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Vector addition word problem question

  1. Jan 22, 2013 #1
    I solved a word problem correctly...But I'm a little confused about what my answer means, I guess I'm having trouble understanding how they got the meaning of the numbers. My question is about step (b).

    Problem:
    The velocity of an aircraft is 200 km/hr due west. A northwest wind of 50 km/hr is blowing.
    a. What is the velocity of the aircraft relative to the ground?
    b. If the pilot's destination is due west, at what angle should he point his plane to get there?
    c. If his destination is 400 km due west, how long will it take him to get there?

    My correct answer:
    G = 200
    Gx = 200*cos(180°) = -200
    Gy = 200*sin(180°) = 0
    Gθ = 180°
    ---
    W = 50
    Wx = 50*cos(315°) = 35.36
    Wy = 50*sin(315°) = -35.36
    Wθ = 315°

    R = √[(-35.46)2+(-164.64)2] = 168.4
    Rx = -164.64
    Ry = -35.36
    Rθ = arctan[(-35.36)/(-164.64)] = 12.12°
    ...And I also solved step C, but it's not relevant.

    The book says the plane is moving 168.4 km/hr at 12.12° south of west...
    I don't understand how they knew what the 12.12° meant...How did they know it was south of west, rather than north-east, which is the direction your average 12.12° on the unit circle would face?
     
  2. jcsd
  3. Jan 22, 2013 #2

    mfb

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    It follows from your definition of the angle as angle between air and ground speed (=W).
     
  4. Jan 22, 2013 #3
    I don't understand... Normally when I do this exact series of equations, I get an angle that I can just use as a normal angle... What is the difference? How am I supposed to know it means "south of west"?
     
  5. Jan 22, 2013 #4

    mfb

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    Somewhere in the solution process, you introduce an angle θ. Check where this is done, and how the angle got defined. You have to stick to that definition in the interpretation of θ afterwards, of course.
     
  6. Jan 22, 2013 #5

    SteamKing

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    Both of the components in your final calculation of Rtheta are negative.
    This should tell you that the vector lies in the fourth quadrant, which is south of west.
     
  7. Jan 22, 2013 #6
    The angle made by 12.12° south of west is 192.12°, wouldn't that be in the 3rd quadrant?
     
  8. Jan 22, 2013 #7

    haruspex

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    I think SteamKing meant that since both components of Rtheta are negative the vector lies in the third quadrant
     
  9. Jan 22, 2013 #8
    So if xcomp = neg, ycomp = neg it's 3rd, xcomp = pos, ycomp = neg it's 4th etc?
     
  10. Jan 22, 2013 #9

    haruspex

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    Yes.
     
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