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Engine with constant power; Time taken to finish track?

  1. Jun 5, 2014 #1

    Nathanael

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    1. The problem statement, all variables and given/known data
    "A funny car accelerates from rest through a measured track distance in time T with the engine operating at a constant power P. If the track crew can increase the engine power by a differential amount dP, what is the change in the time required for the run?"

    2. The attempt at a solution
    I'll just think outloud here so you guys can see my thought process and correct any mistakes that way (I'll get into the math after the reasoning)

    I figured if I could write an equation for the time it takes to move a fixed distance with a constant power (and constant mass) then I should be able to find the answer by differentiating (the time) with respect to power.

    If the power is constant, then the work is linear with respect to time.

    My "line of attack" will be to assume the work being linear causes the kinetic energy to be linear (wrt to time).

    (Edit: section about friction omitted ... Basically I'm ignoring it by assuming it will be common to both cases)

    So, the kinetic energy is increasing linearly w.r.t. time, and so we can write an equation for the velocity as a function of time:
    [itex]v=\sqrt{\frac{2Pt}{m}}[/itex]
    Where P is the power (and m is mass, t is time)

    If we integrate that from 0 to some time "T" and set it equal to "r" (r is the distance of the track) then we should be able to get an equation for the time as a function of the power. (We also know that the distance at time zero is simply zero so we can just take the antiderivative and set it equal to r at some instant T)
    [itex]^{T}_{0}∫v.dt = r = \frac{m}{3P}(\frac{2PT}{m})^{3/2}[/itex]

    Solving for T we get:
    [itex]T=(\frac{9mr^2}{8P})^{1/3}[/itex]

    Then take the derivative of that and you get
    [itex]\frac{dT}{dP}=\frac{9^{1/3}r^{2/3}m^{1/3}}{6P^{4/3}}[/itex]

    Multiply both sides by dP and plug in the above definition of T and I finally get my answer...

    [itex]dT=\frac{TdP}{3P}[/itex]



    But....... This can't be the correct answer because it implies that an increase in power causes an INCREASE in time.... which is counter-intuitive. (The time should decrease.)
     
  2. jcsd
  3. Jun 5, 2014 #2

    BvU

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    A least for solving for T you get a T that decreases with P, so there might be something wrong with your derivative.....
     
  4. Jun 5, 2014 #3

    verty

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    The derivative of ##1 \over P^{1/3}## is?

    PS. He showed his work, or it is an error in the textbook.
     
  5. Jun 5, 2014 #4

    Nathanael

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    Ahh yes thank you for pointing that out. That is where I went wrong (I forgot the factor of -1 from the derivative of [itex]p^{-1}[/itex])

    So my answer would actually be [itex]\frac{-T}{3P}dP[/itex]

    Thank you very much for reading this and paying enough attention to pinpoint my mistake.
     
  6. Jun 5, 2014 #5

    Nathanael

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    It is [itex]\frac{-1}{3P^{4/3}}[/itex]
    Whereas I mistakingly calculated it to be [itex]\frac{1}{3P^{4/3}}[/itex]
    (Well I didn't actually calculate it by itself, but that's essentially the mistake I made)

    I'm not sure what you're referring to, but the textbook's solution is [itex]\frac{-T}{3P}dP[/itex]
     
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