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Solving for Velocity

  1. May 12, 2016 #1
    1. The problem statement, all variables and given/known data
    423,000=5504v+4.37v^3
    2. Questions
    Could someone please advise on how to solve for velocity(v)? Our tutor solved the equation using equation solver software to get 36.9 m/s (133km/h) but he included it in our exam and wanted a manual calculation of the velocity... It's beyond me, I am just trying to gauge how difficult it is to obtain the answer as I find it unusual that he had to use software to get a value but expected us to calculate it manually.
     
  2. jcsd
  3. May 12, 2016 #2

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    This is a cubic equation in v. There is a formula for calculating the roots of a cubic equation, but it is much more complicated than the quadratic formula for sloving quadratic equations.

    https://en.wikipedia.org/wiki/Cubic_function

    Rather than memorizing several pages of formulas, it's probably easier to use a trial and error method to find a root.
     
  4. May 12, 2016 #3
    Now that you have mentioned "trial and error" I'm fairly sure that's the method he wanted us to use. It didn't seem so obvious at the time though! Hopefully it doesn't catch me out again, thanks for the help.
     
  5. May 12, 2016 #4

    gneill

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    Staff: Mentor

    You could also apply an iterative method that will converge to a solution. There are many to choose from. Check out Newton's Method for the "classic" example. It requires a bit of calculus (one derivative). It helps if you have a programmable calculator.
     
  6. May 12, 2016 #5
    let f(v)= 4.37v^3 + 5504v - 423000
    f'(v)= 13.11v^2 + 5504 >0 for all v belong to R.

    => f is monotonically increasing

    f''(v)=26.22v
    => f has a point of inflection at v=0

    you can easily sketch the rough graph for f
    now using the mid point concept
    put v=0 -> f(0)<0
    put v=100 --> f(100)>0
    put v=50 --> f(50)>0
    pt v=25 --> f(25)<0
    put v= 37.5 --> f(37.5)>0
    put v= 31.25 --> f(31.25)<0
    put v= 34.375--> f(34.375)<0
    put v= 35.9375 ---> f(35.975)<0
    put v= 36.71875 --> f(36.71875)<0
    put v= 37.109375 --> f(37.109365)>0
    put v= 36.9140625 --> f(36.9140625) is nearly zero => approximate solution for v= 36.9
     
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