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Drag Force problem

  1. Oct 2, 2008 #1
    1. The problem statement, all variables and given/known data
    You dive straight down into a pool of water. You hit the water with a speed of 6.5 m/s, and your mass is 65 kg.

    Assuming a drag force of the form [tex]F_D=cv=(-1.2*10^4 \tfrac{kg}{s})*v[/tex], how long does it take you to reach 2% of your original speed? (Ignore any effects of buoyancy.)

    2. Relevant equations

    3. The attempt at a solution


    Logically, the answer does not make sense, nor is it correct. Where did I go wrong?
    Last edited: Oct 2, 2008
  2. jcsd
  3. Oct 2, 2008 #2
    You have assumed acceleration is constant. In your equation, [tex] a = \frac{cv}{m} - g [/tex]

    And then you used constant acceleration formulae. However, think about what happens as v changes.
  4. Oct 2, 2008 #3
    Ah, yes. As velocity decreases, the acceleration does as well. This is problematic for me, however, as I have not worked with non-constant accelerations. How should I go about solving this problem?
  5. Oct 2, 2008 #4
    Try this:
    solve for Fnet first, before acceleration.
  6. Oct 2, 2008 #5
    If you mean:

    I can solve for it, but do I use 6.5 as my value for v? I still don't know what to do with the resulting value? Doesn't that still have the same problem of assuming constant acceleration?
  7. Oct 2, 2008 #6
    well think about this:

    [tex] F = m \frac{v}{t} [/tex]

    so [tex] m \frac{v_s}{t} = cv - mg [/tex]

    where v_s is the one which is at 2%
    re arrange to get t.

    I hope I didn't make any careless mistakes I was up all night
  8. Oct 2, 2008 #7


    This is also an incorrect answer =/
  9. Oct 3, 2008 #8
    Shameless bump - still haven't resolved this question :(
  10. Oct 3, 2008 #9
    You have to use partial derivatives.

    FD-mg = ma
    cv-mg = m dv/dt
    integral(dt)|0,t = (dv/ (cv/m)-g)
  11. Oct 3, 2008 #10
    Ahh catamara is totally right, we were on the right track though
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