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Simple Velocity Problem

  1. Sep 8, 2010 #1
    [PLAIN]http://lc.cit.jmu.edu/cgi-bin/plot.png?file=safavifn_jmu_1283993422_6827202_plot.data [Broken]

    1. A bicyclist has the position-versus-time graph as shown. What is the bicyclist's velocity (in m/s) at t = 5 seconds?



    2. Since this is a simple velocity problem, I thought it would be v= 60 m/ 5s



    3. My answer was 12m/s

    I don't know what I'm doing wrong, could it be that I'm reading the graph wrong?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 8, 2010 #2

    rock.freak667

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    velocity is the rate of change of displacement with time or v=ds/dt. So you need to find the gradient of that section from t= 0 to t=5. At what point does it cut the vertical axis? What position does t=5 correspond to?
     
  4. Sep 8, 2010 #3
    So would it be (61-51)/5? Since t=0 is 51m and t=5 is 61m?
     
  5. Sep 8, 2010 #4

    rock.freak667

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    Yes that looks correct to me.
     
  6. Sep 8, 2010 #5
    Thanks.

    How would I do something like t=37s?
     
  7. Sep 9, 2010 #6

    rock.freak667

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    Then you'd need to consider the line segment between t=30 and t=40.

    So get the position at t=37, that will be one point. Then get the position at any time between 30 to 40 (so t=31 for example). Then just use the same gradient formula you used before.
     
  8. Sep 9, 2010 #7
    Yay, I got it.

    Thank you!
     
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