A light plane is flying at 25 degrees east of due magnetic north, while ascending at 15 degrees above the horizon; the plane is flying at 40 m/s. Find the planes celocity true to the delta x, delta y, and delta z?
To find the components of the plane's velocity in the x, y, and z directions, we can break it down into its horizontal (x, y) and vertical (z) components.
Let's start by finding the horizontal components (delta x and delta y).
- Delta x:
The plane is flying at 25 degrees east of due magnetic north, which means it has a 25 degrees angle with respect to the y-axis. We can use trigonometry to find the delta x component.
delta x = velocity * cos(angle)
delta x = 40 m/s * cos(25°)
- Delta y:
Similarly, the plane's angle with respect to the x-axis is 90 degrees minus the angle of 25 degrees. So, the angle is 65 degrees.
delta y = velocity * cos(angle)
delta y = 40 m/s * cos(65°)
Now let's find the vertical component (delta z).
- Delta z:
The plane is ascending at 15 degrees above the horizon, which means it has a 15 degrees angle with respect to the x-y plane.
delta z = velocity * sin(angle)
delta z = 40 m/s * sin(15°)
By calculating these values, you will get the values for delta x, delta y, and delta z, which represent the components of the plane's velocity.