A plane flies due north (90° from east) with a velocity of 100 km/h for 3 hours. During this time, a steady wind blows southeast at 30 km/h at an angle of 315° from due east. After 3 hours, where will the plane’s position be relative to its starting point? Show your work.

Answer 1

East component of wind = (cos 31.5 x 30) = 25.578kph.
South component of wind = (sin 13.5 x 30) = 15.625kph.
The south component slows the plane, so subtract.
(100 - 15.625) = 84.375kph north speed. In 3 hrs, it proceeds (84.375 x 3) = 253.125km.
The east component blows the aircraft towards the east, so in 3 hrs., it will move (25.578 x 3) = 76.734km. in the east direction.
Now you have a right triangle, the hypotenuse representing the plane's path, the height vertical (north) being 253.125km., and the base (east) being 76.734km.
The length of the hypotenuse is the distance from the start, so sqrt.(253.125^2 + 76.734^2) = 254.28km.northeast from start.
The angle from the start will be tanL = (253.125/76.374), atn = 73.2 degrees, north of east