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Vectors - Groundspeed question

  1. May 3, 2013 #1
    1. The problem statement, all variables and given/known data
    A plane is steering N45°E with an airspeed of 525 km/h. The wind is from N60°W at 98 km/h. Find the ground speed and the direction of the plane.


    2. Relevant equations
    Cosine law, sine law, vector addition/subtraction.


    3. The attempt at a solution
    I have attached an image of my diagram (sorry for the messiness!)

    I Solved for R using the cosine law:

    R^2 = 525^2 + 98^2 - 2(525)(98)cos60°
    R^2 = 285229 - 102900cos60
    R^2 = 233779
    R = 483.5 km/h

    As for the angle, theta, I also used the cosine law:

    98^2 = 525^2 + 483.5^2 = 2(525)(483.5)cosθ
    -499793.25 = -507675cosθ
    cosθ = 0.9844
    θ = 10.1°

    Is my method correct?

    Thank you in advance!
     

    Attached Files:

  2. jcsd
  3. May 3, 2013 #2

    SammyS

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    attachment.php?attachmentid=58446&d=1367618081.png

    The wind is from N60°W ...

    That means the angle you label as 60° should be 60° + 45°.
     
  4. May 3, 2013 #3
    Thank you for the reply.

    Okay, so using 60° + 45°, I calculated my new resultant vector as 558.4 km/hr.

    The only issue is when I solve for theta, cosθ = 1.138, which cannot be solved for.

    Is there something I'm missing?
     
  5. May 3, 2013 #4

    SammyS

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    What is the equation you have when solving for cos(θ)?
     
  6. May 3, 2013 #5
    Oh! I'm sorry, I just realized my mistake (substituted the wrong value) :)

    Anyways, my new θ turns out to be:

    98^2 = 525^2 + 558.4^2 - 2(525)(558.4)cosθ
    -577831.5 = -586320cosθ
    cosθ = 0.9855
    θ = 9.8°

    Hopefully my method is correct?

    Thank you for all your help, I really appreciate it :)
     
  7. May 3, 2013 #6

    SammyS

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    Yup. I got 9.759° .
     
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