A jet airplane flying from Darwin, Australia, has an air speed of 260 m/s in a direction 5.0º south of west. It is in the jet stream, which is blowing at 35.0 m/s in a direction 15º south of east. What is the velocity of the airplane relative to the Earth?

Respuesta :

Answer:

Vj,g = - 224.19 i - 31.72 j

magnitude of velocity = 226.42 m/s

Explanation:

Velocity of jet with respect to air = 260 m/s in 5 degree South of west

Velocity of air with respect to ground = 35 m/s in 15 degree south of east

Write these velocities in vector form

Vj,a = 260 (- Cos 5 i - Sin 5 j) = - 259 i - 22.66 j

Va,g = 35 ( Cos 15 i - Sin 15 j) = 33.81 i - 9.06 j

Velocity of jet with respect to air = velocity of jet with respect to ground -    

                                                         velocity of air with respect to ground

Vj,a = Vj,g - Va,g

Vj,g = Vj,a + Va,g

Vj,g = - 259 i - 22.66 j + 33.81 i - 9.06 j

Vj,g = - 224.19 i - 31.72 j

The magnitude of velocity of jet relative to ground or earth is

= [tex]\sqrt{\left ( - 224.19 \right )^{2}+\left ( - 31.72 \right )^{2}}[/tex]

= 226.42 m/s

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