Answer to Question #242461 in Electricity and Magnetism for waleed

Question

 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. after landing in city C, the pilot wishes to return to the origin

along a single straight line. What are the components of the vector representing this displacement?

What should the heading of the plane be?

Answer 1

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

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

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

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

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

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

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

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

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