Question

Instruments in an airplane which is in level flight indicate that the velocity relative to the air (airspeed) is 180.00 km/h and the direction of the relative velocity vector (heading) is 70° east of north. Instruments on the ground indicate that the velocity of the airplane (ground speed) is 150.00 km/h and the direction of flight (course) is 60° east of north. Determine the wind speed and direction. (Round the final answer to two decimal places.)

Answer 1

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 $v_{P/A} = v( cos \theta \ i + sin \theta \ j)$

where:

$v_{P/A}$ = 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:

$v_{P} = v_o( cos \phi \ i + sin \phi \ j)$

where;

$v_p$ = velocity of the airplane

$v_o$ = velocity

∅ = angle of inclination with regard to the base axis;

Then; replacing  $v_o$  = 150 km/h and ∅ = 30°

Therefore, the velocity of the plane in the system is :

$v_p = v_A + v_{P/A}$

$v_A= v_P -v_{P/A}$   --- (1)

$v_A=$ ( 150 cos 30° - 180 cos 20°)i + ( 150 sin 30° - 180sin 20°)j

$v_A=$ (-39.24 km/h)i + (13.44 km/h) j

The magnitude is:

$v_A= (-39.24 km/h)i + (13.44 km/h) j$

$|v_A|^2 = \sqrt{ (-39.24 km/h)^2+ (13.44 km/h)^2}$

$v_A$ = 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  = $tan ^{-1} (2.9)$

θ  =  70.97°

The angle of motion is  70.97° from west of north with a velocity of 41.48 km/h.