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Acceleration lab

  1. Feb 11, 2008 #1
    1. The problem statement, all variables and given/known data

    So basically we allowed a cart to role down an incline plane and we recorded the results with a ticker taper. Now there are a couple of questions that my teacher asked that I do not fully understand.

    This entire section is a bit shady. I do not fully understand what the hell a represents?

    Now, the actual velocity vs. time function should be (this was worked out with calculus):
    v(t) = 331 t + 52 we'll call this the 'accepted value'
    Remember that this is written in slope-intercept format, so the accepted values are:
    a = ____________ mm/s v(0) = 52mm (accepted values)
    Using the interval velocity vs. time graph, Excel worked out the velocity function as
    v(t) = 326 t + 41 we'll call this the 'measured value'
    Remember that this is written in slope-intercept format, so the measured values are:
    a = ____________ mm/s v(0) = 41mm (measured values)
    What should v(0) actually be? _______________
  2. jcsd
  3. Feb 11, 2008 #2
    I have no idea either, is it possible that the units are wrong and it is referring to the acceleration? You can tell that even though your acceleration is reasonably close, about a 2% fractional error by my mental math, that the value of v(0) is completely off, which probably indicates a launcher that launches different from what its predicted to do.
  4. Feb 11, 2008 #3
    so would the acceleration just be the coefficient in front of the t (with possible wrong units?) Also, how would we indicate what the value of V(0) should be?
  5. Feb 11, 2008 #4
    Yes, if a represents acceleration then it is the time derivative of velocity, and the coefficient in front of the t in this case. The value of v(0) is given to you as an accepted value, right? v(0) = 52mm/s. Ah man, I just noticed the units for v(0) are wrong too. Huh? Maybe metaphysics will provide the answer... *Click* Ums online has failed me for the last time.
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