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Homework Help: Ground speed of a plane - vectors

  1. Feb 17, 2009 #1
    1. The problem statement, all variables and given/known data

    A plane is heading due east and climbing at he rate of 80kph. if its speed is 480kph and there's a wind blowing 100kph to the northeast, wha is the groundspeed of the plane

    2. Relevant equations

    3. The attempt at a solution
    w=100cos45+100sin45= 70i+70j
    480= vxi+ 70j+80k
    vxi= 468
    Groundspeed= 468i+70i+70j=538i + 70j = 543kph


    480= vxi+70j
    vxi= 475
    Groundspeed = 475i +70i+70j = 545i + 70j = 549kph
  2. jcsd
  3. Feb 17, 2009 #2


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    Science Advisor

    Re: vectors

    How do you get the "70j" here?

    Again, where did you get the "70 j" as part of the airplane's velocity? If the airplane is going "due east" shouldn't it be 0j?
  4. Feb 17, 2009 #3
    Re: vectors

    yea i thought as much but isn't the y-component of the windspeed part of the airspeed? Can you show me how to solve this problem?
  5. Feb 17, 2009 #4


    User Avatar
    Science Advisor

    Re: vectors

    No, the "airspeed" is the speed through the air and is separate from the wind speed. And you surely can't have thought you should include the j (north-south) component but not the i (north-south) component?

    The velocity relative to the air is vx i+ 80 k and the airspeed is 480 so [itex]vx^2+ 80^2= 480^2[/itex]. Once you have found that, the velocity relative to the ground is (vx+ 70)i+ 70j (the k component is not relevant to moving relative to the ground).
  6. Feb 17, 2009 #5
    Re: vectors

    ok thanks, i got 547.8kph. the answer key is 548.6, i guess is still correct...
  7. Feb 17, 2009 #6


    Staff: Mentor

    Re: vectors

    You lost some precision by the very rough rounding you did early on.
    100 [itex]\sqrt{2}[/itex]/2 is closer to 71 than it is to 70.

    You will always get more precise results if you refrain from rounding until your final result.
  8. Feb 17, 2009 #7
    Re: vectors

    Thanks, it was a complete oversight...duly noted
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