an aircraft is flying northward at 300kmh where steady wind is blowing westward at 80kmh, what is the actual direction of the aircraft over the ground
To determine the actual direction of the aircraft over the ground, we can use vector addition.
Let's break down the velocities into their respective components:
- The velocity of the aircraft flying northward can be represented as (0 km/h, +300 km/h). Since it is flying directly north, there is no component in the horizontal (westward) direction, so the x-component is 0 and the y-component is +300 km/h.
- The velocity of the wind blowing westward can be represented as (-80 km/h, 0 km/h). Since it is blowing directly west, there is no component in the vertical (northward) direction, so the y-component is 0 and the x-component is -80 km/h.
Now, we can add the two vectors together to obtain the resultant velocity:
Resultant velocity = (0 km/h - 80 km/h, +300 km/h + 0 km/h)
= (-80 km/h, +300 km/h)
So, the aircraft's actual direction over the ground is 80 km/h westward and 300 km/h northward.
To calculate the actual direction of the aircraft over the ground, we can use vector addition. We need to find the resultant vector of the aircraft's velocity and the wind's velocity.
Step 1: Draw a diagram representing the situation. Draw a line pointing northward to represent the aircraft's velocity vector and another line pointing westward to represent the wind's velocity vector. The length of the northward line should represent a speed of 300 km/h, and the westward line should represent 80 km/h.
Step 2: Place the tail of the wind's velocity vector at the head of the aircraft's velocity vector, forming a triangle.
Step 3: Use the Pythagorean theorem to find the magnitude of the resultant vector (the diagonal line of the triangle). The magnitude can be calculated using the formula: c = √(a^2 + b^2), where a and b are the lengths of the two sides of the triangle.
In this case, a = 300 km/h (aircraft's velocity) and b = 80 km/h (wind's velocity). Plugging in these values into the formula, we have:
c = √(300^2 + 80^2) = √(90000 + 6400) = √96400 ≈ 310.44 km/h.
Step 4: Use trigonometry to find the angle between the aircraft's velocity vector and the resultant vector. Calculate the angle using the formula: θ = tan^(-1)(b/a), where a and b are the lengths of the two sides of the triangle.
In this case, a = 300 km/h (aircraft's velocity) and b = 80 km/h (wind's velocity). Plugging in these values into the formula, we have:
θ = tan^(-1)(80/300) ≈ 14.04°.
Step 5: Determine the actual direction of the aircraft over the ground. Since the wind is blowing to the west (-x-axis) and the resultant vector is at an angle of 14.04°, we can say that the aircraft is traveling in a direction slightly west of north.
Therefore, the actual direction of the aircraft over the ground is approximately north-west.
Did you draw the diagram?
The direction relative to the ground will be z degrees west of north (or a heading of 360-z) where
tan z = 80/300
Now just find z.