Help me with this vector velocity problem, please.?

Question

An airplane which has a 420 mi/hr ground speed in still air is flying with its nose toward the north. A 60 mi/hr wind is blowing 30 degrees south of west. What are the magnitude and direction of the airplane's velocity relative to the ground?

Answer 1

You need to break the wind into components.

windx = wind cos (angle)
windy = wind sin (angle)

The angle is measured counterclockwise from east (so it's 210).

Then you can add the negative windy to the airplane's velocity to get the total y component. The wind's x component IS the total y component.

Then recombine to get the magnitude of velocity using pythagorus:

v = sqrt (vx^2 + vy^2)

to get the direction, use tan (direction) = vy / vx. Take care, because arctangent is ambiguous, so you have to make sure you have the answer in the right sector.

Answer 2

Lets start with the x:
x = 60cos30 = 51.9615mi/hr
In the y, the wind slows the plane down somewhat:
y = 420 - 60sin30 = 390mi/hr
For magnitude we use the Pythagorean theorem:
Mag = √(390^2 + 51.9615^2) = 393.446mi/hr
Direction would be a little west of north:
Dir = arctan(390/51.9615) = 82.4109degrees north of west
Or:
90 - 82.4109 = 7.5891degrees west of north
Note: Usually a wind is identified by the direction it's coming from, but here it clearly states the direction the wind is blowing.