A plane intends to north with a speed of 250 m/s relative to the ground through a high altitude cross wind of 50 m/s coming from the east. Determine the bearing that the plane should take( relative to due north),
A) the planes speed with respect to the air answer
The plane takes a heading, not a bearing.
So, draw the velocity triangle. You should be able to see that
speed: s = √(250^2+50^2)
heading: tanθ = 50/250
To determine the plane's speed with respect to the air, we need to use vector addition. We can decompose the plane's velocity into two components: one component in the north direction (plane's intended direction) and one component in the east direction (crosswind).
Given:
Speed of the plane relative to the ground = 250 m/s
Speed of the crosswind = 50 m/s
Using Pythagoras' theorem, we can calculate the magnitude of the plane's velocity with respect to the air:
Magnitude of the resultant velocity = √(250^2 + 50^2)
= √(62500 + 2500)
= √65000
≈ 254.95 m/s
Therefore, the magnitude of the plane's speed with respect to the air is approximately 254.95 m/s.
To determine the plane's speed with respect to the air, we need to use vector addition. We have two velocities: the velocity of the plane relative to the ground (250 m/s) and the velocity of the crosswind (50 m/s).
The velocity of the plane with respect to the air is the vector sum of these two velocities. We can use the Pythagorean theorem to find the magnitude of this resultant velocity:
Magnitude of resultant velocity = √(250^2 + 50^2)
= √(62500 + 2500)
= √65000
≈ 254.95 m/s
Therefore, the plane's speed with respect to the air is approximately 254.95 m/s.