Answer:
B) Ground speed: 639 mph; direction 161°
Step-by-step explanation:
165° = 75° South of East = S15°E, therefore;
The component of the speed of the airplane are;
650 × cos(75°) East or x-direction in vector form:[tex]$650 \times \cos \left(75^{\circ}\right) \mathrm{i}$[/tex]
650 × sin(75°) South or negative y-direction in vector form: [tex]$-650 \times \sin \left(75^{\circ}\right) \mathrm{j}$[/tex]
The speed of the airplane is [tex]$650 \times \cos \left(75^{\circ}\right) \mathrm{i}-650 \times \sin \left(75^{\circ}\right) \mathrm{j}$[/tex]
The speed of the wind in vector form is 25i + 25·√3 j
The combined speed of the airplane and the wind is found as follows;
Combined: [tex]$\left(650 \times \cos \left(75^{\circ}\right)+25\right)\left[\dot{i}+\left(25 \cdot \sqrt{3}-650 \times \sin \left(75^{\circ}\right)\right) \mathrm{j}\right.$[/tex], which gives;
The resultant speed of the airplane, [tex]$\left.R=\sqrt{(} 193.23^{2}+(-584.55)^{2}\right)=615.66$[/tex]
The resultant speed of the airplane, [tex]$\mathrm{R}=615.66 \mathrm{mph}$[/tex]
The direction =[tex]$\arctan ((-584.55) /(193.23))=-71.7^{\circ}$[/tex]
The bearing [tex]$=90^{\circ}+71.7^{\circ}=162^{\circ}$[/tex]
Therefore, the closest option is option B, ground speed: 639 mph; direction 161°
Learn more about speed refer:
- https://brainly.com/question/18868816
