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Simple Acceleration Equation Help

  1. Mar 16, 2008 #1
    [SOLVED] Simple Acceleration Equation Help

    g=-9.80 m/s^2
    t= 3.85
    vi= 0.00 m/s
    d=vi(t) + (1/2)gt^2

    d=0 + (1/2)(-9.80m/s^2)((3.85s)^2)

    I'm trying to find the distance. So, this d represents displacement. Since in this case, this is also the distance, would there be any other work to show the sign change, since the d would be going from a vector to a scalar? Or should I have modified this equation beforehand to show d as a scalar, and not have taken into account the gravities negative?

    By the way, sorry, this must seem like a trivial matter really not worth time, but I hate not understanding it down to the tiny details. Thank you very much!
  2. jcsd
  3. Mar 16, 2008 #2
    The negative sign looks fine to me since the displacement is below the axis where you set your initial displacement to be zero.
  4. Mar 16, 2008 #3
    Thanks. So, would I just leave out the negative when expressing it as a distance, and there would be no work needed?
  5. Mar 16, 2008 #4
    That's right. Distance, by definition in physics, cannot be negative. However, most teachers I've come across with in high school tend to blur the distinction between distance and displacement in physics questions. If you want to be technical about it, leave out the negative sign and if the teacher has you bollocked for not putting in the negative sign, tell them the true meaning of distance in physics. :rofl:
  6. Mar 16, 2008 #5
    Haha, excellent. Thank you for the help! :smile:
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