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Time taken to reach 90% of terminal velocity

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data
    A 4.5 kg metal sphere is released in a fluid where k = 10.5 N s^2/m^s. How long does it take to reach 90% of its terminal velocity?

    2. Relevant equations
    Force of drag = kv^2, where k is the drag coefficient
    (I believe we're not considering buoyancy.)

    3. The attempt at a solution
    ƩF = ma
    a = (mg-kv^2)/m
    a = [(4.5)((9.8) - (10.5)v^2]/4.5
    a = 9.8 - 2.3v^2

    At terminal velocity,
    ƩF = 0
    mg = kv^2
    v = √(mg/k)
    v = √[(4.5)(9.8)/10.5]
    v = 2.0 m/s [down]
    90% of this is 1.8 m/s [down].

    So I know I have to find the time taken for the ball to achieve a velocity of 1.8 m/s^2 [down], and a have an equation with both acceleration and velocity. However, acceleration is not constant, so all of my kinematics knowledge (the constant acceleration equations) are useless, so I don't know how to proceed. We're supposed to create a graph to get the answer, but my teacher said there's a way to do this without graphing. I'm trying to find out what this method is. Thanks!
     
    Last edited: Feb 28, 2012
  2. jcsd
  3. Feb 28, 2012 #2
    Disregard what I had here before, misread what time you were trying to find. Not sure what other way to do it.
     
    Last edited: Feb 28, 2012
  4. Feb 28, 2012 #3

    Delphi51

    User Avatar
    Homework Helper

    I don't see how to get it with a graph! I would be very interested in seeing how you do it.

    I did solve it numerically on a spreadsheet, with headings for t, d, v and a.
    You CAN use the constant accelerated motion formulas for short time intervals (the smaller the interval the more accurate they are, and you can make the time interval between rows on the spreadsheet as small as you like; keep decreasing it until the answer no longer changes significantly).

    It should be possible to solve it with calculus, Replace a with dv/dt and it becomes a differential equation. Not an easy one, though. I don't know how to solve it.
     
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