Vector A, a wind velocity vector, has a magnitude of 20 miles per hour and is blowing toward the North. A second wind velocity vector, Vector B, has a magnitude of 50 miles per hour and is blowing toward the Southeast. A third wind velocity vector, Vector C, has a magnitude of 10 miles per hour and is blowing toward the Southwest These three winds emerge (come out of) three separate canyons and merge (meet) at a flagpole carrying a cloth flag. What direction (angle with respect to zero degrees) will the flag be pointing as a result of the three winds? Also, what is the magnitude of the resulting wind velocity vector (expressed in m/s)?
This is how far I got... and is it correct?????
R=20Cos0 N + 50cos135N+50sin135E+10Cos225N+10sin225(-E)
3 answers
Do I just find the angles to all of them?
Most of us say a North wind is FROM the North, but oh well I will pretend to be a mathematician.
North:
A 20
B -50 *.707
C -10 *.707
East:
A 0
B 50*.707
C -10*.707
total N = -22.42
total E = +28.28
so in quadrant 4
tan angle S of E = 22.42/28.28
angle S of E = 38.4
That is all I can say. Being a navigator for me zero is north but in math it is east :) You decide
v = sqrt (22.42^2 + 28.28^2)
North:
A 20
B -50 *.707
C -10 *.707
East:
A 0
B 50*.707
C -10*.707
total N = -22.42
total E = +28.28
so in quadrant 4
tan angle S of E = 22.42/28.28
angle S of E = 38.4
That is all I can say. Being a navigator for me zero is north but in math it is east :) You decide
v = sqrt (22.42^2 + 28.28^2)
Calculate resultant in mi/h and convert to m/s. All angles are CCW from +x-axis.
R = 20[90o] + 50[315o] + 10[225].
R = 20i + 35.36-35.36i + -7.07-7.07i
R = 28.29 - 22.43i = 36.1 mi/h[-38.4o] = 36.1 mi/h[38.1o] S. of E.
R = 36.1mi/h * 1600m/mi * 1h/3600s = 16m/s[38.4o] S. of E.
R = 20[90o] + 50[315o] + 10[225].
R = 20i + 35.36-35.36i + -7.07-7.07i
R = 28.29 - 22.43i = 36.1 mi/h[-38.4o] = 36.1 mi/h[38.1o] S. of E.
R = 36.1mi/h * 1600m/mi * 1h/3600s = 16m/s[38.4o] S. of E.