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Car Work, Acceleration and Power

  1. Mar 16, 2009 #1
    1. The problem statement, all variables and given/known data

    As the driver steps on the gas pedal, a car of mass 1100 kg accelerates from rest. During the first few seconds of motion, the car's acceleration increases with time according to the expression below.

    a = (1.160 m/s^3)t - (0.210 m/s^4)t^2 + (0.240 m/s^5)t^3

    (a) What work is done on the car by the wheels during the interval from t = 0 to t = 3.00 s?

    ____________ J

    (b) What is the wheels' output power at the instant t = 3.00 s?

    ___________ W


    3. The attempt at a solution

    Im really quite confused about this one... Although it seems like a problem where derivatives might be involved to simplify this a little more. Am I on the right track at all? Please Help??
     
  2. jcsd
  3. Mar 16, 2009 #2

    LowlyPion

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    How do you determine work? What is the formula?
     
  4. Mar 16, 2009 #3
    well W = P x t
     
  5. Mar 16, 2009 #4

    LowlyPion

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    What about Force? Can't that give you the work?
     
  6. Mar 16, 2009 #5
    yea, W = F x d
     
  7. Mar 16, 2009 #6

    LowlyPion

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    There you go.

    Now the car is accelerating isn't it? So it's applying a force to the car, the F = m*a kind of thing.

    But it also does it over a distance covered in 3 sec.

    So if there was only a way to figure the distance covered in 3 sec that might prove useful?
     
  8. Mar 16, 2009 #7
    i might be totally going the wrong route here but to get the distance part, can you just like take two anti-derivatives to bring it back to position which would give you distance??

    also for the F=ma, how do you know what the acceleration is? can you just plug 0 and 3 into the equation and take the average of that?
     
  9. Mar 16, 2009 #8

    LowlyPion

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    You're given the acceleration, so yes it looks like you can just integrate twice to determine x position distance at 3. You know initial condition is at rest.

    Edit: Forgot you were trying to get to Work, so you can integrate once then for the Velocity function. Your work then should be equal to the kinetic energy at t=3 right?
     
  10. Mar 16, 2009 #9
    hmm, i did all that but it didnt come out right..
     
  11. Mar 16, 2009 #10

    LowlyPion

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    Forgot you were trying to get to Work, so you can integrate once then for the Velocity function. Your work then should be equal to the kinetic energy at t=3 right?
     
  12. Mar 16, 2009 #11
    hmm maybe, alright well thanks for you help!
     
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