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Trouble Deriving a Constant Acceleration Formula

  1. Sep 9, 2012 #1
    Q: How long does it take a car to cross a 30 m-wide intersection after the light turns green, if it accelerates from rest at a constant 2.00 m/s^2?

    Attempt: V=V0 + at; x-x0/t=v0+at; x-x0=v0+at^2; x=v0+x0+at^2; 30=0+0+2t^2; 15=t^2.

    The square root of 15 ended up being my final answer, when the real answer was closer to 5.48 seconds. I just started physics, so I need to know what I personally did wrong when USING the formula I attempted to use. Giving me a different formula won't be helpful, because it won't tell me what I did wrong, and because my book has a whole chart of formulas I could refer to. If someone could explain to me how I derived that formula incorrectly, it would be a load of help. Thank you!
  2. jcsd
  3. Sep 9, 2012 #2


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    You derived the formula incorrectly by assuming that velocity at time t is displacement/time. That it is not true when velocity is not constant. To compound the problem, your algebra is wrong in equation 3, but it doesn't matter since equation 2 is wrong.The kinematic equations are best derived form the calculus, check out a good web site for this. You might want to refer to the chart of formulas in the book, and you will discover that none of them are the same as the one you derived.
  4. Sep 9, 2012 #3
    Right there; that does not follow. The only way you're going to be able to make the correct step is through integration, which if you're not familiar with isn't going to be possible at all.
  5. Sep 10, 2012 #4


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    You can avoid integration by noting that in a constant acceleration situation, average velocity for any time period equals half the sum of the starting and ending velocities.

    average veloctiy = av = 1/2 (v0 + v1)
    v1 = v0 + at
    av = 1/2 (v0 + (v0 + at))
    av = 1/2 (2v0 + at)
    av = v0 + 1/2 a t

    then distance equals x0 + average velocity x time:

    x = x0 + av t = x0 + (v0 + 1/2 a t) t
    x = x0 + v0 t + 1/2 a t2
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