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Velocity-Time graph issue

  1. Dec 18, 2006 #1
    1. The problem statement, all variables and given/known data

    1. Plot the following motion of an object on a d-t graph.
    Position (m north of the CN tower) Time (s)
    0 0
    50 5
    150 10
    300 15
    500 20
    750 25
    1050 30

    1. Describe the motion of the object.
    2. Is the object accelerating at a uniform rate? To find out, add a third column to the chart above and calculate the average velocity during each interval.
    3. Plot velocity vs time now on another graph. What do you notice? Does the object exhibit uniform acceleration? Explain how you know.
    4. Calculate the acceleration of the motion.
    5. Calculate the area under the velocity-time graph. Compare your answer to the value you expected to obtain. Explain your comparison.


    2. Relevant equations



    3. The attempt at a solution

    Since there's charts and graphs involved I have my answer in html format here.

    http://www.freewebs.com/dazed42/physics.htm

    The problem is, the area under the velocity-time graph is supposed to give the displacement. I know the displacement is 1050m, but the area of my velocity-time graph yields only 900. What gives?
     
  2. jcsd
  3. Dec 18, 2006 #2
    why are you solving for the area under the graph? explain.
     
  4. Dec 18, 2006 #3
    "The displacement of the object can also be determined using the velocity-time graph. The area between the line on the graph and the time-axis is representative of the displacement"

    http://www.glenbrook.k12.il.us/gbssci/phys/CLass/1DKin/U1L6e.html
     
  5. Dec 18, 2006 #4
    I think I may have found the problem. The /average/ velocity for seconds 25-30 is 60 m/s.

    The problem makes sense if the final velocity is 70 m/s (at 30s). This makes a triangle with base 30 and height 70. The area of which is .5 x 30 x 70 = 1050m. But I don't know how to arrive at that final velocity of 70 m/s via the formulas.
     
  6. Dec 18, 2006 #5
    well, your problem was that you had a position vs. time graph, not a velocity vs. time graph.your position vs. time graph was quadratic; therefor, your velocity vs. time graph will be linear IE the triangle.
     
  7. Dec 18, 2006 #6
    well, the fucntion for position is x(t)=1/2at^2 + v0t + x0
    so, that measn that the time differential for the position fucntion is the velocity function. v(t)=at+v0
     
  8. Dec 18, 2006 #7
  9. Dec 18, 2006 #8

    cristo

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  10. Dec 18, 2006 #9
    Here's the Velocity-Time graph with the area under the curve measured.

    [​IMG]

    And here's the Distance-Time graph from Question 1
     

    Attached Files:

  11. Dec 18, 2006 #10
  12. Dec 18, 2006 #11

    cristo

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    Don't worry, you've not done anything wrong. The fact that the question says
    hints at the fact that you will get an answer different to what is expected.

    Since you are only given the information about the displacement and time from which to calculate the average velocity, you had to make one assumption about the intial velocity when the object was at position x=0, namely that the velocity at t=0 is zero. However, suppose the intial velocity was not zero. How would this affect the question?

    Can you find the initial velocity such that the distance is what you expected? [hint: s=ut+(at^2)/2]

    edit: sorry, the hint was meant for the second bit-corrected
     
    Last edited: Dec 18, 2006
  13. Dec 18, 2006 #12
    Ohhh, that would make sense Cristo. Unfortunately I don't know how to use that equation. I know what "a" and "t" are, but what are "s" and "u"?

    Also, did I calculate the correct acceleration? 2 m/s^2?

    Thanks a lot man

    Edited to add: If the initial velocity is 10m/s then the rest of the pieces fall into place. I only say that because I /should/ be getting 70, but I /am/ getting 60, therefore an extra initial velocity of 10 fixes things up. But you see I arrived at that logically, not mathematically.
     
    Last edited: Dec 18, 2006
  14. Dec 18, 2006 #13

    cristo

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    Sorry, should have defined my notation! s is displacement, u is initial velocity.

    Yes


    Not quite: Using the equation: s=ut+(at^2)/2. We require s=1050, t=30, a=2 and we want to find u.

    Putting the numbers in, we obtain, u=5m/s

    If you plot this graph, you will obtain a velocity of 65m/s at t=30. This is due to the fact that the graph has an intercept, so the area under is not a triangle (it's a triangle on top of a rectangle). Your logical argument misses this point, as on calculating the area, we would then have the 900m from the triangle (the same as in your graph, but shifted up) and the resulting 150m will come from a rectangle of height 5m/s and length 30s. (Try it for yourself: plot a v-t graph like before, but increase each value of v by 5m/s. Then calculate the area under the graph).
     
    Last edited: Dec 18, 2006
  15. Dec 19, 2006 #14
    Sweet. I owe you one, thanks.
    That was a pretty tricky question, considering the only formula we've been given so far is v = d/t in this unit.

    :\
     
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