Asked by Deb
An aircraft maintains a speed of 500 miles per hour in a southwestern direction. The velocity of the jet stream is constant 50 miles per hour from the west. Find the resultant vector, the speed, and actual direction of the aircraft.
Answers
Answered by
Henry
R = 500mi/h @ 225o + 50mi/h @ 0o
X = 500*cos225 + 50 = --303.6/303.6 mi/h.
Y = 500*sin225 = -353.6 mi/h.
tanAr = Y/X = -353.6/-303.6 = 49.35o =
Reference Angle.
A = 49.35 + 180 = 229.35o,CCW = 40.65o
West of South = Direction.
V = X/cosA = -303.6/cos229.35=466 mi/h.
X = 500*cos225 + 50 = --303.6/303.6 mi/h.
Y = 500*sin225 = -353.6 mi/h.
tanAr = Y/X = -353.6/-303.6 = 49.35o =
Reference Angle.
A = 49.35 + 180 = 229.35o,CCW = 40.65o
West of South = Direction.
V = X/cosA = -303.6/cos229.35=466 mi/h.
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