Answer to Question #102632 in Electricity and Magnetism for Ali hamza

Question

A commuter airplane takes the route shown in Figure 3.20. First, it flies from the origin of the coordinate system shown to city A, located 175 km in a direction 30.0° north of east. Next, it flies 153 km 20.0° west of north to city B. Finally, it flies 195 km due west to city C. Find the location of city C relative to the origin.

Answer 1

The route of  airplane look like one can see in figure.

OA=175 km, AB=153 km, BC=195 km

$\vec{OA}=(OA\cdot cos30\degree)\cdot \hat {i}+(OA\cdot sin30\degree)\cdot \hat j=175km\cdot \frac{\sqrt{3}}{2} \hat i+ 175km\cdot \frac{1}{2} \hat j=\\= 151.55km\cdot \hat i+87.5km \cdot \hat j$

$\vec{AB}=AB\cdot cos(90\degree+20\degree)\cdot \hat i+AB\cdot sin(90\degree+20\degree)\cdot \hat j=\\=AB\cdot cos(110\degree)\cdot \hat i+AB\cdot sin(110\degree)\cdot \hat j=153km\cdot (-0.342)\cdot \hat i+153km\cdot 0.9397\cdot \hat j=-52.3km\cdot \hat i +143.8km\cdot \hat j$

$\vec {BC}=-BC\cdot \hat i=-195km\cdot \hat i$

$\vec {OC}=\vec {OA}+\vec {AB}+\vec {BC}$

$\vec {OC}=(151.55-52.3-195)km\cdot \hat i+(87.5+143.8)km \cdot \hat j=-95.8km\cdot \hat i+231.3km \cdot \hat j$

$OC=\sqrt{(95.8km)^2+(231.3km)^2}=250km$

$\alpha=sin^{-1}(\frac{95.8}{250})=22.5\degree$

Answer: City C located 250 km in a direction $22.5 \degree$ west of nord.