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Determining power output at different time intervals

  1. Dec 18, 2016 #1
    1. The problem statement, all variables and given/known data
    A 49.0 kg sprinter, starting from rest, runs 58.0 m in 9.90 s at constant acceleration.
    What is the sprinter's power output at 1.90 s , 3.00 s , and 5.70 s ?

    2. Relevant equations
    ##d=v_it + \frac{1}{2}at^2##
    ##F=ma##
    ##W=Fd##
    ##P=\frac{W}{t}##

    3. The attempt at a solution
    I got the acceleration from given data:
    ##58=(0)(9.9) + \frac{1}{2}(a)(9.9)^2 \longrightarrow a \approx 1.184 m/s^2##
    Then I got the force:
    ##F=(49)(1.184) \approx 58 N##
    Then I tried to calculate the distance traveled at different time intervals, starting with at ##t=1.90##:
    ##d=\frac{1}{2}(1.184)(1.9)^2 \approx 2.14##
    Finally, I tried to get Work and then Power from that:
    ##W=Fd=(58)(2.14)=123.9 J \longrightarrow P=\frac{W}{s}=\frac{123.9}{1.9} \approx 65.2 W##
    However, this is not correct, the correct answer is 130 W. What am I doing wrong here?
     
  2. jcsd
  3. Dec 18, 2016 #2

    nrqed

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    You need to use the formula Power = F v (force times velocity) and use the velocities at the given times. (Basically, what you calculated are average power over the given time intervals, while they want the instantaneous powers at the given times)
     
  4. Dec 18, 2016 #3
    Ah okay, so using ##v_f = v_i + at## I get ##v_f = (1.184)(1.9) \approx 2.25## and then ##P = Fv = 58(2.25) \approx 130 W## is correct. If anyone sees this, could they explain why this formula works when the previous one doesn't? Still trying to wrap my head around that.
     
  5. Dec 18, 2016 #4

    nrqed

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    Think about the equation ## P = F \frac{d}{t} ##, this is really ## P = F \frac{\Delta x}{\Delta t} ##, this gives a power output between two instants separated by a time ## \Delta t## and while the object moved a distance ##\Delta x##. This involved a choice of two instants so one essentially calculates an average power output. If the velocity is constant, the power obtained this way is the same as the instantaneous power at every instant because the power output is constant. if the velocity is not constant, to get the instantaneous power output at a given instant one must take a limit:

    $$P_{inst} = lim_{\Delta t \rightarrow 0} ~~ F ~\frac{\Delta x}{\Delta t} = F \, v_{inst} $$
     
  6. Dec 18, 2016 #5
    Thanks a bunch, that really clarifies it!
     
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