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Position time graph describing constant velocity

  1. Oct 2, 2007 #1
    a. Q: Find the position at 20s. A: I'm sure it's 56m.

    b. Q: Find the maximum displacement
    A- 28-0 m = 28m
    B- 28-28 m = 0m
    C- 56-28m = 28 m
    D - 0 - 56m= -56m. This is the one I am worried about, is it correct?

    c. Q: Find the velocity in section A of the graph
    A: v=m=rise/run = (28-0)m/(8-0)s = 3.5m/s

    d. Q: Find the velocity in section D of the graph
    A: v=m = rise/run = (0-56)m/(40-20)s = -2.8m/s Is the negative sign correct? Is it negative because the bicycle is going back in the direction of its point of reference?

    e. Q: What is happening in section B?
    A: The bicycle has remains stopped. No motion.

    f. Q: Calculate the average velocity for the whole trip.
    A. V= 3.5 m/s
    B. V= 0 m/s
    C. V= 5.6 m/s
    D. V= -2.8 m/s

    Vave = (3.5)+(0)+(5.6)+(-2.8) m/s = 6.3 m/s / 4 = 1.575 m/s = 1.58 m/s (significant digits)
    Does this seem like an accurate average velocity? Should I have included the result of b which is zero velocity, because using that number greatly reduces my total average velocity...
  2. jcsd
  3. Oct 2, 2007 #2


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    Staff Emeritus
    Science Advisor

    The above is all correct.
    I'd use a weighted average here. So, the average velocity = (3.5*8+0*7... etc)/40 (do you see what I'm doing there? You need to weight the average velocity since the four components are not all equal time intervals)

    Apart from that, well done! :smile:
  4. Oct 2, 2007 #3


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    Science Advisor

    For part D, the answer should be +56 m. The maximum displacement is simply the furthest distance the point gets from zero; at a glance, you can see that this is +56m.

    Yes, velocity is negative when the object is going backwards.

    The easy way to get average velocity is (total displacement)/(total time). In this case, total displacement is zero, so the average velocity is zero. This is equivalent to the "weighted average".
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