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One more thing with motion problem

  1. Jan 23, 2006 #1
    I'm referring to the same problem as before: https://www.physicsforums.com/showthread.php?t=107835

    But now my question is how do I figure out parts c and e? My textbook says to just take the slope of the position-time curve. But if t = 1 in part c, how do I setup the slope? It'll just be something over 1, or that something. My problem is if I take the slope of the change in position over 1, I just get my answer to part a again, the average velocity or 71.3. I'm stuck. Please help?
     
  2. jcsd
  3. Jan 23, 2006 #2

    eep

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    the instantaneous velocity function is given by the time-derivative of the position function.

    do you have an expression for the velocity at some time t?
     
  4. Jan 23, 2006 #3
    So does that mean that I take the derivative of x = 2.30^5 and plug in 1s and 2s, respectively for parts c and e?
     
  5. Jan 23, 2006 #4

    eep

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    yup! and the function for instantaneous acceleration is the time-derivative of the velocity function. One could also take the slope of the position function at t=1.0, since that's what the formula for the instantaneous velocity is giving you.
     
  6. Jan 23, 2006 #5
    Thanks! So basically..

    For the instantaneous velocity take the first derivative, and for the instantaneous acceleration take the second derivative? (Of x = 2.30t^5)
     
  7. Jan 23, 2006 #6

    eep

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    Correct. Of *any* position function.
     
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