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Kinematic question involving differential equation

  1. Apr 22, 2010 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m falls from rest; the resistance of the air when the speed is v is kv2 where k is a constant. Show that after a time t, the velocity v and the time t are related by the equation t = V/2g ln[(V + v )/(V - v)] where V2
    = mg/k

    2. Relevant equations

    3. The attempt at a solution
    ma = mg - kv2
    a = g - k/m v2
    dv/dt = g - k/m v2
    dv = g - k/m v2dt
    v = gt - k/m v2t
    t = v/(g - k/m v2)

    But I don't know how to continue from here.
    I have been studying differential equation on my own, so I am not very strong in the foundation
  2. jcsd
  3. Apr 22, 2010 #2


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    Homework Helper

    This is a seperable differential equation. But it is enough if you replace the given v(t) relation back into the original equation.

  4. Apr 22, 2010 #3
    It is quite complicated to solve the differential equation, but if you find v(t) from the given "solution" (which you want to prove), given as t(v), you can put this v(t) into your equation and see the all the terms cancel out. This should leave you with something like 1=1.
    Just to kinda clairify.
  5. Apr 22, 2010 #4
    I'll get you started

    Hi there,
    OK, let me get you started on this, and I'll leave it in a place where you should be able to finish it from there. I have already gone through and done the proof, but I cannot simply give it to you as it will be removed shortly after if I do, so instead I'll get you a little over half way there, explaining what I'm doing along the way, and then I'll tell you what you need to do to finish the proof.

    Let's start with what you had:

    ma = mg - kv2
    re-write a as dv/dt (dv/dt is the infinitesimally small change in velocity with respect to an infinitesimally small change in time, which is the acceleration at a given point in time)
    m(dv/dt) = mg - kv2
    mdv = (mg - kv2)dt

    Now since we know that V2 = mg/k lets factor out k from the right hand side of the equation.

    mdv = k(mg/k - v2)dt
    mdv = k(V2 - v2)dt

    dv/(V2 - v2) = (k/m)dt
    The denominator on the left hand side is called the difference of two squares and can be re-written as:

    dv/[(V +v)(V - v)] = (k/m)dt

    Now in order to integrate the left hand side it will be far easier to do if you separate the terms in the denominator such that you only have an integration with respect to one over V+v with a constant in the numerator summed with another integral with only V-v in the denominator with again a constant in the numerator.

    How you do this is called separation by partial fractions. Basically if you have something like

    1/[(x + a)(x +b)] you set this equal to how you'd like it separated, for instance using my example, I would write this as:

    1/[(x + a)(x + b)] = A/(x + a) + B/(x + b) where A is not a and B is not b, necessarily, as they could be equal, but they are not meant to be the same value. The right hand side of this equation is what you want your equation to look like as it is far easier to integrate than one with the product of two terms each having the variable of integration within those terms in the denominator. So to solve for A and B you multiply each side by the denominator on the left hand side to get:

    1 = A(x + b) + B(x + a), now let x = -b
    1 = A(-b + b) + B(-b + a)
    1 = B(a - b)
    B = 1/(a - b) and a and b are both constants, so that worked well. Now lets do the opposite so we can find A. We have to let x = -a and we get:

    1 = A(-a + b) + B(-a + a)
    1 = A(b - a)
    A = -1/(a - b) a and b are still constants, so all is still well. Now let's set the original fraction equal to what we found.

    1/[(x + a)(x + b)] = [(-1/(a - b))/(x + a)] + [(1/(a - b))/(x + b)]

    Using partial fractions is what you need to do to the equation below (derived from above, and I don't mean from God, but further back in this reply) in order to separate the terms in the denominator on the left hand side so that you can integrate each side and solve for t.

    When you get done you will have a k and an m term in there you need to get rid of, but you can because you know that V2 = mg/k, so you can plug that in and get rid of everything except V,g and the natural log of that fraction involving the velocity, v.

    dv/[(V +v)(V - v)] = (k/m)dt

    So this is where I leave you. Try using partial fractions to separate the left hand side into two different fractions with only one v term in each denominator, you can basically just do what I did with the partial fractions, integrate, re-arrange a little and get rid of the k and m that will still be there when you are done by using V2 = mg/k and solve for t and that should give you the same thing that you need to prove is true.

    You also need to know some common (and by common I mean well known ones, not a base 10 log, although the same operations apply to both the common and natural log) log operations in order to combine stuff after integrating, like the ln(a) - ln(b) = ln(a/b).

    Hopefully I have given you enough that you can take it from here and solve the proof on your own and also I hope I haven't given you too much of the procedure so that it is as easy as simply performing a couple of easy steps and you have the proof.

    If what I did is something you haven't seen before, yet you understand what I have shown you, then you have learned something new, and that's always a good thing. Now just take it to its conclusion and solve for the proof.

    Good luck and many smiles,
    Craig :smile:
  6. Apr 23, 2010 #5
    Thanks for the detailed answer
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