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Homework Help: Power Relative to Time Question

  1. Jul 25, 2010 #1
    1. The problem statement, all variables and given/known data
    A funny car accelerates from rest through a measured track distance in time 57 s with the engine operating at a constant power 290 kW. If the track crew can increase the engine power by a differential amount 1.0 W, what is the change in the time required for the run?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jul 25, 2010 #2
    Go on then, what's your idea of how to do it?
  4. Jul 25, 2010 #3
    I know that you have to take the integral of the P=dW/dt, but i don't know where to go from there.
  5. Jul 25, 2010 #4
    Well, you know that power is the rate of doing work. So you can work out how much work has been done in accelerating. Yes?

    Now you also know that work = Force * distance.

    You don't know what the distance is - but you do know it is the same for both runs so there's a good chance it will cancel - so just call it s.

    We also know that s =1/2 * a * t^2 and f = m * a (Newtons equations)
    So s=1/2 * f/m * t^2
    m and s are constants so f * t^2 must be a constant.

    i.e f1 * t1 ^2 = f2 * t2^2 ( calling the force and time for each run 1 and 2)

    So if we multiply each side by s, we get
    s * f1 * t1^2 = s * f2 * t2^2

    but s* f is the work done in each run, so now we have a simple formula relating the work done to the time of the run ... and we know the work done.
  6. Jul 25, 2010 #5
    I appreciate the help. I still dont understand it. Would it be too much to ask for a step by step solution? Physics is a very hard subject for me.
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