An airplane flies at 400 mph with a direction of 120° relative to the air. The plane experiences a wind that blows 40 mph with a direction of 50°.

Part A: Write each of the vectors in linear form. Show all necessary calculations. (6 points)

Part B: Find the sum of the vectors. Show all necessary calculations. (2 points)

Part C: Find the true speed and direction of the airplane. Round the speed to the thousandths place and the direction to the nearest degree. Show all necessary calculations.

The necessary calculations for vector arithmetic are handily carried out by a suitable calculator. The attached shows a TI-84 work-alike, but there are many other possible choices.

The necessary calculations for doing this "by hand" (without a calculator that does vectors or complex numbers) are ...

a∠b = a·cis(b) = (a·cos(b) +i·a·sin(b))

a +i·b = √(a²+b²)∠arctan(b/a) . . . . . with attention to signs and quadrant

__

Ordinarily, bearings are measured clockwise from North, and angles in the x-y or complex plane are measured counterclockwise from +x (East). If you use bearing angles without any translation, the resulting (complex plane) coordinates from rectangular/polar conversion are ...

(North + i·East)

Coordinates in this system can be added and subtracted in the usual way. When the result is converted back to a distance and direction, that direction can be considered to be a bearing angle. In short, we can use bearing angles in our math with no special treatment, provided we are consistent in that usage.

__

A.

The airplane speed vector in <north, east> coordinates is ...

400∠120° = 400cos(120°) +i·400·sin(120°) = <-200, 346.410>

Similarly, the wind speed vector is ...

40∠50° = 40cos(50°) +i·40sin(50°) = <25.712, 30.642>

__

B.

The sum of the vectors is found from the sum of their components:

plane + wind = <-200 +25.712, 346.410 +30.642> = <-174.288, 377.052>

__

C.

Converting this vector in linear form back to a magnitude and direction gives ...

magnitude = √((-174.288)² +377.052²) = √172544.644542

magnitude ≈ 415.385 . . . mph

direction = arctan(-377.052/174.288) . . . . . 90° < θ < 180°

direction ≈ 115°

_____

Additional comment

As we discussed above, the vector coordinates used here are <N, E> coordinates. If you translate the bearing angles to Cartesian plane angles (measured CCW from +x), the corresponding coordinates will be <E, N> coordinates. That is, rectangular coordinates will be reversed from those shown here. If you're filling in an answer box, you may want to be aware of that fact.