An aircraft travels through the air with a velocity of 200m/s due east. It is found to be travelling over the ground with a velocity of 220m/s in a direction of 20(degree) south of east.The difference between the two is due to thw wind velocity.
What is the wind velocity?
draw the figure. For wind speed, use the law of cosines:
speed^2=200^2+220^2-2*200*220Cos20
make certain from your diagram you understand that.
Now, the hard part, the direction. Label the diagram, ABC, where A is at the origin, and you go CLOCKWISE around the triangle. You need angle B or C, then convert that to axis angles.
Lets do angle B.
Law of sines:
220/sinB=windspeedabove/sin20
solve for angle B. Now, that is the angle in the triangle. The direction then is, since it is measured against an axis, the direction is 90-B South of East. Check that carefully.
To find the wind velocity, we can use vector addition.
The velocity of the aircraft with respect to the ground can be found by adding the velocity of the aircraft in still air to the wind velocity.
Let's assume the wind velocity is represented by vector W, and the velocity of the aircraft in still air is represented by vector A. The velocity of the aircraft with respect to the ground, represented by vector G, is given as:
G = A + W
We are given the magnitude and direction of the velocity of the aircraft with respect to the ground, which is 220 m/s in a direction 20 degrees south of east. In vector form, this can be represented as:
G = 220 m/s at 20 degrees south of east
To find the wind velocity, we need to find the vector A, the velocity of the aircraft in still air.
We are also given the magnitude and direction of the velocity of the aircraft through the air, which is 200 m/s due east. In vector form, this can be represented as:
A = 200 m/s due east
To find vector W, the wind velocity, we can subtract vector A from vector G:
W = G - A
Now, let's calculate vector W using the given values:
W = 220 m/s at 20 degrees south of east - 200 m/s due east
To subtract the vectors, we need to break them down into their x and y components.
The x component of vector G is given by Gx = G * cos(theta), where theta is the angle south of east. Therefore:
Gx = 220 m/s * cos(20 degrees) = 207.97 m/s
The y component of vector G is given by Gy = G * sin(theta), where theta is the angle south of east. Therefore:
Gy = 220 m/s * sin(20 degrees) = 74.74 m/s
The x component of vector A is given by Ax = A * cos(0 degrees), since it is due east. Therefore:
Ax = 200 m/s * cos(0 degrees) = 200 m/s
The y component of vector A is given by Ay = A * sin(0 degrees), since it is due east. Therefore:
Ay = 200 m/s * sin(0 degrees) = 0 m/s
Now, we can subtract the x and y components to get the x and y components of vector W:
Wx = Gx - Ax = 207.97 m/s - 200 m/s = 7.97 m/s
Wy = Gy - Ay = 74.74 m/s - 0 m/s = 74.74 m/s
Therefore, the wind velocity can be represented as:
W = 7.97 m/s due east + 74.74 m/s due south
So, the wind velocity is approximately 7.97 m/s due east and 74.74 m/s due south.
To find the wind velocity, we need to understand how the aircraft's velocity relative to the ground is affected by the wind. The difference between the aircraft's velocity through the air and its velocity over the ground is due to the wind velocity.
Let's break down the given information:
- The velocity of the aircraft through the air is 200 m/s due east.
- The velocity of the aircraft over the ground is 220 m/s in a direction 20 degrees south of east.
To find the wind velocity, we can use vector addition. We'll first calculate the horizontal component (east-west) of the wind velocity, and then the vertical component (north-south) of the wind velocity. Finally, we can combine these two components to find the magnitude and direction of the wind velocity.
Step 1: Calculate the Horizontal Component of the Wind Velocity
The horizontal component of the aircraft's velocity through the air is 200 m/s due east (positive direction).
Since the aircraft's velocity over the ground has a component in the east direction, it means that the wind is pushing the aircraft in the west direction (opposite to the east movement). The magnitude of the horizontal component of the wind velocity can be calculated as:
Horizontal Component of Wind Velocity = Horizontal Component of Aircraft's Velocity through the Air - Horizontal Component of Aircraft's Velocity over the Ground
Horizontal Component of Aircraft's Velocity through the Air = 200 m/s
Horizontal Component of Aircraft's Velocity over the Ground = 220 m/s * cos(20°)
Step 2: Calculate the Vertical Component of the Wind Velocity
The vertical component of the aircraft's velocity through the air is 0 m/s (since the direction is due east).
The aircraft's velocity over the ground also has a southward component. It means that the wind is pushing the aircraft northward (opposite to the south movement). The magnitude of the vertical component of the wind velocity can be calculated as:
Vertical Component of Wind Velocity = Vertical Component of Aircraft's Velocity through the Air + Vertical Component of Aircraft's Velocity over the Ground
Vertical Component of Aircraft's Velocity through the Air = 0 m/s
Vertical Component of Aircraft's Velocity over the Ground = 220 m/s * sin(20°)
Step 3: Combine the Horizontal and Vertical Components
To find the magnitude and direction of the wind velocity, we can use the Pythagorean theorem and inverse tangent function.
Magnitude of Wind Velocity = √(Horizontal Component of Wind Velocity)^2 + (Vertical Component of Wind Velocity)^2
Direction of Wind Velocity = tan^(-1)(Vertical Component of Wind Velocity / Horizontal Component of Wind Velocity)
Now, let's calculate the wind velocity:
Horizontal Component of Wind Velocity = 200 m/s - 220 m/s * cos(20°)
≈ 200 m/s - 206.99 m/s
≈ -6.99 m/s (opposite to the east direction)
Vertical Component of Wind Velocity = 0 m/s + 220 m/s * sin(20°)
≈ 0 m/s + 74.92 m/s
≈ 74.92 m/s (northward)
Magnitude of Wind Velocity = √(-6.99 m/s)^2 + (74.92 m/s)^2
≈ √48.8601 m^2/s^2 + 5614.6064 m^2/s^2
≈ √5663.4665 m^2/s^2
≈ 75.23 m/s
Direction of Wind Velocity = tan^(-1)(74.92 m/s / (-6.99 m/s))
≈ tan^(-1)(-10.7273)
≈ -81.32°
Therefore, the wind velocity is approximately 75.23 m/s, directed 81.32 degrees west of north.