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Newtonian mechanics w/ drag - please help

  1. Sep 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider a particle of mass m whose motion starts from rest in a constant gravitational field. If a resisting force proportion to the square of the velocity (i.e. kmv^2) is encounterd, show that the distance s the particle falls in accelerating from v0 to v1 is given by


    2. Relevant equations

    3. The attempt at a solution
    I am continually frustrated by these types of problems because I struggle with coming up with a DE that I can solve.

    my''= km(y')^2 -mg
    y''= k(y')^2 - g

    so I have a non linear equation
    I wanted to make a substitution u = y'
    so that u'=ku^2 - g, but this doesn't help at all.

    Can someone please point me in the right direction? Thanks.
  2. jcsd
  3. Sep 18, 2009 #2


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    Try the substitution v = y'. You should get a differential equation in v which you should be able to separate and then integrate. Once you have v(t) you integrate once more and you have y(t).
  4. Sep 18, 2009 #3
    Ok, yes I see that now.

    mv'=kmv^2 - mg

    dv/dt = kv^2 - g

    dv/(kv^2 - g) = dt , (1/k)*dv/(v^2 - g/k) = dt

    and I used an integration table to get this:

    (1/k)*1/(2*sqrt(g/k))*ln[(v-sqrt(g/k))/(v+sqrt(g/k))] = t + c

    Only I have a problem when I go to apply v(0) = 0 in that the natural log turns to ln(-1).
  5. Sep 19, 2009 #4


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    The problem is that you did the indefinite integral first then tried to put in the integration constant. Put the limits of integration (t={0,t}, v={0,v}) in, then integrate. That takes care of the integration constant automatically and is a good habit to acquire.

    Also, I would suggest that you recast the original equation in terms of the terminal velocity vT which you can easily get from the original diff. eq. if you consider what "terminal velocity" means. It makes life easier.
  6. Sep 19, 2009 #5
    I don't know what is wrong with doing the integrals and then adding a constant to one side. Both integrals give constants, but I subtracted one to the other side to get a "composite constant."

    I know that you don't have a constant for definite integrals, but I'm not quite seeing how using those limits you suggested make it easier. I've always previously done IVPs by getting constants and solving for them with the initial conditions. Could you elaborate a bit? I really appreciate your help.
  7. Sep 19, 2009 #6


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    Logarithms are tricky. We are doing physics, not math. For example, if you say


    where x and a have dimensions of length, that may be mathematically correct, but is not physically correct because the argument of a logarithm must be dimensionless when you do physics. Compare with

    [tex]\int_{0}^{x}\frac{dx}{x+a}=ln(x+a)-ln(a)=ln \big( \frac{x+a}{a} \big)[/tex]

    which is what you would have got have got, had you set

    [tex]C = - ln(a)[/tex]

    That's what I meant when I said that putting in the limits of integration "takes care of the integration constant automatically and is a good habit to acquire."
    Last edited: Sep 19, 2009
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