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Particle Position from a V(m/s) vs T(s) Graph

  1. Jan 12, 2008 #1
    [SOLVED] Particle Position from a V(m/s) vs T(s) Graph

    A particle starts from Xo=10 m and To=0 and moves with the velocity graph shown in Figure Ex2.11.

    a. What is the object's position at T=2s, 3s, and 4s?
    b. Does this particle have a turning point? If so, at what time?

    I've gotten the answer to b. part. The particle turns at T=3s when the velocity is =0.
    A. part is giving me problems. I've got the answer to t=2s by figuring the problem like this, the particle starts at 10m when Xo=0 and To=0. So, 10m + 1(8)+2(4)=26m.
    But, when T=3s and when T=4s, my answers are wrong.

    The answers in the back of the book give for a. part: When T=3s is 28m, and when
    T=4s is 26 m.

    I've attached the questions and the graph, where am I going wrong? Thanks for any help received!
     

    Attached Files:

  2. jcsd
  3. Jan 12, 2008 #2

    Doc Al

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    Staff: Mentor

    Note that the velocity is continually changing. In the interval from T = 0 to some other time, what's the average velocity? Use that to find the displacement.
     
  4. Jan 12, 2008 #3

    kdv

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    You got the correct answer for 2 seconds by pure luck. The coorect approach is completely different from what you did. To find the position at any time, you must add the position at 0 second to the displacement between 0 and the time where you want to find the position of the object.

    For example,
    [tex] x(2 s) = x(0) + \Delta(x) [/tex] between 0 and 2 seconds

    The key point is that the displacement in a velocity graph is given by the area under th ecurve . Find the area under the curve between 0 and 2 seconds and you will find that it is 16 meters. Added to x(0) = 10 m gives 26 meters for the position at 2 seconds.

    Now find the area under the curve between 0 and 3 seconds and add it to 10 meters and you will find the position at 3 seconds.

    And so on. The only thing you must be careful about is that once the graph goes below the horizontal axis, you must count the are as being negative (after 3 seconds).
     
  5. Jan 12, 2008 #4
    Is this where I would use displacement=1/2 X b X h to get the area? then add this back to 10m? I would have no problem getting the area under the curve if I had the equation I was looking at and not a diagram (just use integration), but in looking at the graph it is intimidating and confusing. If I take 1/2 X b X h at the t=2s, I get 1/2 X 2 X 8, which is 8, I'm not getting the 16. If for the area at t=3s, I get 1/2 X 3 X 12, which I do get 18. So, I'm still missing something, rather misunderstanding what I should do. I do appreciate all of your help though in trying to get me to understand the concept.
     
  6. Jan 12, 2008 #5

    kdv

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    Consider from 0 to 2 seconds. The formula 1/2 times b times h would work if you had a triangle but you don't. It's hard to explain with words but break the area under the curve into two pieces: you have a triangle on top of a rectangle.


    The rectangle is 2 s wide and 4m/s high so it has an area of 8 meters.

    The triangle on top of the rectangle has a width of 2 seconds (the base) and a height of 12-4 = 8 m/s. SO it has an area of 1/2 * 2 * 8 = 8 m also (by pure luck it has the same area as the rectangle.

    So a total area of 16 meters between 0 and 2 seconds. So position at 2 seconds = 16 + 10 = 26 meters

    Between 0 and 3 seconds you can use 1/2*b*h because you have a rectangle.
     
  7. Jan 12, 2008 #6
    Okay, that is what I was missing--never took geometry, just a small amount of trig. I basically began college in algebra, then pre-calc classes, Cal 1, and now cal 2 and PHY 213.

    I've got the idea now. It seems as though all of Physics problems are word problems, some dealing with graphs, and so every problem seems unique. I've got several problems to get completed over the weekend, so I'll be utilizing this forum quite often, especially since Physics is so new to me.

    I'll use the triangle equation and the trapezoid equation I think for t=4s? Looks like a trapezoid where the velocity makes it's turn at t=3s.

    Thanks again!
     
  8. Jan 12, 2008 #7

    kdv

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    You are welcome. The fundamental point here is that the displacement s given by the area under the curve in a velocity graph. And you must decompose any area into rectangles and triangles.

    For 0 to 4 seconds, what you must do is to consider separately the part of the are which is above the horizontal axis and the part below. The part above is from 0 to 3 seconds so the area of that part is simply what you got from 0 to 3 seconds. From 3 to 4 seconds, you again have a triangle so you use 1/2*b*h but this time you assign a negative value to your answer (so it gives -2 meters for the area) which you must add up to the area under the curve between 0 and 3 seconds for the total displacement.


    Best luck
     
  9. Jan 12, 2008 #8

    Doc Al

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    Staff: Mentor

    While kdv gave an excellent explanation, which you would do well to study carefully, you can also make use of [itex]\Delta x = v_{ave}\Delta t[/itex]. For example, from T = 0 to T = 2s the velocity goes from 12 m/s to 4 m/s. Thus the average velocity in that interval is right in the middle (since the plot of v is a straight line) at 8 m/s. So the change in x is (8 m/s)*(2 s) = 16 m. Thus the position at T = 2s is the starting position ([itex]x_0 = 10[/itex] m) plus 16 m, or x = 26 m.

    This is, of course, completely equivalent to what kdv has explained. But sometimes looking at things from a slightly different angle helps understanding sink in deeper.

    [Edit: Thanks to kdv for pointing out my initial error.]
     
    Last edited: Jan 12, 2008
  10. Jan 12, 2008 #9

    kdv

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    Yes, this is a very good way to do it too! (just a small correction: the velocity from 0 to 2 seconds is the middle value between 12 and 4 m/s so it's actually 8 m/s therefore the area under the curve is + 16 meters as obtained previously).

    I agree that this is actually faster than the method I explained. As long as the OP is aware that this works only for straight line graphs (and that the area under the curve approach is more general) I agree that Doc Al's method is actually faster in this example.


    Regards
     
  11. Jan 12, 2008 #10

    Doc Al

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    Staff: Mentor

    Oops! :redface: Thanks. (I'll correct my error in the other post.)
     
  12. Jan 12, 2008 #11
    Thanks for all of the great help! I'm working on my other Physics homework now. I got the answer for all three now thanks to you guys. I'm definitely keeping this information in my notes to look back on when another problem such as this comes along.

    Thanks again for all of your help, Matt.
     
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