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Homework Help: Vectors and direction/bearings.

  1. Sep 28, 2010 #1
    1. The problem statement, all variables and given/known data
    Jane walks 400m due South, then 250m in the direction S 85 degrees W, and finally 180m in the direction N 20 degrees E. Determine the magnitude and direction of her resulting displacement, relative to her starting point.


    2. Relevant equations
    None, I don't think.


    3. The attempt at a solution
    Well, I've gotten as far as trying to determine the magnitude of the second (250m) vector. My question is though, as she's heading south, does that mean that the vector is negative? Also, I'd also like to check with you very brainy, very lovely folks - the angle I've used for the
    ||AB||cos theta i etc is -95 degrees, as I've gone in the clockwise direction from the x-axis (90+90+85 = 265 degrees, then subtracting that from 360 degrees to give the negative 95 degrees).

    I was sick when we did vectors in class, so I apologise if I've made some very glaring mistakes, but any help would be greatly appreciated.
     
  2. jcsd
  3. Sep 28, 2010 #2

    HallsofIvy

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    Science Advisor

    There is no such thing as a "negative" (or "positive") vector. The indvidual components may be positive or negative. Strictly speaking whether they are positive or negative depends upon how YOU want to set up YOUR coordinate system but I suspect you are using the "standard"- north is the positive y-direction, east is the positive x-direction.

    Also, I don't think you are "trying to determine the magnitude of the second (250m) vector." You are given that- it is 250. I think you are trying to determine the components.

    In the standard coordinate system, going "400 m due south" would be <0, -400> or -400j. If you draw a line "south 85 degrees west" and then draw vertical and horizontal (north-south and east-west) lines you have a right triangle with angle 85 degrees (If you vertical line goes south from the first point and west east from the last point). Since she is going south and west, both components will be negative. The components will be <-250 sin(85), -250 cos(85)> or -250 sin(85)i- 250 cos(85)j. If your horizontal and vertical lines are west from the first point and north from the last, then the angle will 90- 85= 5 degrees but you also will reverse "opposite" and "near" sides. Since sin(85)= cos(5), that's the same thing.

    If you are measuring angles counter-clockwise from the positive x-axis (east), which is the standard, then your angle will be 90+ 90+ 5= 185 degrees. And, in that standard, the x component is always cosine and y sine. Here, that would be <250 cos(185), 250 sin(185)>. Since cos(180+ 5)= -cos(5) and sin(180+ 5)= -sin(5) that's exactly what I had before.

    Measuring "in the clockwise direction from the (positive) x-axis" would be 90+ 85, not 90+ 90+ 85. The first 90 takes you from east to south and then you have the "85 degrees west"
     
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