Posted by Anonymous on Thursday, December 13, 2012 at 8:08pm.
The pilot of an aircraft wishes to fly due west in a 46.8 km/h wind blowing toward the south. The speed of the aircraft in the absence of a wind is 204 km/h.
(a) In what direction should the aircraft head?
° north of west
(b) What should its speed relative to the ground be?
- physics - Henry, Saturday, December 15, 2012 at 8:49pm
a. X = -204 km/h.
Y = -46.8 km/h.
tanA = Y/X = -46.8/-204 = 0.22941.
Ar = 12.9o South of West.
The plane should head 12.9o North of West.
b. A = 180 + Ar = 180 + 12.9 = 192.9o
V = X/cosA = -204/cos192 = 209.3 km/h.
- physics - Henry, Thursday, October 16, 2014 at 7:44pm
a. Vpa = Vpg + Va = -204 km/h
Vpg - 46.8i = -204
Vpg = -204 + 46.8i
Tan Ar = Y/X = 46.8/-204 = -0.22941
Ar = -12.9o = Reference angle.
A = -12.9 + 180 = 167.1o CCW = 12.9o N.
of W. = Direction.
b. Vpg = X/Cos A = -204/Cos167.1 = 209.3 m/s. = Velocity of plane relative to gnd.
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