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U-substitution for finding v(t)

  1. Oct 17, 2007 #1
    1. The problem statement, all variables and given/known data
    A .25kg ball is dropped from rest. Air resistance is f= -.05v
    a. find v(t)
    b. What is the terminal velocity of the ball?
    c. At what time does the ball reach 50% of its terminal velocity? 90%?

    2. Relevant equations
    F=ma
    F= mg - .05v

    3. The attempt at a solution
    Began to solve the equation knowing that I have to separate the terms, moves all the velocity terms to one side, then integrate.

    ma = mg - .05v
    a = mg/m - .05v/m
    a = g - .2v

    dv/dt = g - .2v
    dt = dv/(g-.2v)

    now im just wondering to integrate that terms on the right. U-substitution for the denominator?
     
  2. jcsd
  3. Oct 17, 2007 #2
    [tex]\int{\frac{du}{a+k u}}=\frac{ln(a+k u)}{k}[/tex]
    Integral tables are your friend ^_^
     
  4. Oct 17, 2007 #3
    so just to make sure im correct, [tex]\int\frac{dv}{g-.2v}[/tex] is [tex]\frac{ln(g-.2v)}{.2} + constant[/tex]
     
    Last edited: Oct 17, 2007
  5. Oct 17, 2007 #4
    alright so i assumed that was correct and went on with the problem, but ran into trouble at c.

    ΣF = ma = mg - .05v
    a = mg/m - .05v/.25
    a = g - .2v
    dv/dt = a

    dv/dt = g - .2v
    dv/(g - .2v) = dt
    ∫ dv/(g - .2v) = ∫dt
    ln(g - .2v)/.2 = t - c
    ln(g - .2v) = .2(t - c)
    e[tex]^{.2(t - c)}[/tex] = g - .2v
    e[tex]^{.2(t - c)}[/tex] - g = -.2v
    -[e[tex]^{.2(t - c)}[/tex] - g]/.2 = v(t)

    Ball starts at rest, v = 0 when t = 0
    [e[tex]^{-.2c}[/tex] - g]/-.2 = 0
    -5[e[tex]^{-.2c}[/tex] - g] = 0
    -5e[tex]^{-.2c}[/tex] + 5g = 0
    -5e[tex]^{-.2c}[/tex] = -5g
    e[tex]^{-.2c}[/tex] = g
    ln(g) = -.2c
    -ln(g)/.2 = c
    c = -11.42

    v(t) = -5e.2(t + 11.42) + 49.05

    What is the terminal velocity?
    ΣF = ma = mg - .05v = 0
    .05v = .25g
    v = 5g = 49.05 m/s

    At what time does the ball reach 50% of its terminal velocity?
    v(t) = -5e.2(t + 11.42) + 49.05
    v(t) = 24.525 m/s

    24.525 - 49.05 = -5e[tex]^{.2(t + 11.42)}[/tex]
    4.905 = e[tex]^{.2(t + 11.42)}[/tex]
    ln(4.905) = .2(t + 11.42)
    ln(4.905) - 2.284 = .2t
    [ln(4.905) - 2.284]/.2 = t

    t = -3.47 sec
    t is negative. whattttt????
     
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