Asked by Maddie
an airplane is flying a compass of 340 degrees at 325 mph. A wind is blowing with the bearing 300 degrees at 30 mph. Find the component form of the velocity, and find the actual ground speed and direction of the plane.
Answers
Answered by
Henry
V = 325mph[340o] + 30mph[300o]
X = 325*cos340 + 30*cos300 = 320.4 mph =
Hor. component.
Y = 325*sin340 + 30*sin300 = -137.1 mph
= Ver. component.
TanA = Y/X = -137.1/320.4 = -42802
A = -23.17o,CW = 336.8 CCW.
V = 320.4/cos336.8 = 348.6mph[336.8o]
X = 325*cos340 + 30*cos300 = 320.4 mph =
Hor. component.
Y = 325*sin340 + 30*sin300 = -137.1 mph
= Ver. component.
TanA = Y/X = -137.1/320.4 = -42802
A = -23.17o,CW = 336.8 CCW.
V = 320.4/cos336.8 = 348.6mph[336.8o]
Answered by
Suman
V= <325 cos110, 325 sin110>
W=<40cos130, 40sin130>
Resultant =V+W
Actual ground speed = magnitude of resultant = √(-136.87)^2+(336.04)^2 = 362.85 mph
W=<40cos130, 40sin130>
Resultant =V+W
Actual ground speed = magnitude of resultant = √(-136.87)^2+(336.04)^2 = 362.85 mph
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