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A jet airliner moving initially at 3.00 3 102 mi/h due east enters a region where the wind is blowing 1.00 3 102 mi/h in a direction 30.0° north of east. (a) Find the components of the velocity of the jet air- liner relative to the air, SvJA. (b) Find the components of the velocity of the air relative to Earth,

Respuesta :

Answer:

a) Vaw = 2.19 mi / h  θ = 346.8º b) Vmgx = 0.866 mi / h   Vawy = -0.5 mi / h

Explanation:

We using the sum of vectors, we use the indexes ‘a’ for the plane, ‘g’ for the land ‘w’ for the wind,

          [tex]v_{ag}[/tex] = [tex]v_{aw}[/tex] + [tex]v_{wg}[/tex]  

          [tex]v_{aw}[/tex] = [tex]v_{ag}[/tex] + [tex]v_{wg}[/tex]

To facilitate the calculation we decompose with respect to xy coordinate system

         [tex]v_{wgx}[/tex] =  [tex]v_{wg}[/tex] cos 30

         [tex]v_{wgy}[/tex]=  [tex]v_{wg}[/tex] sin30

       

         Vwgx = Vwd cos 30  

         Vwgx = Vwd sin30  

         Vmgx = 1.00 cos 30  

         Vmgx = 0.866 mi / h  

         Vmgy = 1.00 sin30  

         Vwgy = 0.5 mi / h  

Let's find the resulting components  

         Vawx = 3.00 -0.866  

         Vawx = 2,134 mi / h  

         Vawy = 0 - 0.5  

         Vawy = -0.5 mi / h

Let's use Pythagoras' theorem  

        Vaw2 = Vawx2 + Vawy2  

        Vaw = Ra Vawx2 + Vawy2  

        Vaw = ra 2,134 2 + 0.52  

        Vaw = 2.19 mi / h  

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