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Linear Algebra - Vectors

  1. Oct 30, 2009 #1
    1. The problem statement, all variables and given/known data

    A plane flies at 400 km/h [south 30 degrees east]. The wind is blowing from the east at 100 km/h. Find the resulting speed of the plane.

    2. Relevant equations

    Not sure

    3. The attempt at a solution

    I've drawn a diagram similar to this:

    |\
    | \
    | \
    |30\
    | \
    | \ 400 km/h
    | \
    |_____\
    100km/h

    I thought this was just a simple question of finding the left side of the triangle, but I don't seem to get the answer. Any hints is greatly appreciated.

    Thank you in advance.
     
    Last edited: Oct 30, 2009
  2. jcsd
  3. Oct 30, 2009 #2

    Mark44

    Staff: Mentor

    You have drawn a 30-60-90 right triangle. If the hypotenuse is 400 km/hr, the base would be 200 km/hr, not 100 km/hr.

    Decompose the plane's velocity into a south component and and east component.
     
  4. Oct 30, 2009 #3
    Hmm so would this work?

    use 30 degrees and 400 km/h to find the speed going straight south

    cos (30) * (south_speed/400)
    south_speed = cos(30) * 400 = 346

    Then use speed going straight south and incorporate the 100km/h wind blowing west, use Pythagorean theorem and find the resultant speed(hypotenuse)?

    346^2 + 100^2 = 130000^2
    = 360.5

    Does that seem right?
     
  5. Oct 30, 2009 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Set up a vector for the plane velocity: P = < ?, ?>
    Write a vector for the wind velocity: W = <?, ?>

    The resultant velocity is V = P + W and its length is the groundspeed.
     
  6. Oct 30, 2009 #5
    Can you please show me how? I was only only taught how to do it when the points were given.
     
  7. Oct 30, 2009 #6

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    Think of the plane as at the origin with the x direction East and y direction North. The direction described as "South 30 degrees East" is the polar coordinate angle -60 degrees. So the plane velocity is given to you in polar coordinates, r = 400 and [itex]\theta = -\pi/3[/itex]. Use the polar coordinate equations to get P = <x, y> for the airplane. And you should easily be able to write the wind vector since you are given its length 100 and direction "from the East".

    By the way, I didn't say your answer was incorrect. I'm just trying to show you how you should set up such problems.
     
    Last edited: Oct 30, 2009
  8. Oct 31, 2009 #7

    Mark44

    Staff: Mentor

    Yes, this is pretty close. I get a slightly larger figure since I didn't round off what you're calling south_speed, which I get as 346.41 km/hr (approx.).
     
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