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Kinematics 4 -- Calculate the velocity from the position versus time graph

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  1. Jul 11, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-7-11_21-51-59.png

    2. Relevant equations
    vav = total distance traveled/ total time taken

    3. The attempt at a solution
    I have a problem in part c .
    How am I supposed to solve it?
    Is there any standard way to solve it?
     

    Attached Files:

  2. jcsd
  3. Jul 11, 2017 #2

    TSny

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    Take an arbitrary time to. How can you graphically determine the average velocity for the time interval between t = 0 and t = to? [Edit: Can you interpret the average velocity as the slope of some line?]

    It might help to first take a specific example. Suppose you take t0 = 10 s. How would you get the average velocity for the time interval t = 0 to t = 10 s? Does this average velocity equal the instantaneous velocity at t = 10 s?
     
    Last edited: Jul 11, 2017
  4. Jul 11, 2017 #3

    cnh1995

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    Irodov?

    One standard way could be using numerical methods. Newton's forward interpolation or Lagrange's interpolation will work.
    You can find the displacement-time relationship for three intervals: i)from t=0 to t=10s ii)from t=10 to t=14.5s iii) from t=14.5 to t=20s.
    Once you have the s-t relationship, you can find t0.
     
  5. Jul 11, 2017 #4
    As TSny indicates, you can pretty much determine the answer for part c by eye, just by looking at the graph.
     
  6. Jul 12, 2017 #5
    Thanks.
    I got it.
    I never interpreted average velocity as the slope of some line. Thanks for this insight.
    So, what I have to find out is the time moment at which the line whose slope defines the instantaneous velocity passes through the origin. This time moment turns out to be 16 s.
    Yes

    Thanks to all for replying.
     
  7. Jul 12, 2017 #6

    TSny

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    Yes. Good work.
     
  8. Jul 12, 2017 #7
    Thanks, TSny
     
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