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A plane is steering at S65°W at an air speed of 625 km/h. The wind is from the NW at 130 km/h.
Find the ground speed and track of the plane. Include a vector diagram in your solution.
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The solution by the other tutor (@MowMow) is INCORRECT.
It is incorrect, just because the direction of the wind is determined/identified and used incorrectly in his post.
I came to bring a correct solution.
Explanation
"S65°W" refers to a bearing or direction, meaning South 65 degrees West.
This notation indicates a direction starting from South and then rotating 65 degrees towards the West."
S O L U T I O N
So, S65°W is the direction of 270° - 65° = 205° in the standard coordinate plane.
The direction of the wind "from NW" is 135° + 180° = 315° in the standard coordinate plane.
We are given two vectors
the airspeed of the plane = 625*(cos(205°), sin(205°)) km/h; (1)
the wind speed vector = 130*(cos(315°), sin(315°)) km/h. (2)
Let the groundspeed of the plane be (x,y).
The groundspeed is the sum of the vectors (1) and (2)
x = 625*cos(205°) + 130*cos(315°) = 625*(-0.90630778703) + 130*(+0.70710678118) = -474.5184853 km/h;
y = 625*sin(205°) + 130*sin(315°) = 625*(-0.42261826174) + 130*(-0.70710678118) = -356.0602951 km/h.
So, the groundspeed magnitude is = = 593.2509812 km/h.
The angle of the vector with the x-direction of the coordinate plane is
a = + = + = + = + =
= 3.14159265 + 0.643732 = 3.78532465 radians = 216.883127 degrees.
ANSWER. The groundspeed magnitude is about 593.251 km/h.
The direction of the groundspeed is about 216.883 degrees counterclockwise from positive direction of x-axis,
or 90 - 36.883 = 53.117 degrees from South to West (S53.117°W).
Solved.