Vector A, a wind velocity vector, has a magnitude of 20 miles per hour and is blowing toward the North. A second wind velocity vector, Vector B, has a magnitude of 50 miles per hour and is blowing toward the Southeast. A third wind velocity vector, Vector C, has a magnitude of 10 miles per hour and is blowing toward the Southwest These three winds emerge (come out of) three separate canyons and merge (meet) at a flagpole carrying a cloth flag. What direction (angle with respect to zero degrees) will the flag be pointing as a result of the three winds? Also, what is the magnitude of the resulting wind velocity vector (expressed in m/s)? Note: Convert all variables to the metric (mks) system before performing your calculations.

Question

Vector A, a wind velocity vector, has a magnitude of 20 miles per hour and is blowing toward the North. A second wind velocity vector, Vector B, has a magnitude of 50 miles per hour and is blowing toward the Southeast. A third wind velocity vector, Vector C, has a magnitude of 10 miles per hour and is blowing toward the Southwest These three winds emerge (come out of) three separate canyons and merge (meet) at a flagpole carrying a cloth flag. What direction (angle with respect to zero degrees) will the flag be pointing as a result of the three winds? Also, what is the magnitude of the resulting wind velocity vector (expressed in m/s)? Note: Convert all variables to the metric (mks) system before performing your calculations.

Answer 1

add the vectors:

Using N as direction 000
R=20Cos0 N + 50cos135N+50sin135E+10Cos225N+10sin225(-E)

now add the N components, and the east components.

change miles/hr to m/s, I would do it at the end.

Answer 2

To find the resulting direction and magnitude of the wind velocity vector at the flagpole, we can use vector addition.

First, let's convert the given magnitudes from miles per hour to meters per second, as requested:

Magnitude of Vector A: 20 miles per hour = 20 * 1609.34 meters / (3600 seconds) = 8.94 m/s (approximately)
Magnitude of Vector B: 50 miles per hour = 50 * 1609.34 meters / (3600 seconds) = 22.35 m/s (approximately)
Magnitude of Vector C: 10 miles per hour = 10 * 1609.34 meters / (3600 seconds) = 4.47 m/s (approximately)

Now, let's break down the given wind velocity vectors into their horizontal (x) and vertical (y) components.

Vector A: Blowing North
The direction of Vector A is straight up along the positive y-axis. Therefore, its x-component is 0, and its y-component is 8.94 m/s.

Vector B: Blowing Southeast
To find the components of Vector B, we need to calculate the horizontal and vertical velocities.

Vertical component (along the y-axis):
Using the given magnitude (22.35 m/s), we can find the vertical component:
Vertical component = Magnitude * sin(45 degrees) = 22.35 * sin(45) = 15.84 m/s (approximately)

Horizontal component (along the x-axis):
Using the given magnitude (22.35 m/s), we can find the horizontal component:
Horizontal component = Magnitude * cos(45 degrees) = 22.35 * cos(45) = 15.84 m/s (approximately)

Therefore, the components of Vector B are:
x-component: 15.84 m/s (eastwards)
y-component: 15.84 m/s (northwards)

Vector C: Blowing Southwest
To find the components of Vector C, we need to calculate the horizontal and vertical velocities.

Vertical component (along the y-axis):
Using the given magnitude (4.47 m/s), we can find the vertical component:
Vertical component = Magnitude * sin(45 degrees) = 4.47 * sin(45) = 3.16 m/s (approximately)

Horizontal component (along the x-axis):
Using the given magnitude (4.47 m/s), we can find the horizontal component:
Horizontal component = Magnitude * cos(45 degrees) = 4.47 * cos(45) = 3.16 m/s (approximately)

Therefore, the components of Vector C are:
x-component: -3.16 m/s (westwards)
y-component: 3.16 m/s (northwards)

Now, we can add the x-components and the y-components separately to find the resultant components.

Resultant x-component = 0 (from Vector A) + 15.84 m/s (from Vector B) + (-3.16 m/s) (from Vector C)
= 12.68 m/s (approximately, towards the east)

Resultant y-component = 8.94 m/s (from Vector A) + 15.84 m/s (from Vector B) + 3.16 m/s (from Vector C)
= 28.94 m/s (approximately, towards the north)

To find the magnitude of the resultant vector, we use the Pythagorean theorem:

Magnitude of Resultant Vector = sqrt((Resultant x-component)^2 + (Resultant y-component)^2)
= sqrt((12.68 m/s)^2 + (28.94 m/s)^2)
= sqrt(160.6624 + 837.2836)
= sqrt(997.946)
= 31.61 m/s (approximately)

Finally, to find the direction (angle with respect to zero degrees) of the resultant vector, we can use trigonometry:

theta = atan(Resultant y-component / Resultant x-component)
= atan(28.94 m/s / 12.68 m/s)
= atan(2.280 )

Using a calculator or reference table, we find:
theta ≈ 63.49 degrees

Therefore, the flag will be pointing at an angle of approximately 63.49 degrees with respect to zero degrees, and the magnitude of the resulting wind velocity vector is approximately 31.61 m/s.