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Vectors - Find resultant displacement

  1. Feb 6, 2008 #1
    1. The problem statement, all variables and given/known data

    While exploring a cave, a spelunker starts at the entrance and moves the following distances. She goes 75 m north, 250 m east, 135 m at an angle 30° north of east, and 125 m south. Find the resultant displacement from the cave entrance.
    Magnitude


    2. Relevant equations

    I do not quite understand vectors all that much. I am new to this and I do not know where to begin with that problem.

    3. The attempt at a solution

    I've tried drawing a picture but it doesn't seem right so I'm stuck there.
     
  2. jcsd
  3. Feb 6, 2008 #2

    Kurdt

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    Do you know about vector components and vector addition?
     
  4. Feb 6, 2008 #3

    HallsofIvy

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    The only "hard" one is that "135 m at an angle 30° north of east". If you draw a picture of that, you should see a right triangle with hypotenuse of length 135 and angle 30° The "opposite side" (North) is 135 sin(30)= 135/2= 67.5 m. The "near side" (East) is 135 cos(30)= 135 sqrt(3)/2= 116.9 m

    Now make a list of East-West and North-South distances treating "East" and "North" as positive, "West" and "South" as negative:
    East-West North-South
    0 75
    250 0
    67.5 117
    0 -125
    Finally add up each of those. If you like, once you have a total for "East-West" and "North-South" you can use the Pythagorean theorem to get the straight line distance and the arctan to get the angle so you can write the answer in the same form as the vectors given in the problem.
     
  5. Feb 6, 2008 #4
    Thank you
     
    Last edited: Feb 6, 2008
  6. Feb 6, 2008 #5
    I did what you said. I got 324.492 for the straight line distance and 78.08 degrees for the angle but when i input those values it said I was wrong. Any help on what I'm doing wrong?
     
  7. Feb 7, 2008 #6

    Kurdt

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    How are you working out the angle? The resultant displacement can also be treated as a right angled triangle with the east-west and north-south components being the two non-hypotenuse sides. how would you then work the angle out?
     
  8. Feb 7, 2008 #7
    I did it wrong but I ended up getting the displacement. But when I try to use trig to get the angle it comes out wrong.
     
  9. Feb 7, 2008 #8

    Kurdt

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    If you show us what you're doing to get the angle then we can help.
     
  10. Feb 7, 2008 #9
    Okay for I got 361.6275447 for the displacement. I know that this is a right triangle so I use arctan (y/x) y being 360.585526 and x being 27.43432879. Which I calculated when adding all the x-components and y-components. I did that but the angle did not work when I input it on the site where I do my work. The displacement however worked. The angle is supposed to be North of East which I don't understand.
     
  11. Feb 7, 2008 #10

    Kurdt

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    Your numbers seem slightly out for the components of the vectors. Your method for obtaining the angle is correct.
     
  12. Feb 7, 2008 #11
    I don't know what to do. I've tried doing it over and over again but it comes out incorrect.
     
  13. Feb 7, 2008 #12

    Kurdt

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    Ok so you added the x (west-east) components: 250 + 135*cos(30)

    and the y (south-north) components: 75 + 135*sin(30) - 125

    So what do you get for each?
     
  14. Feb 7, 2008 #13
    I figured it out. For the angle we need North of East and what I was getting was South of East so what I did is subtract 90, since it's a 90 degree triangle, from the angle I was getting and I got the angle for North of East.
     
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