Answer:
33.5 km at [tex]63.4^{\circ}[/tex] south of west
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
[tex]a_x = 15 km\\a_y = 0[/tex]
2) 30 km south: the components of this part of the motion are
[tex]b_x = 0\\b_y = 30 km[/tex]
Therefore, the components of the net displacement are
[tex]r_x = a_x + b_x = 15 + 0 = 15 km\\r_y = a_y + b_y = 0 +30 = 30 km[/tex]
So, the magntidue of the net displacement is
[tex]r=\sqrt{r_x^2+r_y^2}=\sqrt{15^2+30^2}=33.5 km[/tex]
while the direction is:
[tex]\theta=tan^{-1}(\frac{r_y}{r_x})=tan^{-1}(\frac{30}{15})=63.4^{\circ}[/tex]
south of west.
