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Distance From a Velocity-Time Graph

  1. Sep 28, 2011 #1
    I need to know how to determine distance from a velocity time graph. The graph lines are all linear with changes in direction and slope, but no curves.

    I've tried finding the area under the lines but that doesn't seem to be it...
     
  2. jcsd
  3. Sep 28, 2011 #2
    Velocity=distance/ time. multiply by the time stamp at whatever point you are on on the graph.
     
  4. Sep 28, 2011 #3
    You know that Velocity is a change in position over change in time, correct?

    Using that, you should be able to figure out the position. (At least, that's how I do it.)
     
  5. Sep 28, 2011 #4
    What did you do when you tried finding the area under the graph?
     
  6. Sep 28, 2011 #5
    completing each time increment as a triangle and using 1/2b*h.
     
  7. Sep 28, 2011 #6
    Could you post a picture of the graph?
     
  8. Sep 28, 2011 #7
    maybe, lemme try
     
  9. Sep 28, 2011 #8
    well uh no actually. idk how D: could i get an e-mail? i think i could e-mail it
     
  10. Sep 28, 2011 #9
  11. Sep 28, 2011 #10
    it's #7 i need help with!
     
  12. Sep 28, 2011 #11
    The blue graph right?

    What ares did you use 1/2b * h on (what time)?
     
    Last edited: Sep 28, 2011
  13. Sep 28, 2011 #12
    i posted a pdf of the question sheet. you can see the graph on the second page. i'm fairly sure i found the area correctly, but wouldn't that give me the displacement? and i need distance... look at number 7 on the pdf. plz :/
     
  14. Sep 28, 2011 #13
    What did you find for displacement?
     
  15. Sep 28, 2011 #14
    i got -54
     
  16. Sep 28, 2011 #15
    Okay, and for what parts did you use 1/2b*h?
     
  17. Sep 28, 2011 #16
    because from times 1-3 seconds the line goes from 0 to -18, i did 1/2 the base (y axis time) plus the height of -18. this got -36.
    time interval 3-4 seconds have a constant velocity of -18 m/s so i added -18 to the first -18 getting -36. then i found the are from 4-5 by multiplying 1/2 the base of 1 second, by the height of -9, getting an additonal -4.5. Sum of all these? -54!
     
  18. Sep 28, 2011 #17
    i meant times the height! sorry i'm a confusing typer !
     
  19. Sep 28, 2011 #18
    please ignore the first time i said "this got -36" its a typo...
     
  20. Sep 28, 2011 #19
    You found the area for time 4-5s wrong.

    SJXDh.png

    You found the red part, but not the blue part. The blue part is a rectangle.

    Also, I think for this problem you can assume displacement and distance are meant to be the same thing.
     
    Last edited: Sep 28, 2011
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