Question

A pilot wants to fly on a bearing of 60.8. By flying due east he finds that a 59 mph wind blowing from the south puts him on course. Find ground speed of the plane

Answer 1

68

Explanation:

We can use vector addition to solve this problem. Let's consider two vectors: the velocity of the plane relative to the ground, which we want to find, and the velocity of the wind relative to the ground, which we know is due south with a magnitude of 59 mph.

Let's call the ground speed of the plane "v" and the angle between the plane's velocity and the due east direction "θ". The bearing of 60.8 is the angle between the plane's velocity and due north, so the angle between the plane's velocity and due east is (90 - 60.8) = 29.2 degrees.

Now we can use trigonometry to find the components of the plane's velocity relative to the ground:

v(cosθ, sinθ) = (v cosθ, v sinθ)

The plane's velocity relative to the wind is due east, so its x-component is v cosθ = 59 mph. Therefore, we can solve for v:

v = 59 / cosθ

v = 59 / cos(29.2) ≈ 67.55 mph

So the ground speed of the plane is approximately 67.55 mph. Round it up to 68.