FInd displacement of airplane - vectors question

Question

An airplane leaves the airport travelling N 55 degrees W at 580km/hour. After 1.5 hours the airplane then turns north and travels 2 hours at 610km/hour. What is the displacement(magnitude and direction) of the airplane after a 3.5 hour trip

Answer 1

N 55 degrees W means:

The distance traveled is the speed multiplied by the time so (580)(1.5) = 870 km

Then, the direction changes to north and the distance traveled north is (610)(2) = 1220 km

So that makes the problem turn ito this:

You can do this problem a couple of ways. The traditional "physics" method is to break each vector into its x-component and y-component, add all the x-components and add all the y-components and then put the resultant back together using the Pythagorean Theorem.

You can also use the Law of Cosines. Using alternate interior angles, the supplement to angle θ is 55° making angle θ = 125°

Using the Law of Cosines:

d² = 870² + 1220² - 2(870)(1220)cos(125°)

d² = 756900 + 1488400 - 2122800cos(125°)

d² = 756900 + 1488400 - (-1217588)

d² = 3462888

d = 1860.88 km

You can then use the Law of Sines to find angle α which will be the same as the direction of the displacement vector west of north (alternate interior angles again).