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Final velocity of an object accelerated across a given distance with constant power

  1. Aug 10, 2007 #1
    1. The problem statement, all variables and given/known data
    An object of mass M begins with a velocity of 0 m/s at a point. A power input of P watts goes directly to kinetic energy until the object has traveled a distance of X meters. What is the final velocity of the object?

    So, we have constant variables
    M = mass
    X = distance that power will be input
    P = power level

    And also
    V = final velocity, after traveling distance X (to be solved)

    2. Relevant equations

    E = 1/2MV[tex]^{2}[/tex]
    X = [tex]\int[/tex]V(t) dt - V(t) is V as a function of time.
    E = P*t

    3. The attempt at a solution

    Find V as a function of E (easy)
    V = [tex]\sqrt{2E}[/tex]/M

    Find V as a funciton of time t (use equation)
    V = [tex]\sqrt{2P*t}[/tex]/M

    Now take the equation for X
    X = [tex]\int[/tex]V(t) dt

    And Find X as a function of t directly, knowing the V(t) function
    X = [tex]\int[/tex][tex]\sqrt{2P*t}[/tex]/M dt Edit: Should be, and was calculated as ([tex]\int[/tex] [tex]\sqrt{2P*t}[/tex] dt)/M

    Integrate (remember that sqrt(2P) is a constant)
    X(t) = (2/3)*[tex]\sqrt{2P}[/tex]*t[tex]^{3/2}[/tex]/M

    Now change to t in terms of X
    t(X) = ((3/2)/[tex]\sqrt{2P}[/tex]*M*X)[tex]^{2/3}[/tex]

    And finally slide that into the V(t) equation
    V(X) = [tex]\sqrt{2P}[/tex]*((3/2)/[tex]\sqrt{2P}[/tex]*M*X)[tex]^{1/3}[/tex]/M

    Simplify (whew!)
    V(X) = 3[tex]^{1/3}[/tex]*2[tex]^{-1/6}[/tex]*P[tex]^{1/3}[/tex]*X[tex]^{1/3}[/tex]*M[tex]^{-2/3}[/tex]

    So, V correlates directly with the cube root of X, the cube root of P, and M^(-2/3), with a weird constant.


    Am I right? Am I not? If not, where did I go wrong?
     
    Last edited: Aug 10, 2007
  2. jcsd
  3. Aug 10, 2007 #2

    learningphysics

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    Your general solution is fine, but I think you made some calculation errors... your m should be under the square root... so that affects your final exponent on m... also I'm getting that the 2 cancels out (ie the power of 2 is 0)...

    I'm getting: [tex]3^{1/3}*p^{1/3}*x^{1/3}*m^{-1/3}[/tex] as the final velocity. Though I may have screwed up...
     
    Last edited: Aug 10, 2007
  4. Aug 10, 2007 #3
    Which specific step does the miscalculation occur? I can't quite see what you mean.

    Also, a more general question about energy: What will a constant power input do to something with a velocity with no immediate context, like a rocket in space?

    On the one hand, I think acceleration should remain constant, because it's change in momentum is relative only to itself, but on the other hand, a constant rate of fuel consumption should mean a a constant power input. The two don't seem to be compatible. What gives?
     
    Last edited: Aug 10, 2007
  5. Aug 10, 2007 #4
    Integrate (remember that sqrt(2P) is a constant)
    X(t) = (2/3)*LaTeX graphic is being generated. Reload this page in a moment.*tLaTeX graphic is being generated. Reload this page in a moment./M

    Now change to t in terms of X
    t(X) = ((3/2)/LaTeX graphic is being generated. Reload this page in a moment.*M*X)LaTeX graphic is being generated. Reload this page in a moment.


    here?

    edit: the step where u isolated t.
     
  6. Aug 10, 2007 #5
    I see where I went wrong with the 2^(-1/6), it should just be 3^(1/3) for the constant, but I still don't see the problem with the M. It was transformed from M to M^(2/3), then square rooted to M^(1/3), which was over M, which is M^(-2/3).
     
  7. Aug 10, 2007 #6

    learningphysics

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    Here is the mistake... the M should be under the square root.
     
  8. Aug 10, 2007 #7
    Ah, you're indeed right. I see now. And it makes sense if you look at units too.

    Okay, Thanks learningphysics!
     
  9. Aug 10, 2007 #8

    learningphysics

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    Acceleration isn't constant for constant power...
     
  10. Aug 10, 2007 #9

    learningphysics

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    you're welcome.
     
  11. Aug 10, 2007 #10
    Can anyone answer the third post of the topic, where I ask about conservation of momentum V. conservation of energy in the case of the rocket?
     
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