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 [tex]v_{P/A} = v( cos \theta \ i + sin \theta \ j)[/tex]
where:
[tex]v_{P/A}[/tex] = 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:
[tex]v_{P} = v_o( cos \phi \ i + sin \phi \ j)[/tex]
where;
[tex]v_p[/tex] = velocity of the airplane
[tex]v_o[/tex] = velocity
∅ = angle of inclination with regard to the base axis;
Then; replacing [tex]v_o[/tex] = 150 km/h and ∅ = 30°
Therefore, the velocity of the plane in the system is :
[tex]v_p = v_A + v_{P/A}[/tex]
[tex]v_A= v_P -v_{P/A}[/tex] --- (1)
[tex]v_A=[/tex] ( 150 cos 30° - 180 cos 20°)i + ( 150 sin 30° - 180sin 20°)j
[tex]v_A=[/tex] (-39.24 km/h)i + (13.44 km/h) j
The magnitude is:
[tex]v_A= (-39.24 km/h)i + (13.44 km/h) j[/tex]
[tex]|v_A|^2 = \sqrt{ (-39.24 km/h)^2+ (13.44 km/h)^2}[/tex]
[tex]v_A[/tex] = 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 = [tex]tan ^{-1} (2.9)[/tex]
θ = 70.97°
The angle of motion is 70.97° from west of north with a velocity of 41.48 km/h.
