Using the sides given in the triangle shown on the right, you
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Answer:
Velocity of the plane relative to the ground
= [(-145.98î) + (15j)] knots
Magnitude = groundspeed = 146.75 knots
Direction = -5.9°
Step-by-step explanation:
This is a relative velocity question.
Let the velocity of the airplane relative to the wind be V(aw)
Velocity of the airplane relative to the ground = Va
The velocity of the wind relative to the ground = Vw
The relative velocity theory enables us o relate these as thus,
V(aw) = Va - Vw
Va = V(aw) + Vw
V(aw) = (-120î) knots (120 west)
For Vw, the wind is said to be blowing from 120°, reading this on a three figure bearing as angle from the north,
This gives the real direction of the wind to be 150° from the positive x-axis as shown in the attached image.
Vw = (30cos 150°)î + (30sin 150°)j
Vw = (-25.98î + 15j) knots
Va = V(aw) + Vw = (-120î) + (-25.98î + 15j)
Va = (-145.98î) + (15j)
Magnitude of Va = √[(-145.98)² + (15)²] = 146.75 knots
Direction = tan⁻¹ (-15/145.98) = -5.9°
Hope this Helps!!!
