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Power and velocity as a function of time

  1. Sep 11, 2009 #1
    1. The problem statement, all variables and given/known data

    A steam-engine is traveling along a rail, with a constant power output of [tex]1.5MW[/tex], regardless of its velocity.

    1. What is the mass of the steam-engine if it is known that it accelerates from [tex]10\frac{m}{s}[/tex] to [tex]25\frac{m}{s}[/tex] in [tex]6 sec[/tex]?
    2. Describe the velocity of the steam-engine as a function of time.
    And the other questions follow from these two.

    2. Relevant equations
    [tex]P=\vec F \cdot \vec V=\frac{\Delta W}{\Delta t}=constant[/tex]


    3. The attempt at a solution
    1.

    [tex]P=\frac{\Delta W}{\Delta t}=\tfrac{1}{2}\frac{(v_f^2-v_0^2)}{\Delta t}m[/tex]

    [tex]m=\frac{2P\Delta t}{v_f^2-v_0^2}[/tex]
    //

    2.

    [tex]F=\frac{P}{v}[/tex]

    [tex]a=\tfrac{P}{m}v^{-1}[/tex]

    [tex]\dot v=\tfrac{P}{m}v^{-1}[/tex] Here we have a simple differential equation.

    [tex]\frac{dv}{dt}=\tfrac{P}{m}\frac{dt}{dx}[/tex]

    [tex]\frac{dv}{dx}\frac{dx}{dt}=\tfrac{P}{m}\frac{dt}{dx}[/tex]

    [tex]v\cdot dv=\tfrac{P}{m}dt[/tex]

    [tex]\tfrac{1}{2}v^2=\tfrac{P}{m}\cdot t+C[/tex]

    Now how do I get from here to the final solution? Omitting the [tex]+C[/tex] completely provides a function [tex]v(t)[/tex] that is indeed a solution to the differential equation. However, it assumes that [tex]v_0=0[/tex]
    I can't quite find a way to incorporate the initial values into the solution.

    Well, I did find a way, but it doesn't quite make sense mathematically, I'd love a thorough explanation as to why the following is correct:

    [tex]\tfrac{P}{m}\equiv \tau[/tex]
    [tex]v(t)=\sqrt{2\tau\cdot t+v_0^2}[/tex]

    Taking the time-derivative provides us with:
    [tex]\dot v(t)=\frac{1}{2\sqrt{2\tau\cdot t+v_0^2}}\cdot 2\tau[/tex]
    The above must also be equal to: [tex]\tau v^{-1}[/tex] in order to satisfy the differential equation.
    [tex]\tau v^{-1}=\tau \frac{1}{\sqrt{2\tau\cdot t+v_0^2}}[/tex]

    So the differential equation is satisfied.

    Checking that against the data in question #1, shows that it holds, and it makes sense, since at [tex]t=0\rightarrow v=v_0[/tex], however, it seems as though we used the integration constant quite haphazardly, so I'd love an explanation. :)

    Is it just a matter of isolation [tex]v(t)=\sqrt{2\tau\cdot t+2C}[/tex]

    And then finding, using the initial values of the problem that [tex]2C=v_0^2[/tex] ?


    With thanks in advance, Anatoli.
     
    Last edited: Sep 11, 2009
  2. jcsd
  3. Sep 11, 2009 #2

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    The problem asks you to find the velocity as a function of time. Assume that that "time" starts (t = 0) when the steam-engine is traveling at 10 m/s. Put that in your equation and solve for C. It would be clearer to you what is going on if you had limits to your integration of the diff. eq., namely v0 to v on the left and t0 to t on the right.
     
  4. Sep 12, 2009 #3
    Oof! Big mistake on my part, thanks a lot for helping clear that up. :)
    I assumed [tex]v_0=0[/tex] so no wonder that showed up in my result!
     
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