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Introductory Vector Problem- Airplane emergency landing

  1. Jan 19, 2017 #1
    1. The problem statement, all variables and given/known data

    A plane leaves the airport in Galisteo and flies 180 km at 67.0 degrees east of north; then it changes direction to fly 255 km at 49.0 degrees south of east, after which it makes an immediate emergency landing in a pasture.

    When the airport sends out a rescue crew, how far should this crew fly to go directly to this plane


    3. The attempt at a solution
    I assume I need the angle as measured counterclockwise from the x-axis,
    90deg - 67deg =23deg

    Ax = 180cos23 =165.69 km
    Ay = 180sin23 =70.33 km

    Bx = 255cos49 =167.295 km
    By = -255sin49 = -192.45 (because I have my vector pointed in the negative y direction along the positive x-axis)

    Ax+Bx = 332.99 km
    Ay+By = -122.12 km

    R= sqrt [(332.99^2)+(122.12^2)]
    R= 310 km

    I'm not sure where I've gone wrong. I also used Ax=180sin67 and Ay=180cos67, which give the same numbers (because they're complimentary angles?). I didn't round my numbers when I calculated everything; I put in the exact numbers when I added and took the square root.
    Mastering Physics says I'm wrong, but doesn't provide a hint as to why.
     
  2. jcsd
  3. Jan 19, 2017 #2

    Orodruin

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    This number must be at least 333 ...
     
  4. Jan 19, 2017 #3
    Thank you! I rechecked my math, and somehow I must not have entered it into my calculator correctly. This may be a silly question, but how would I know it's supposed to be at least 333? Is it because 333 was the largest squared number and it was added to the other number?
     
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