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Change of variable dif equation problem

  1. Jul 24, 2005 #1
    Driving your 1000 kg car on I-5 at 100 km/h, you suddenly noticed a large group of animals. You slam on a break and this provides a stopping force equal to v^3 plus the friction force of 10v. Set up and solve a differential equation for velocity. You need to find a suitable change of variable.
     
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  3. Jul 24, 2005 #2

    Zurtex

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    What have you attempted?
     
  4. Jul 24, 2005 #3
    So far? Nothing...I'm really looking for a place to start =(
     
  5. Jul 24, 2005 #4

    Pyrrhus

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    Start off with Newton's 2nd Law

    [tex] \sum_{i=1}^{n} F_{i} = m \frac{dv}{dt} [/tex]
     
  6. Jul 24, 2005 #5
    Alright, so I starting with Newton's equation...I'm thinking:
    F = m*V = 1000*V
    then I'm assuming F = Stopping force which is: F = V^3 +10V = 1000V
    Integrating both sides...
    (V^4)/4 + 5V^2 = 500V^2
    Then solving for V, I got 13.416
    But this totally seems wrong to me...
     
  7. Jul 24, 2005 #6

    Pyrrhus

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    You cannot integrate it like that :surprised

    [tex] v^3 + 10v = m \frac{dv}{dt} [/tex]

    [tex] \frac{dt}{m} = \frac{dv}{v^3 + 10v} [/tex]
     
  8. Jul 24, 2005 #7
    So integrating that I got
    t/m = .1 ln|v| - .05 ln|v^2 + 10|
    I wasn't really sure how to integrate the right hand side on paper, so I used my calculator
    Solving for v, using m = 1000:
    e^(t/50) = v/((v^2) + 10)
    I used a variable g to = e^(t/50) which led to:
    gv^2 +10g = v
    I tried using the quadratic after that:
    (1 +/- [1-40g]^.5 )/2g = v
    now I'm totally confused
     
    Last edited: Jul 24, 2005
  9. Jul 24, 2005 #8
    If you think about this physically for a sec
    you'll see you need a minus sign.
    dv/dt really aught to be getting smaller
    not bigger.

    mv' = -(v^3 + 10v)

    this is seperable as above.
    [tex] -\frac{dv}{v^3 + 10v} = m dt [/tex]

    to integrate it we can breakit into partial fractions
    [tex] \frac{1}{x^3 + 10x} = \frac{A}{x} + \frac{Bx + C}{x^2 + 10} [/tex]
    where we pick A, B, C to make the equality hold.

    Once you've got the easier to integrate sides.
    Integrate using the initial conditions.
    [tex] \int_{v_0}^v \ldots dv = \int_0^t \ldots dt [/tex]

    then invert the formula to solve for v.


    [or if we want to look for a change of vars..]
    mv' + v^3 + 10v = 0.
    so
    m(vv') + v^4 + 10v^2 = 0

    [tex] \frac{m}{2} \frac{d}{dt}(v^2) + (v^2)^2 + 10 (v^2) = 0 [/tex]

    which looks awefully amenable to the substitution u=v^2..
     
    Last edited: Jul 25, 2005
  10. Jul 25, 2005 #9

    Pyrrhus

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    You're right Qbert the aceleration should be negative, but you make a slight mistake, It's [itex] \frac{dt}{m} [/itex].
     
  11. Jul 25, 2005 #10
    Following that...
    since m = 1000
    (500)u d/dt + u^2 + u = 0

    Is it a legal operation to factor a "u" out of this equation?
     
  12. Jul 25, 2005 #11
    doh! hoisted by my own petard.
    good catch.
     
  13. Jul 25, 2005 #12
    no way! (in this context, d/dt standing alone is meaningless.)

    you have to solve the ODE: m du/dt = -(u^2 + 10u).
     
  14. Jul 25, 2005 #13
    Okay, so I solving m du/dt = -(u^2 + 10u):
    t/500 + C1 = 1/10 ln(u+10) - 1/10 ln(u)
    t/50 + C2 = ln((u+10)/u) where C2 = 10*C1
    e^(t/50 + C2) = Ce^(t/50)-1 = 10/u
    u = 10(Ce^(t/50))^-1
    v^2 = 10(Ce^(t/50))^(-1) -----> v = +/-[10(Ce^(t/50))^-1]^(1/2)
    But v should just be the negative right?
     
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