Answer to Question #114678 in Mechanics | Relativity for Michael Mateus

Question #114678
6. An airplane whose air speed is 650km/h is supposed to fly in straight path 35° north of
east. But a steady 95 km/h wind is blowing from north. In what direction should the
plane head and what is the plane resultant speed?
Expert's answer

This problems is related to the notion of addition of vectors.

The resultant velocity "\\vec{v}_{rel} = \\vec{v}_{wind} + \\vec{v}_{plane}." We know that the wind counteracts to the movement of

the plane, so the resultant velocity will be smaller than the initial velocity "v_{plane}" . Also, we conclude that

the angle in question should be larger than "35^\\circ" .

The angle between "\\vec{v}_{wind}" and "\\vec{v}_{res}" should be "\\alpha = 90^\\circ + 35^\\circ = 125^\\circ." We may calculate the angle "\\beta" between "\\vec{v}_{res}" and "\\vec{v}_{plane}" using the law of sines

"\\dfrac{v_{plane}}{\\sin\\alpha} = \\dfrac{v_{wind}}{\\sin\\beta}."

Therefore, "\\sin\\beta = \\dfrac{v_{wind}}{v_{plane}}\\sin\\alpha = \\dfrac{95}{650} \\sin125^\\circ \\approx 0.12, \\quad \\beta = 6.9^\\circ." The third angle "\\gamma" between "v_{wind}" and "v_{plane}" is "\\gamma = 180^\\circ - \\alpha-\\beta = 180^\\circ - 125^\\circ-6.9^\\circ= 48.1^\\circ."

Therefore, we may calculate "v_{res}" using the law of cosines

"v_{res}^2 = v_{wind}^2 + v_{plane}^2 - 2v_{wind}v_{plane}\\cos\\gamma."

"v_{res}\\approx 590\\,km\/h."

The angle in question is "90^\\circ - \\gamma = 90^\\circ-48.1 = 41.9^\\circ" north of east.

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