An aircraft travels at 250km/h in still air. If a wind of 50km/h blows steadily from the southwest and the pilot wishes to fly due north. On the return journey the wind speed and direction are unaltered. Find the new course and ground speed
To find the new course and ground speed, we need to consider the effect of the wind on the aircraft.
Let's consider the first leg of the journey, where the airplane is flying due north.
Since there is a wind blowing from the southwest, it will create a drift that affects the aircraft's course. The angle between the direction the aircraft is pointing (north) and the direction it is actually moving due to the wind (resultant ground track) is known as the wind angle.
To find the wind angle, we can use trigonometry. The velocity triangle can be drawn as follows:
/|
V / |
/ | W
/θ |
/____|
Where V is the velocity of the aircraft (250 km/h) and W is the velocity of the wind (50 km/h).
We can calculate the wind angle (θ) using the tangent function:
tan(θ) = W / V
θ = arctan(W / V)
θ = arctan(50 / 250)
θ ≈ 11.31°
Therefore, the wind is blowing at an angle of approximately 11.31° with respect to the north.
To find the ground speed, we can use the Pythagorean theorem:
Ground Speed = sqrt(V^2 + W^2)
Ground Speed = sqrt(250^2 + 50^2)
Ground Speed = sqrt(62500 + 2500)
Ground Speed = sqrt(65000)
Ground Speed ≈ 254.95 km/h
Therefore, the ground speed is approximately 254.95 km/h.
Now, let's consider the return journey where the wind speed and direction are unaltered.
Since the wind angle remains the same (11.31°), the aircraft needs to align itself with the wind (southwest) in order to cancel out its effect and fly due north.
To find the new course, we need to subtract the wind angle from 180° (since the wind is coming from the southwest).
New Course = 180° - 11.31°
New Course ≈ 168.69°
Therefore, the new course is approximately 168.69° and the ground speed is approximately 254.95 km/h.
To find the new course and ground speed, we need to break down the velocity vectors into their components.
Let's assign the following directions:
North: N
South: S
East: E
West: W
Given:
Aircraft speed in still air = 250 km/h
Wind speed = 50 km/h
Wind direction = Southwest
First, let's find the components of the wind velocity:
The wind is blowing from the southwest, which is 45 degrees clockwise from the south direction. Therefore, we can split the wind velocity into two components:
Wind component in the north-south direction = Wind speed × cos(45°)
Wind component in the east-west direction = Wind speed × sin(45°)
Calculating the components:
Wind component in the north-south direction = 50 × cos(45°)
= 50 × 0.7071
≈ 35.355 km/h (rounded to 3 decimal places)
Wind component in the east-west direction = 50 × sin(45°)
= 50 × 0.7071
≈ 35.355 km/h (rounded to 3 decimal places)
Now, let's calculate the ground speed and course for the journey north.
ground speed = aircraft speed in still air + wind component in the north-south direction
ground speed = 250 + 35.355
≈ 285.355 km/h (rounded to 3 decimal places)
course = the angle between the direction of flight and the north direction
Since the pilot wishes to fly due north, the course will be 0 degrees.
Therefore, the new course is 0 degrees, and the ground speed is approximately 285.355 km/h.
On the return journey, we assume that the wind speed and direction remain the same.
The wind is still blowing from the southwest, so the wind component in the north-south direction and the east-west direction remain the same.
For the return journey, the aircraft is now flying due south, opposite to the initial direction. Hence, the course will be 180 degrees.
Therefore, the new course for the return journey is 180 degrees, and the ground speed remains approximately 285.355 km/h.