CHAPTER 3 - 21

# CHAPTER 3 - 21 - x y z θ v D North East θ h D 1 D 2 D 3...

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Unformatted text preview: x y z θ v D North East θ h D 1 D 2 D 3 up 70. If v PW is the velocity of the airplane with respect to the wind, v PG the velocity of the airplane with respect to the ground, and v WG the velocity of the wind with respect to the ground, then v PG = v PW + v WG , as shown in the diagram. We have two unknowns: v PG and θ (or α ). If we use the law of sines for the vector triangle, we have v PW /sin (90° + β ) = v WG /sin α , or sin α = ( v WG / v PW )sin (90° + β ) = [(120 km/h)/(680 km/h)] sin (90.0° + 35.0°) = 0.145, or α = 8.31°. Thus we have θ = α + β = 8.31° + 35.0° = 43.3° N of E . 71. If v CG is the velocity of the car with respect to the ground, v MG the velocity of the motorcycle with respect to the ground, and v MC the velocity of the motorcycle with respect to the car, then...
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CHAPTER 3 - 21 - x y z θ v D North East θ h D 1 D 2 D 3...

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