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Acceleration, Solve for time (Algebra Based)

  1. Sep 7, 2008 #1


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    1. The problem statement, all variables and given/known data
    22. The average person passes out an
    acceleration of 7g (that is, seven times the
    gravitational acceleration on Earth).
    Suppose a car is designed to accelerate at
    this rate. How much time would be
    required for the car to accelerate from rest
    to 60.0 miles per hour? (The car would need
    rocket boosters!)

    2. Relevant equations

    3. The attempt at a solution
    v1 = 60
    v0 = 0
    a = 7 * 9.8 => 68.6m/s^2

    a = change in velocity/change in time

    So, 68.6m/s^2 = 60mph/t
    68.6 * t = 60mpg
    t = 60/68.6
    = 0.87 seconds

    I am not sure if this is correct or not.
    The answer seems within reason since a very high powered sports car accelerates from 0-60 miles per hour in around 3 seconds, so 0.87 seconds would line up well with the exaggeration of the speed of the car mentioned at the end of the word problem.

    Any help is appreciated :).
  2. jcsd
  3. Sep 7, 2008 #2
    You are mixing English and SI units.
  4. Sep 7, 2008 #3


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    Care to elaborate on where I went wrong and what I can do to correct my problem?
  5. Sep 7, 2008 #4
    What he/she meant was that you are using G's (or 9.8m/s2) which is the acceleration on earth in metric, and then Miles per hour, which is SI.

    To fix this, take the MPH and multiply by .44 (or to be more exact, .44704) to get meters/second

    By the way, the way to get the .44 is roughly 1600 meters/3600seconds = .44 (mile/hour hour=60 minutes=60 seconds. so 602)

    This should give you some help, or at least point you in the correct direction.

    Your equation is correct though.
    Last edited: Sep 7, 2008
  6. Sep 7, 2008 #5


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    Alright, if I multiply 60mi/h by .44, I receive 26.4m/s. I divide this by 68.6m/s^2 to get an answer of .385s? Both the numerator and denominator have 3 significant digits, so I stop at .385 correct?

    I see my mistake now. Thank you both for pointing that out to me.
  7. Sep 7, 2008 #6
    No problem Dig,
    That seems correct to me.

    Good luck in physics this year.
  8. Sep 7, 2008 #7


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    Thank you so much :).
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