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Explain to me Step Functions for VT graphs

  1. Mar 2, 2013 #1
    I am sorta confuse about this step functions. In the picture below, the top is the distance time graph and the bottom is velocity time graph. The arrow pointing is what is confusing me. Its not suppose to be a step up (answers in my book) but i calculated the slopes

    First curve line = 30
    Second straight diagonal line = 60
    Second curve line = 35

    So I put it in my velocity time graph but the part where the arrow is pointing, isn't that correct though?

    But the correct answers show that the first line on the velocity time graph is connected to the horizontal second line.

    I am so confuse...

    Both graphs have same intervals.

    http://imgur.com/52DSmKv

    http://imgur.com/52DSmKv
     
  2. jcsd
  3. Mar 2, 2013 #2

    Delphi51

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    No graphs are visible!
     
  4. Mar 2, 2013 #3
    i posted the link

    http://imgur.com/52DSmKv
     
  5. Mar 2, 2013 #4
    The problem is with the first curved line. The average velocity from 0 to 6 is 30, but the velocity is varying from zero at t = 0 to a value higher than 30 at t = 6, in order for the average to be 30. If the velocity is varying linearly with time in this region, then the velocity at t = 6 has to be 60 in order to the average to be 30. Then there won't be any step.
     
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