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Airplane problem. 2d3d motion

  1. Oct 14, 2008 #1
    Hello everyone,
    as my name suggests I love physics. Currently i am trying to get into physics again and find myself stuck with this problem.
    I would really appreciate if you gave me advice on just how to approach this problem, coz I really want to solve it myself.

    Problem
    You are flying in a light airplane spotting traffic for a radio station. Your flight carries you due east above a highway. Landmarks below tell you that your speed is 57.0 m/s relative to the ground and your air speed indicator also reads 57.0 m/s. However, the nose of your airplane is pointed somewhat south of east and the station's weather person tells you that the wind is blowing with speed 20.0 m/s.

    In which direction is the wind blowing? Express your answer as an angle measured east of north.

    End.

    Please guys I just need a starting point. Could it be that I could add or substract the velocities vectorially and then use the cosine rule in order to finally get the degrees?
    I believe that since the nose is pointing south of eaast there must be a 45 degree angle.
    Am I right?

    wait. or could it be that i have to approach this problem from the frame of reference of the earth then add the vel. vectors and obtain the degrees? since there is a steady wind?
     
    Last edited: Oct 15, 2008
  2. jcsd
  3. Oct 15, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi iluvphys! Welcome to PF! :smile:

    Hint: be systematic … in physics, everything is equations, so don't try to do things by "feel".

    In this case, give everything names, so that you can put them into equations.

    Use vAW for the velocity of the airplane relative to the wind, vAG for the velocity of the airplane relative to the ground, and vWG for the velocity of the wind relative to the ground.

    Then write a formula for vAW vAG and vWG, and then draw it as a vector triangle (with arrows!), and solve. :smile:
     
  4. Oct 15, 2008 #3
    Hi, thanks for your help.
    Well since Vag equals Vaw I think I can treat them the same.
    I have drawn a vector triangle as you suggested and have come to believe that the resultant Velocity = Vag+Vwg. =>> Vag+Vwg = 57 + 20 = 77.
    There is also an angle 45° since the nose is pointing south east and therefore I am not sure whether I can just add these vectors like that. I mean is vector addition really that simple?
    Or do I have to add them using pythagoras: sqrt(57^2+20^2)?
    Its just a little too confusing and I though this example would help me get into physics.
     
  5. Oct 15, 2008 #4

    tiny-tim

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    Hi iluvphys! :smile:
    Nooo … it is pointing "somewhat south of east" … that is only telling you that it is pointing to the right rather than the left.
    I don't understand this …

    can you please describe the vector triangle you have drawn? :smile:
     
  6. Oct 15, 2008 #5
    Hi,
    I am sorry if i am bothering you.

    I really assumed that the angle was 45 but now that you mention it you are right.
    I am kind of lost. So my triangle doesnt make any sense :(
    Thats why I started adding the velocities but that cant be the case.

    Basically all I know is that the velocity wrt earth/air of the plane is 49 m/s. And the velocity wrt earth of the wind is 21 m/s.
    I dont know how I can derive an equation when I dont know the angle or how the velocity vectors correspond to each other.
    So how can I derive equations which tell me where the wind comes from and what angle it is?
    I am really confused. Please help.
     
  7. Oct 15, 2008 #6

    tiny-tim

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    vector triangle

    Hi iluvphys! :smile:

    The triangle has two sides of 57 (one is due east), and one side of 20.

    So the angle of the 20 side is … ? :smile:
     
  8. Oct 15, 2008 #7
    Hello sir,
    I used the cosine rule and got 20.21°

    I assumed that you said one side is due east means that it looks like the vector picture i attached.
    Is my answer to your question acceptable?
    Thanks again for your kind help-
     

    Attached Files:

    Last edited: Oct 15, 2008
  9. Oct 15, 2008 #8

    tiny-tim

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    Don't call me sir!!

    Yes, it's 20.21º …

    though you could have got it quicker by splitting the isoceles triangle in two, to get a right-angled triangle with sides 10 and 57. :wink:

    And can you now answer which direction the wind is blowing? :smile:
    hmm … there's one wrong with the picture …

    you haven't put the letters A G and W on …

    if you don't do that, you can't be sure that the arrows are the right way round. :wink:
     
  10. Oct 16, 2008 #9
    Hi tiny-tim,
    i have spent quite some time on this problem and thanks to your help got most of it.
    So is the answer to which direction the wind is blowing not 20,21?
     
  11. Oct 16, 2008 #10

    tiny-tim

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    No, 20.21º is the angle between the plane's actual speed and its speed relative to the wind.

    This is why it is so important to label the triangle!
     
  12. Oct 16, 2008 #11
    can i use the cosine again and get the angle?
    Or how do I know what the angle is? Do i have to use cosine again?

    Then I would get 31.54?
     
    Last edited: Oct 16, 2008
  13. Oct 16, 2008 #12
    I have been trying for days now and I cant seem to have gotten the grip of it yet.

    Could you please show me how you got your answer? I really tried figuring it out but i am almost hopeless. I have 20.2 but what do i do next?
     
    Last edited: Oct 16, 2008
  14. Oct 16, 2008 #13
    Could the answer be 180-31.54= 148?
     
    Last edited: Oct 16, 2008
  15. Oct 16, 2008 #14

    tiny-tim

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    How did you get 31.54º? :confused:

    You have a triangle AWG … which angle were you trying to measure?
     
  16. Oct 16, 2008 #15
    I added 20.2 and tan-1 (20/57)!

    Really could you please tell me how you did it and got your answer.
    Because I have lost hope and i really need a light bulb over my head please.
    Thank you really for your help.
     
  17. Oct 16, 2008 #16

    tiny-tim

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    iluvphys, you don't seem to be trying. :frown:

    You have a triangle AWG … which side represents the speed of the wind?
     
  18. Oct 16, 2008 #17
    Finally....
    The answer i was lookin for is 20.21/2.

    Thank you very much for your patience. It was really absurd because i was doing all the relativity questions with ease and then there comes a problem like this... :)

    Thank you very much again tiny-tim!
     
    Last edited: Oct 16, 2008
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