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Work, Energy And Power Problem

  1. Feb 12, 2013 #1
    1. The problem statement, all variables and given/known data
    The operation manual of a car, mass 760kg, claims that the car, starting from rest, can reach a speed of 15m.s in first gear in 3 seconds. The acceleration of the car is constant and friction can be ignored. If the car operates under these conditions, calculate:

    9.1) The power output in first gear
    9.2) The maximum distance covered in 3 seconds

    2. Relevant equations

    delta y=vf+vi/2 *t

    3. The attempt at a solution

    9.1) ek = 1/2mv^2
    = 1/2*760*15^2
    = 85500 J

    p = w/t
    = 85500/3
    = 28500 W

    9.2) delta y = vf+vi/2 *t
    = 15/2 *3
    = 22.5m

    Could Someone Please Check?
  2. jcsd
  3. Feb 12, 2013 #2


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    Distance looks good, calc for power is the average power delivered by the engine over that time and distance, not the instantaneous power delivered at the 3 second mark, which is higher, since instantaneous power varies from zero at the start to a max at the given time. I don't know if the problem is asking for average power or instantaneous power.
  4. Feb 12, 2013 #3


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    Since it asks for 'power output', peak power seems the more likely interpretation to me. I.e., what power is it capable of. But I agree the question is poorly worded.
  5. Feb 13, 2013 #4
    So whats the difference between instantaneous and average power?
  6. Feb 13, 2013 #5


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    The average power is the equal to the rate that work is done over a given time period, that is, [itex]P_{avg} = \Delta W/\Delta t [/itex].

    The instantaneous power is limiting value of the average power as the time interval Δt approaches zero, that is, [itex]P_{instant} = dW/dt [/itex], which, in this example for constant force, is equal to [itex] Fds/dt[/itex], or [itex]Fv_{instant}[/itex].

    The average power is 28.5kW, the instantaneous power at t=0 is 0, and at t=3, the instantaneous power is ___?___?
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