Answer:
Explanation:
From the given information:
The coordinate axis is situated in the east and north direction.
So, the north will be the y-axis and the east will be the x-axis
Similarly, the velocity of the plane in regard to the air in the coordinate system will be ![v_{P/A} = v( cos \theta \ i + sin \theta \ j)](https://tex.z-dn.net/?f=v_%7BP%2FA%7D%20%3D%20v%28%20cos%20%5Ctheta%20%5C%20i%20%2B%20sin%20%5Ctheta%20%5C%20j%29)
where:
= velocity of the plane in regard to the air
v = velocity
θ = angle of inclination of the plane with respect to the horizontal
replacing v = 180 km/ and θ = 20° in above equation, then:
The velocity of the airplane in the coordinate system as:
![v_{P} = v_o( cos \phi \ i + sin \phi \ j)](https://tex.z-dn.net/?f=v_%7BP%7D%20%3D%20v_o%28%20cos%20%5Cphi%20%5C%20i%20%2B%20sin%20%5Cphi%20%5C%20j%29)
where;
= velocity of the airplane
= velocity
∅ = angle of inclination with regard to the base axis;
Then; replacing
= 150 km/h and ∅ = 30°
Therefore, the velocity of the plane in the system is :
![v_p = v_A + v_{P/A}](https://tex.z-dn.net/?f=v_p%20%3D%20v_A%20%2B%20v_%7BP%2FA%7D)
--- (1)
( 150 cos 30° - 180 cos 20°)i + ( 150 sin 30° - 180sin 20°)j
(-39.24 km/h)i + (13.44 km/h) j
The magnitude is:
![v_A= (-39.24 km/h)i + (13.44 km/h) j](https://tex.z-dn.net/?f=v_A%3D%20%28-39.24%20km%2Fh%29i%20%2B%20%2813.44%20km%2Fh%29%20j)
![|v_A|^2 = \sqrt{ (-39.24 km/h)^2+ (13.44 km/h)^2}](https://tex.z-dn.net/?f=%7Cv_A%7C%5E2%20%3D%20%5Csqrt%7B%20%28-39.24%20km%2Fh%29%5E2%2B%20%2813.44%20km%2Fh%29%5E2%7D)
= 41.48 km/h
The airplane is moving at an angle of the inverse tangent to the abscissa and ordinate.
The angle of motion is:
tan θ = 39.24/13.44
tan θ = 2.9
θ = ![tan ^{-1} (2.9)](https://tex.z-dn.net/?f=tan%20%5E%7B-1%7D%20%282.9%29)
θ = 70.97°
The angle of motion is 70.97° from west of north with a velocity of 41.48 km/h.