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Two dimensional kinematics

  1. Oct 9, 2009 #1
    1. The problem statement, all variables and given/known data
    a flying saucer manueverign with a constant acceleration is observed with the positions and velocities shown below. what is the saucer's acceleration?

    2. Relevant equations
    a = [tex]\Delta[/tex]v / t

    3. The attempt at a solution
    so a time is not given.. is there another way to find acceleration?
    ** actually if someone knows a site that gives a clear and concise tutorial on two-dimensional kinematics that would be great :). I know there is one attached to this site.. but i do better with actual examples and more visual.
     

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  3. Oct 9, 2009 #2

    tiny-tim

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    Hi indietro! :smile:

    (I can't see the picture yet, but …)

    The standard trick for finding dv/dt from v and s without involving t is to use the chain rule …

    dv/dt = dv/dx dx/dt = v dv/dx :wink:
     
  4. Oct 9, 2009 #3
    oo ok for the picture it is a x-y graph that shows a point at (0.0) with a vertical vector (200[tex]\hat{j}[/tex] m/s) and a second point at (2000, 1000) with a south-east vector (200[tex]\hat{i}[/tex] - 100[tex]\hat{j}[/tex] m/s)

    for the chain rule: is the v the final velocity?
     
  5. Oct 9, 2009 #4

    tiny-tim

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    v = dx/dt
     
  6. Oct 9, 2009 #5
    so i have a question: for an x-y graph showing the trajectory, what does [tex]\vec{r}[/tex] (starts at origin and goes to a point on the trajectory) tell me? the velocity at that point? or does it only tell me the direction of velocity at that point?
     
  7. Oct 9, 2009 #6
    sorry but im really confused as how to relate an xy-graph, a vx graph, a vy graph and acceleration. Like what does each tell me, how can i find position after a certain time, how can i find acceleration?
     
  8. Oct 10, 2009 #7

    tiny-tim

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    Hi indietro! :smile:

    (just got up :zzz: …)
    r only tells you the position, and the direction of the tangent of the trajectory tells you the direction of the velocity.
    Why are you using a graph? Does the question tell you to?

    If it doesn't, then forget graphs, and just use equations. :smile:
     
  9. Oct 10, 2009 #8
    yes the question gives me all the information in the form of an vx graph and vy graph ...:(
     
  10. Oct 10, 2009 #9

    tiny-tim

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    hmm … I still can't see your picture.

    I'll have to wait until I can see it.

    (I've reported it, so hopefully it'll come up soon :redface:)
     
  11. Oct 10, 2009 #10

    tiny-tim

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    oooh, I see it now! :tongue2:​

    ok, that isn't a graph (a graph would be a continuous curve) …

    it's just a diagram, defining the two velocities in a picture instead of in words.

    So you don't have to use a "graph method" …

    just write the velocities as 200j and 200i - 100j, and carry on from there. :smile:
     
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