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Homework Help: Finding Velocity from Acceleration

  1. Mar 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Particle (Mass = m) falling through a viscous liquid due to gravity.

    Experiences a drag force that is proportional to its speed.
    Magnitude of drag force .
    F = ku = +ve Constant x velocity.

    At time t = 0, z = position = 0, initial speed = u_0.

    2. Relevant equations

    Acceleration found to be

    du/dt = g - (k/m)u

    3. The attempt at a solution

    So to find the velocity this needs to be integrated, the way I attempted to do it was to separate the variables and substitute in
    α = mg / k
    (du/dt) = g(1-u/α)
    This lead to the integral

    ∫(α du/ α-u) = ∫g dt
    α∫(1/-u) du = g ∫dt

    -α ln (α-u) = gt + c

    ln (α-u) = -(gt + c )/α
    (α-u) = exp -(gt + c)/α

    u = α - exp[-(gt+c)/α]

    Would that be correct up to there, because when I carry on to find c and to substitute for α it gets messy, so I'm unsure if I've done it right.

  2. jcsd
  3. Mar 14, 2012 #2

    du/dt = g - (k/m)u

    Above is correct equation. Solve the homogeneous equation by separation of variables. Then visualize a particular solution and add solutions together. The particular solution is 'particularly' easy to see. Evaluate constant by initial condition.
  4. Mar 14, 2012 #3
    For your solution


    a,b,d are constants.
  5. Mar 14, 2012 #4

    Okay so I ended up with

    u(t) = [U_0 - (gm/k)] exp(-kt/m) + (gm/k)

    Is that anywhere near correct?
  6. Mar 15, 2012 #5
    That is what I got. You can 'test' your solution. Let t go to infinity and the velocity is mg/k which agrees with a force balance on the object. At t=0, the solution is V=V0.

    Good work!
  7. Mar 16, 2012 #6
    Awesome, thanks mate.

    If I were to need to find the position I would need to integrate that, right?

    Just that I need in a specific form, and I can't seem to get it in that form.
  8. Mar 16, 2012 #7
    Yes, the integral of velocity over time provides displacement. Displacement is the area under the velocity-time curve. With an indefinite integral, you'll get another constant of integration that you evaluate from the initial condition at t=0 which the problem stated is z=0.
  9. Mar 16, 2012 #8
    Yeah okay, so this is what I ended up with as the position

    z(t) = [( m exp(kt /m)(g m-k U_0)) /k + gmt) ]/k + c

    Which is quite hard to understand when wrote on here.

    But is that correct, trying to work out when I'm going wrong as I can't get it in the right form.
    If that is right at least I know I should be able to write it in that form.

    Cheers again.
  10. Mar 17, 2012 #9
    You essentially have:

    z(t) = integral[((Vo - mg/k)exp(-kt/m) + mg/k)*dt]

    Vo-mg/k and mg/k are constants so you would get after integration

    z(t) = (-m/k)[Vo - mg/k]exp(-kt/m) + mgt/k + C

    with C the constant of integration that is evaluated when t=0.

    Check your integration.

    Also, your parentheses do not balance.
    Last edited: Mar 17, 2012
  11. Mar 17, 2012 #10
    When deriving formulas/functions, it is prudent to plug in the units to ensure you have not made a careless error. If the units are not correct, you have an error.
  12. Mar 17, 2012 #11
    Don't know how I missed the whole constant thing, feel a bit of an idiot now.

    I'm getting

    (V0 - mg/k)+ mg/k ∫exp(-kt/m)

    [(V0 - mg/k)+ mg/k] [-m/k exp(-kt/m) + c]

    Am I heading the right way now?
  13. Mar 17, 2012 #12
    You still have an error. Write it like this and integrate:

    z(t)= (V0 - mg/k)*integral(exp(-kt/m))dt + integral(mg/k)dt
  14. Mar 17, 2012 #13
    Ah right, now I'm getting answers that match up now, I didn't think to do it that way, thank you for the advice.
    Problem is now though the form it is wanted in is extremely similar to what I'm getting, or rather we're getting, except the last term is


    Rather than


    Am I missing something really simple?
  15. Mar 17, 2012 #14
    OK, when you integrate you pick up another constant of integration that must be evaluated. See if that helps.
  16. Mar 18, 2012 #15
    Okay I got it, cheers for the help mate.

    I have one last question to do, and would appreciate it if you could perhaps give me a nudge in the right direction with it.

    It leads on from the last question.

    Position of a particle acting under a constant force F in the same z direction is given by

    z = u0 t + (F/2m)t^2

    With the same initial conditions.
    It asks me to show that the position of a particle falling under the influence of gravity through a viscous fluid can be approximated by that of a particle acted upon by a uniform force F, for sufficiently small values of t.
    It then asks for an expression for F.

    All I can think of is t≈0, and then to substitute this in.
  17. Mar 20, 2012 #16
    Talking to people in class it apparently uses the approximation for exp, I'm so confused.
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