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# 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?
1
2020-05-12T09:59:10-0400

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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