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Mechanics question

  1. Apr 25, 2007 #1
    On another day the Cessna leaves Taupo airport when there is an Easterly wind (i.e a wind blowing from direction East) of speed 38 km/h. If the plane starts out directly above Taupo with an airspeed v=137 km/h, pointing in a direction θ = 26 ° East of North, how far is it from Taupo after 50 min of flying at constant altitude?

    i tried to do this problem but i cannot get a correct answer.. i tried all different ways but still i cant get it..
    can u help please.
     
    Last edited: Apr 25, 2007
  2. jcsd
  3. Apr 25, 2007 #2

    Danger

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    Gold Member

    Your question refers to what pilots call a 'wind vector triangle'. It belongs in the homework section, though. Welcome to PF.
     
  4. Apr 26, 2007 #3

    Mentz114

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    You need to do some geometry. Draw a set axes, and mark them N, S, E and W. Now draw a line ( starting at the origin) at 28deg E of N with length 137 ( any units you like).
    From the tip of this line draw another going E with length 38 ( same units).
    Now connect the tip of this line with the origin ( Taupo).
    The answer is the length of this line times 50 Km.
     
  5. Apr 27, 2007 #4
    yup got it.

    i got the answer

    thanks a lot for your help.

    thank u
     
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