A plane is flying 250 m/s west with a crosswind of 25 m/s north affecting the flight. What is the magnitude and direction of the plane?
To calculate the magnitude and direction of the plane, we can use vector addition.
First, let's break down the velocity of the plane and the velocity of the crosswind into their horizontal (west/east) and vertical (north/south) components.
The plane's velocity can be broken down as follows:
- Horizontal component: 250 m/s west
- Vertical component: 0 m/s (since the plane is not moving up or down)
The crosswind's velocity can be broken down as follows:
- Horizontal component: 0 m/s (since the crosswind is affecting the plane's north/south movement, not east/west)
- Vertical component: 25 m/s north
Now, let's add the horizontal and vertical components separately.
For the horizontal direction:
The plane's horizontal component is 250 m/s west, and since the crosswind does not affect the plane's east/west movement, its horizontal component remains 0 m/s. Therefore, the total horizontal velocity of the plane is 250 m/s west.
For the vertical direction:
The plane's vertical component is 0 m/s, and the crosswind's vertical component is 25 m/s north. To calculate the total vertical velocity, we add these two components together: 0 m/s + 25 m/s = 25 m/s north.
Finally, we can use the Pythagorean theorem to find the magnitude of the resulting velocity. The magnitude is calculated as the square root of the sum of the squares of the horizontal and vertical components:
Magnitude = √(horizontal component^2 + vertical component^2)
Magnitude = √(250^2 + 25^2) = √(62500 + 625) = √63125 ≈ 251.24 m/s
So, the magnitude of the plane's resulting velocity is approximately 251.24 m/s.
To determine the direction, we can use trigonometry. The direction is generally given in terms of an angle east or west of north. In this case, the plane is flying west and the crosswind is affecting its northward movement.
To calculate the direction angle (θ), we can use the tangent function:
θ = arctan(vertical component / horizontal component)
θ = arctan(25 m/s / 250 m/s) = arctan(0.1) ≈ 5.71 degrees
Since the plane is flying west with a crosswind from the north, the resulting direction angle is 5.71 degrees west of north.
Therefore, the magnitude of the plane is approximately 251.24 m/s, and its direction is approximately 5.71 degrees west of north.
Vp = -250m/s + i25m/s, Q2.
Tan A = Y/X = 25/-250 = -0.10000
A = -5.71o = 5.71o N. of W.= 174.3o,CCW
Vp = X/Cos A = -250/Cos(174.3)=251 m/s