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Relative Velocity with plane and wind

  1. Mar 21, 2013 #1
    1. The problem statement, all variables and given/known data
    An aircraft flew due east fro P to Q at u1 km/h.Wind speed from south west was v km/h.On the return journey from Q to P,due west, the aircraft's speed was u2 km/h, the wind velocity being unchanged.If the speed of the aircraft in still air was x km/h, x>v,show by resolving the perpendicular to PQ.or otherwise,that
    u1 - u2 = v√2


    2. Relevant equations



    3. The attempt at a solution[/b/

    Can somebody help me with this?I think I made a simple mistake.

    vpw = (xcosa)i - (xsina)j

    vw = (v√2/2)i + (v√2/2)j

    vp = (xcosa + v√2/2)i +(xsina + v√2/2)

    since the j component must be zero sina = 2v/2x ............. cosa =√(4x^2 -2v^2)/2x

    vp =x√(4x^2 -2v^2)/2x + v√2/2 = (v√2 + √(4x^2 -2v^2))/2 = u1


    return journey

    since the plane is travelling in the opposite direction

    vpw = (-xcosa)i - (xsina)j

    vp = (-xcosa + v√2/2)i +(-xsina + v√2/2)j

    since j is zero cosa =√(4x^2-2v^2)/2x

    u2=vp= (v√2 - √4x^2 -2v^2)/2 i

    u1 - u2 = (√(4x^2 - 2v^2) +√(2v^2))/2 - (√(4x^2 -2v^2) - √(2v^2))/2

    If I simplify it doesn't work out.Should I have multiplied u2 by minus 1 since its going in the minus i direction?
     
  2. jcsd
  3. Mar 21, 2013 #2

    Simon Bridge

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    Probably the easiest way to check your working is to draw out the vector triangles and do the geometry.
     
  4. Mar 21, 2013 #3
    Sorry im not exactly sure what you mean.I get the correct answer if I use the vector as opposed to the magnitude of the vector.
     
    Last edited: Mar 21, 2013
  5. Mar 21, 2013 #4

    Simon Bridge

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    The solution involves adding and subtracting vectors - there are two main ways of doing this. You have been resolving each vector into components N-S and E-W.
    Have you tried sketching the vectors out using head-to-tail?
     
  6. Mar 22, 2013 #5
    I just figured where I went wrong thanks anyway.
     
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