is actually better but no physicist ever uses it anymore after a while and just manipulates the dv and dt as separate things)
  8. Jan 28, 2008 #7
    I'm going to read this in just a second, but since you're still up i might as ask you something else heh.

    Super ball problem, and theres a step where we are supposed to:
    "We can factor the momentum conversation equation out of the energy conservation equation."

    Isn't that like saying, "We can factor velocity from acceleration?" (F = dp/dt, and A = dv/dt)?
  9. Jan 28, 2008 #8


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    I'm not entirely sure what you mean but probably you have something like

    [tex]\tfrac12 m_1 v_1^2 = \tfrac12 m_2 v_2^2 = \tfrac12 (m_2 v_2) v_2 [/tex] (energy)

    [tex]m_1 v_1 = m_2 v_2 [/tex] (momentum)

    Then you can use the second to write the first as
    [tex]\tfrac12 m_1 v_1^2 = \tfrac12 (m_1 v_1) v_2 [/tex]
    and cancel out [itex]\tfrac12 m_1 v_1[/itex] from both sides, giving
    [tex]v_1 = v_2 [/tex]

    So basically, you "divide" one of the equations out of the other one.
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