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Homework Help: Terminal velocity of a body of mass

  1. Sep 24, 2008 #1
    A small body of mass m located near the Earth’s surface falls from rest in the Earth's gravitational field. Acting on the body is a resistive force of magnitude kmv, where k is a constant and v is the speed of the body.

    a) Write the differential equation that represents Newton's second law for this situation.

    my answer: F = ma
    a = dv/dt
    F = mg - kmv
    ma = mg - kmv
    a = g - kv
    dv/dt = g - kv

    I'm pretty sure I got this one right.

    b) Determine the terminal speed vT of the body.

    F = 0
    g - kv = 0
    g = kv
    vT = g / k

    I think I got this one right too.

    c) Integrate the differential equation once to obtain an expression for the speed v as a function of time t. Use the condition that v = 0 when t= 0.

    What I did so far:

    dv/dt = g - kv
    dt = dv / (g - kv)

    So I would integrate dt = dv / (g - kv). The problem is I don't know how. Can someone please explain how to do this kind of integration, why I would do it that way?
  2. jcsd
  3. Sep 24, 2008 #2


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    Often times if you have something like:

    dv/dt = v,

    you'll think to yourself, "oh, the equation is separable", and you'll go:

    dv/v = dt

    and then "integrate"

    to get ln(v) = t + C

    notice, however, that you integrated the left side wrt v and the right hand side wrt t, which makes NO sense! So what is actually going on here? Why does this nonsensical operation of splitting dv/dt actually work? It works because you are implicitly using the chain rule as follows

    [tex] \frac{dv}{dt} = v [/tex]

    [tex] \frac{1}{v}\frac{dv}{dt} = 1 [/tex]

    notice that the left hand side looks like something that *has been* differentiated using the chain rule, and we can write it as that something:

    [tex] \frac{d}{dt}(\ln{v}) = 1 [/tex]

    Now we can integrate BOTH sides wrt time (which makes waaay more sense)

    [tex] \int { \frac{d}{dt}(\ln(v)) \, dt } = \int{\, dt} [/tex]

    [tex] \ln(v) = t + C [/tex]

    In your more complicated example, it is important to realize that this is what is going on, otherwise you'll get stuck.

    We have

    [tex] \frac{1}{g-kv}\frac{dv}{dt} = 1 [/tex]

    We want the left hand side to look like something that has been differentiated by the chain rule. It SORT OF does. The only problem is that the derivative of ln(g-kv) wrt time actually differs from what's on the left hand side by a factor of -k. Once you get that sorted, it's the same procedure.
  4. Sep 24, 2008 #3

    Thanks for the explanation. I'm starting to understand this more.

    What do you mean that the left differs by a factor of -k?

    I have the answer. It's -(1/k)ln(g - kv) = t + C

    Where did that come from? Where did the other k come from?

    I thought doing this problem would help me understand how these weirder intergrations work, but it's confusing me even more.
  5. Sep 24, 2008 #4


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    alright, here goes:

    The goal is to recognize that the left hand side can be expressed as an exact derivative, so that integrating it becomes trivial.

    the equation has a term 1/(g-kv) multiplied by v'(t), which suggests that it is the result of differentiating ln(g-kv) with respect to time using the chain rule. If that were true, then the left hand side would reduce to the time derivative of the natural log of g-kv, and integrating it would be easy (integral of a derivative is just the function itself...so you're just left with the log term). But that's NOT quite right. Check it out:

    [tex] \frac{d}{dt} (\ln(g-kv)) = \frac{1}{g-kv} \frac{d}{dt}(g-kv) \frac{dv}{dt} [/tex]

    [tex] = \frac{1}{g-kv} (-k) \frac{dv}{dt} [/tex]

    Notice that this doesn't QUITE match the left hand side of our differential equation. There's that extra factor of (-k). Which means that the left hand side of our equation is this thing divided by -k. THAT's where the 1/(-k) comes from.
  6. Sep 24, 2008 #5
    I stared at the problem and answer while waiting for an answer and had the same train of thought. That was very helpful. Thanks a lot.
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