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Homework Help: Non-constant acceleration kinematics

  1. Sep 10, 2008 #1
    1. The problem statement, all variables and given/known data
    The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared, and is given (in SI units) by a = -3.00v^2 for v > 0. If the marble enters this fluid with a speed of 1.50 m/s, how long will it take before the marble's speed is reduced to half of its initial value?

    2. The attempt at a solution
    I know that this question appeared here a while ago (found an archive to this exact one from two years ago) but I wasn't too sure with some of the help they gave so I'm asking again if someone would kindly help. I know that it involves a=dv/dt so that I end up with something like (v2-v1)=-3v^2(t-0). That may be wrong, I'm in a high school AP class and am just starting with some basic derivations in my calc class, so it is very possible that there's a flaw in all that (and in some of my terminology or whatnot) but if I'm on the right track my main concern would be the v^2. I saw on another post that someone did

    dv/v^2 = -kdt

    -1/v = -kt +C

    but I'm not sure how they got there (besides it looking like they took the integral)
    Again, things may be wrong and I wouldn't be surprised, but any help would be appreciated, thanks.
  2. jcsd
  3. Sep 10, 2008 #2


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    Homework Helper

    Welcome to PF.

    The solution you are referencing is using acceleration as dv/dt. Since what you are looking for is a value for the difference in time that should be found by employing the difference in velocity of the given problem. You separate the V terms to the V side. (This eliminates I think your main concern.) And then integrate both sides.

    That yields the relationship V to dv and t to dt.

    -1/V = -k/t + C

    Now using this relationship, subtract from one known Vo the other known V1, and with the initial t being 0 you can solve for t1

    (-1/1.5) - (-1/ .75) = (-3*0+c) - (-3t1+c) = 3*t1

    That's at least how I see the problem.
  4. Sep 10, 2008 #3
    Thanks for the help, but I'm sorry to say I'm still a little bit lost on that...
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