An aifplane is flying on a compass heading (bearing) of 170* at 460 mph. A wind is blowing with the bearing 200* at 80mph. a) Find the component form of the velocity of the airplane. b) Find the actual ground speed and direction of the airplane.
An airplane flying at 550 mph has a bearing of N 58 E. After flying for 1.5 hours, how far north and how far east has the plane traveled from its point of departure?
d = 550mi/h[N58oE] * 1.5h = 825mi[N58oE]825*Cos58 = 437 mi. N.
825*sin58 = 700 mi E.
To solve this problem, we will use vector addition to find the component form of the velocity of the airplane and then use trigonometry to find the ground speed and direction.
a) To find the component form of the velocity of the airplane, we need to consider the effect of the wind on the airplane's velocity. We can break down the velocity of the airplane into two components: one component in the direction of the wind and one component perpendicular to the wind.
Let's assume the x-axis points east and the y-axis points north. The given compass heading of the airplane is 170°, which means the angle formed with the positive x-axis is 170°. Therefore, the component of the airplane's velocity in the x-direction (u_x) can be found using cosine:
u_x = magnitude of velocity * cos(angle)
= 460 mph * cos(170°)
Similarly, the component of the airplane's velocity in the y-direction (u_y) can be found using sine:
u_y = magnitude of velocity * sin(angle)
= 460 mph * sin(170°)
b) To find the actual ground speed and direction of the airplane, we need to combine the effect of the wind and the airplane's velocity.
The wind is blowing with a bearing of 200°, which forms an angle of 200° with the positive x-axis. Let's call the wind velocity vector as (v_x, v_y). Using the given values, we can calculate the components of the wind velocity:
v_x = magnitude of wind velocity * cos(angle)
= 80 mph * cos(200°)
v_y = magnitude of wind velocity * sin(angle)
= 80 mph * sin(200°)
Now, we can find the total velocity of the airplane by adding the velocity of the airplane (u_x, u_y) and the velocity of the wind (v_x, v_y):
Velocity of the airplane (total) = (u_x + v_x, u_y + v_y)
Finally, to find the ground speed and direction, we calculate the magnitude and angle formed by the total velocity vector.
Ground speed = magnitude of the total velocity vector
Direction = angle formed by the total velocity vector with the positive x-axis
By following these steps, you can find the component form of the velocity of the airplane (a) and the actual ground speed and direction of the airplane (b) using vector addition and trigonometry.
The way you have worded the problem, with the word "bearing",
460 mph at bearing 460 mph is the actual ground speed and direction of the airplane.
I suspect that they meant 460 mph at 170 to be the air speed and heading of the place, and that they want you to solve for the ground speed and direction by adding the wind velocity vector.
In that case:
ground speed components are:
Vy (north component) = 460 cos 170 + 80 cos 200 = -528.19 mph
Vx(east component) = 460 sin170 + 80 sin200 = 52.52 mph
Ground speed = 530.79 mph
bearing = arctan Vx/Vy = 174.2 degrees