Help with trig problem … plane is flying at a certain height and speed, whats the bearing and ground speed?
This problem is listed in my book, but they do a terrible job of explaining how to come up with the answer. Please show me how to solve this problem.
The bearing of an airplane is 315° with an air speed of 500 mph. If the wind is blowing at a bearing of 60° at 25 mph, find the ground speed and the bearing (to the nearest tenth of a degree) of the actual course of the plane.
1 you first lay an coordinate system on the wind rose. The east-west direction is the x-axis, and the north-south direction is the y-axis.
The plane goes right to north-west at 45 degrees (measured from the -x axis) at 500 mph. This is a vektor v(plane) = (-500/√2/+500/√2). (Because tan 45 deg =1, and the amount of the vektor = √((-500/√2)^2+(500/√2)^2) = 500, are you familiar with vektors?).
The wind blows at 30 deg (measured from the +x axis). The wind vektor is v(wind) = (21.66/12.5)
(Because tan 30 deg is 12.5/21.66, and the amount of the vektor is √(21.66^2+12.5^2) = 25. I can explain it to you how to get to these numbers, if you want).
Now add the wind vektor to the plane vektor:
v(plane)+v(wind)= ((-500/√2 + 21.66) /((500/√2) + 12.5) = (-331,89/366.05)
The speed of the plane, that is the amount of the resulting vektor(plane+wind) is √(-331,89^2 + 366.05^2) =494.1048 mph.
The resulting direction of the plane is tan alpha = 366,05/331,89 = 1.1029, alpha = 47,8 degrees measured from the -x axis. This makes it 270 + 47.8 = 317.80 degrees.
