In still air, an airplane is flying at 100 mph. It encounters a wind blowing toward the west at 50 mph. What should be the plane's compass heading for its course to be N30ºW?

Question

In still air, an airplane is flying at 100 mph. It encounters a wind blowing toward the west at 50 mph. What should be the plane's compass heading for its course to be N30ºW?

Answer 1

If the plane is flying on heading NθE, then we want

100sinθ-50 = -v/2
100cosθ = v √3/2

100sinθ = 50-v/2
100cosθ = v√3/2

(50-v/2)^2 + 3/4 v^2 = 100^2
v = 115.14

So, cosθ = .997
θ = 4.3º

and the heading is N4.3ºE

Answer 2

Could you explain the (50-v/2)^2 + 3/4v^2 = 100^2 part? I'm a little confused.

Answer 3

square the previous equations and add them up. Recall that

sin^2θ + cos^2θ = 1

Answer 4

To determine the plane's compass heading, we need to consider the combined effect of the airplane's speed and the wind's speed.

First, let's break down the given course, N30ºW. This means the plane should be heading 30 degrees west of north.

Next, we need to find the resultant velocity, which is the vector sum of the airplane's velocity and the wind's velocity. Since the wind is blowing toward the west, its velocity will be negative.

Given that the airplane is flying at 100 mph (let's assume this is the plane's airspeed or its speed in the absence of wind) and the wind is blowing at 50 mph toward the west, we can set up a coordinate system with east as the positive x-axis and north as the positive y-axis.

The airplane's velocity can be represented as 100 mph along the positive x-axis (east), and the wind's velocity can be represented as -50 mph along the negative x-axis (west).

By adding these vectors, we get the resultant velocity vector which represents the plane's velocity through the air.

To find the compass heading, we need to determine the angle between the resultant velocity vector and the positive y-axis (north).

Using trigonometry, we can find this angle by applying the inverse tangent function (arctan) to the y-component of the resultant velocity vector divided by the x-component of the resultant velocity vector.

Let's calculate:

x-component of resultant velocity = airplane velocity + wind velocity
x-component = 100 mph + (-50 mph) = 50 mph to the east (positive x-axis)

y-component of resultant velocity = 0 mph (since the wind is blowing only in the x-axis)
y-component = 0 mph

Now, finding the angle:

angle = arctan(y-component / x-component)
angle = arctan(0 mph / 50 mph)
angle = arctan(0)
angle = 0 degrees

Therefore, the compass heading for the plane to maintain a course of N30ºW in these conditions is 0 degrees or due north.