The ground speed and direction of the plane is 323 kph with a directional bearing of S 22° E
To find the ground speed of the plane, we need to know what a vector is
A vector is a variable that has both magnitude and direction.
Since the airplane is headed due south with an airspeed of 300 kph, its vector is V = (-300 kph)j.
Also, the wind is blowing from the west at 120 kph. Its vector is v = (-120 kph)i
Now, the ground speed of the plane V' is the resultant vector of the airspeed and wind speed.
A resultant vector is the sum of two or more vectors.
So, V' = v + V
V' = (-120 kph)i + (-300 kph)j
So, its magnitude V' = √(x² + y²) where
So, V' = √[(-120 kph)² + (-300 kph)²]
= √[14400 kph² + 90000 kph²]
= √(104400 kph²)
= 323.11 kph
≅ 323 kph
Its direction, Ф = tan⁻¹(y/x)
Ф = tan⁻¹(-300 kph/-120 kph)
Ф = tan⁻¹(2.5)
Ф = 68.2°
From the South-line we have α = 90° - Ф = 90° - 68.2° = 21.8° ≅ 22°
So, the directional bearing of the plane is S22°E.
So, the ground speed and direction of the plane is 323 kph with a directional bearing of S 22° E
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Answer:
A on edge 2022
Step-by-step explanation:
323 kph and 22 E
