Answer:
You should choose airspeed 158.11 km/h at 18.4° west of south
Explanation:
The distance to the air port is 450 km due to south
You should to reach the airport in 3 hours
→ Velocity = distance ÷ time
→ Distance = 450 km , time = 3 hours
→ The velocity of your plane = 450 ÷ 3 = 150 km/h due to south
A wind is blowing from west 50 km/h
We need to know what heading and airspeed you should choose to
reach your destination
At first we must find the resultant velocity of your plane and the wind
The south and west are perpendicular, then the resultant velocity is
→ [tex]v_{R}=\sqrt{(v_{p})^{2}+(v_{w})^{2}}[/tex]
→ [tex]v_{p}=150[/tex] km/h , [tex]v_{w}=50[/tex] km/h
→ [tex]v_{R}=\sqrt{(150)^{2}+(50)^{2}}=158.11[/tex] km/h
To cancel the velocity of the wind, the pilot should maintain the velocity
of the plane at 158.11 km/h
The direction of the velocity is the angle between the resultant velocity
and the vertical (south)
→ The direction of the velocity is [tex]tan^{-1}\frac{50}{150}=18.4[/tex]°
The direction of the velocity is 18.4° west of south
You should choose airspeed 158.11 km/h at 18.4° west of south
