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Relative Motion in 2D - a plane & wind speed.

  1. Feb 10, 2009 #1
    1. The problem statement, all variables and given/known data
    A plane attains an airspeed of 620 km/hr. The plane sets out for a destination 1085 km to the north (a 0° course) but discovers that the plane must be headed 20° east of north to fly there directly. The plane arrives in 2.00 h.

    What is the wind speed?
    What is the direction of the wind?

    2. Relevant equations

    Vf = Vi + a*t
    x = vi*t + 0.5a*t^2
    vf^2 = vi^2 + 2a*x
    V in x-dir'n = |Vi|cosΘ
    V in y-dir'n = |Vi|sinΘ

    3. The attempt at a solution

    I found the north component of velocity as 1850km/2.00h = 542.5km/h

    But after that I am completely lost.

    Then, I think to find the direction of the wind u need inverse-tan of the y-dir'n velocity & x-dir'n velocity of the wind??? But I can't do that without getting the first part first.

    Please, help. Thx.
    Last edited: Feb 10, 2009
  2. jcsd
  3. Feb 10, 2009 #2
    by the way 1850km/2.00h = 542.5m/s is not right...
    1850km/2.00h = 542.5m/H... you need use hour
    then do this 545.5-620=-74.5km/h (wind speed)
    the wind must be negative
  4. Feb 10, 2009 #3
    Ah, yes, that was a typo. I meant to say 542.5km/h. Thx for pointing out my mistake.

    Btw, in your calculation of wind speed you subtracted 620km/h from 545.5km/h... It's supposed to be 542.5km/h, right? Then that means wind speed is -77.5km/h??
  5. Feb 10, 2009 #4
    yeah...sorry i am doing so many problems...
  6. Feb 10, 2009 #5
    also please consider the type of air plant... because sometime negative diraction wind actually giving the force to air plane...
  7. Feb 10, 2009 #6


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    Homework Helper

    Not so fast there Amelia Earhart.

    It's true that the plane had to fly a longer distance, but the wind speed would have been -77 if the pilot had stayed on a totally Northern course. But he didn't.

    Draw a picture and along his flight heading he went in the air 1240km, but he ended up only 1085 km due north.

    What wind vector would produce that result?
  8. Feb 10, 2009 #7
    Ok, so I found his NE distance to be 1154.6km by Pythagoras.
    Then: 1154.6km / 2.00h = 577.3km/h (the speed due NE).

    Next: 577.3km/h - 620km/h = -42.7km/h as the wind speed... but it's WRONG. What am I doing wrong?
  9. Feb 10, 2009 #8
    where you get 1154.6?
  10. Feb 10, 2009 #9
    1850/2=542.5??? it is 1085
  11. Feb 10, 2009 #10


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    Homework Helper

    For one thing the hippopotamus is 1240 for your distance vector on the N-20-E heading. 2 hours at 620km/h = 1240.

    Then during the time of his flight the wind moved the plane to the North axis.

    What is that distance?

    That distance divided by the 2 hours is your average wind speed in km/h.
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