Answer:
33.5 km at
south of west
Step-by-step explanation:
To solve the problem, we take the following convention:
- West is positive x-direction
- South is positive y-direction
Using this convention, we can describe the two parts of the motion as follows:
1) 15 km west: the components of this part of the motion are

2) 30 km south: the components of this part of the motion are

Therefore, the components of the net displacement are

So, the magntidue of the net displacement is

while the direction is:

south of west.