Finding resultant direction

Question

An airplane is traveling 583 m/s at 68.1° degrees north of west when a wind starts to blow. The velocity of the wind is 110 m/s and it is blowing toward 19.0° east of north. What is the resultant direction of the plane, assuming due east is 0 degrees, due north is 90 degrees, due west is 180 degrees and due south is 270 degrees? (Remember, no units for degrees.)

Answer 1

First identify compass degrees in the math-angle (CCL from E) so airplane is heading at 68.1 degrees N of W which is 180-68.1 = 107.9 degrees. The wind is given as 19 degrees E of N or 90-16 = 74 degrees

Find x component of resultant: v_x = v_px + v_wx = (583 m/s)cos(107.9°) + (110 m/s)cos(74°)

Do the same for y components (using sin instead of cos)

Find magnitude of resultant: sqrt(v_x²+v_y²) = |v|

Find direction of resultant θ = tan^-1(v_y/v_x) (add 180° if v_x<0)

Resultant is the vector sum of the velocities of the plane in still air and the wind

Please consider a tutor. Take care.

Answer 2

The plane travels at 583 @ 111.9 degrees based on East = 0

The wind starts blowing 110 @ 71 degrees based on East = 0

Start by drawing the plane's speed vector, then at the tip of that arrow, draw the wind vector. Using east as 0 conveniently let's you use a normal xy graphing method. Add the two vectors:

X1: cos(111.9)*583=-217.5

Y1: sin(111.9)*583=540.9

X2: cos(71*110)=35.8

Y2: sin(71*110)=104

Using inverse sin and cos you can do the rest and come up with:

670 m/s at an angle of 105.7 degrees (using E as 0)