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Two Dimensions

  1. Jun 4, 2007 #1
    A bird watcher meanders through the woods, walking 0.45 km due east, 0.70 km due south, and 2.75 km in a direction 43.0° north of west. The time required for this trip is 2.50 h.
    (a) Determine the magnitude and direction (relative to due west) of the bird watcher's displacement. Use kilometers and hours for distance and time, respectively.
    (b) Determine the magnitude and direction (relative to due west) of the bird watcher's average velocity.

    Heres what I have done so far:

    X component=.45+2.7cos43=-2.425
    Y component=.70+2.7sin43=-2.541

    2.7/2.5=1.08
    square root of (-2.425^2)+(-2.541)^2=3.512<<<<... answer is wrong

    I also tried using trig tan-1(2.425/2.541)


    OR WOULD it be 2.7cos43-.45 & 2.7sin43-.70
     
  2. jcsd
  3. Jun 4, 2007 #2

    Doc Al

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    Get your signs straight. Pick +X to be East and +Y to be North. (When you're done you can reword the answer with respect to due West, like they want it.) Redo this.

    Did you draw out the path taken? That will help keep your signs in line.
     
  4. Jun 4, 2007 #3
    yes it would be a right triangle, I think. So I would have .45*sin43<x component and -.70*cos43<y component and I have 2.75 for the hypoteneuse. Also I know the total time was 2.5 hrs. so I'm guessing I need to use a kinematics equation to solve for x?
     
  5. Jun 4, 2007 #4

    Doc Al

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    The path is not a right triangle. To find the x-component of the displacement, just add the x-component for each of the three "legs" of his path. Do the same for the y-component.

    Show the components you get for each leg including the correct sign.
     
  6. Jun 4, 2007 #5
    This is where I struggle is the component method: would it be .45sin43+0=.307 & -.70cos43+0=-.512 Then would I take the square root of (.307^2)+(-.512^2)
    or do I do it like this: .45sin43+-.70cos43+2.75cos43=1.806
     
  7. Jun 4, 2007 #6

    Doc Al

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    Let's start over. Do this: Find the x & y components of the displacement for each of the three "legs" of the path.
    Do it systematically, step by step.
     
  8. Jun 4, 2007 #7
    Thats what I've been trying to do, my teacher hasn't spent very much time on these problems and i barely understand the component method
     
  9. Jun 4, 2007 #8

    Doc Al

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    You'll understand it better once you try to solve the problem. Do the first step right now: 0.45 km due east

    What's the x-component of the displacement?
    What's the y-component of the displacement?

    (Draw the step on a diagram!)
     
  10. Jun 4, 2007 #9

    Doc Al

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  11. Jun 4, 2007 #10
    x-comp: .307
    y-comp:-.512
     
  12. Jun 4, 2007 #11

    Doc Al

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    No. How did you get those numbers? The motion is due east, which means completely in the +X direction.

    Did you draw a diagram?

    The only time angles are involved (for calculating components) is when the vector makes some angle with the x-axis.
     
  13. Jun 4, 2007 #12
    yes I drew a diagram, I just thought that if you had an angle i would have to find the x and y components using the component method....I got the .307 and /.512 using the vector length X cos or sin of the angle measure.
     
  14. Jun 4, 2007 #13

    Doc Al

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    Draw a diagram just for the first step. 0.45 km due east

    The diagram should just be an arrow pointing to the right (+x direction). No angles needed.
     
  15. Jun 4, 2007 #14
    okay ive got that...then I should have an arrow pointing downward due south from the end of the previous vector? then i drew an arrow from the end of that vector 43 degrees north of west
     
  16. Jun 4, 2007 #15

    Doc Al

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    Yes, but draw each step separately so you can find the components. Then just add the x-components together to find the x-component of the total displacement. Same thing for the y-components.
     
  17. Jun 4, 2007 #16
    0.45 km + 0.70 km + and 2.75 km=3.9 km
     
  18. Jun 4, 2007 #17

    Doc Al

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    If you're adding x-components, only the first term is correct.

    If something goes due south, what's its x-component?
     
  19. Jun 4, 2007 #18
    .45<x-component so y=0
    if something goes due south its x-component=0 so it would be just -.70
     
  20. Jun 4, 2007 #19

    Doc Al

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    OK, but be careful not to add x- & y-components together: keep them separate. Add up all the x-components to get the x-component of the total; then do a similar, but separate, calculation for the y-component.
     
  21. Jun 4, 2007 #20
    isn't there only one x-component so it would =.45
    and one y component=.70<<these answers seem too obvious am I missing something?
     
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