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Homework Help: Differentiating Power and Time

  1. Oct 23, 2011 #1
    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? (Use any variable or symbol stated above as necessary.)

    2. Relevant equations
    I think its:
    P = dW/dT
    P = Fvcos(Phi) = F dot v
    D = 1/2aT^2

    3. The attempt at a solution
    Sorry this is my first post, please be lenient on me...
    Okay, so far I started with

    P = dW/dT = F [dot] V

    F = (ma)

    V = aT

    so P = ma(aT)cos(Phi)

    P = [mTa^2] cos(Phi)

    P = mT[(2d)/T^2]^2 cos(Phi)

    P = mT[(4d^2)/T^4] cos(Phi)

    P = [4md^2/T^3] cos(Phi)

    dP/dT = [8md/3T^2] [-sin(Phi)]

    dT = [[3T^2(dP)/8md] [-sin(Phi)]

    I know its wrong, but I need some guidance because I'm really confused...
  2. jcsd
  3. Oct 23, 2011 #2

    I like Serena

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    Homework Helper

    Welcome to PF, napaul! :smile:

    Your formulas for P are correct.
    However, I'm afraid your formula for D is wrong.

    Similarly your formula for V is wrong.

    Your formula for D and V only hold when acceleration a is constant, which in this case it is not.

    The proper formulas are:
    [tex]dv = adt \quad \text{or} \quad v=v_0 + \int_0^t a dt[/tex]
    [tex]dx = vdt \quad \text{or} \quad D=\int_0^T v dt[/tex]

    You can get "a" as a function of "v" from F=ma and P=Fv=constant.
    From there you should solve dv=adt...

    Btw, you can leave Phi out of your equations since a car would always accelerate in the direction of its speed.
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