Mechanics question

  • Thread starter aks_sky
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  • #1
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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.
 
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Answers and Replies

  • #2
Danger
Gold Member
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Your question refers to what pilots call a 'wind vector triangle'. It belongs in the homework section, though. Welcome to PF.
 
  • #3
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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.
 
  • #4
55
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yup got it.

i got the answer

thanks a lot for your help.

thank u
 

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