Air Navigation and Course Correction. Semester exam in 45 minutes.

[pablot93]

Homework Statement

A pilot wishes to fly 2875 km to an airport 18 degrees east of north from his departure point(for the sake of the problem and simplicity, use the origin). He wants to make the trip in 3 hours and 20 minutes. The pilot finds that he actually has to make a change in his air heading to compensate for a 130 km/hr wind coming from due south. What should be his...
A)Ground Speed
B)Air Speed
C)Magnetic Compass heading

Homework Equations

Cosine Law
c^2 = a^2 + b^2 - 2ab cosC

The Attempt at a Solution

I know you make a parallelogram, with his ground speed and the wind speed, and then use the cosine law. From there, use the sine law. I just need someone to walk me through it please.
I have my physics exam in an hour and I'm still kind of iffy on these types of problem from the beginning of the semester.

 

[Feldoh]

A good place to start would be to see how much of the distance the wind speed would propel the plane in 3 and 1/3 hours.

Whatever is left is what the plane's velocity must compensate for, but remember that the wind is only going north which means you're going to need to break the distance into components.

 

[kerimek]

As a scientist, it is important to approach this problem systematically and use the appropriate equations to find the solution. Let's break down the steps to solve this problem:

1. First, let's draw a diagram to represent the situation. The pilot's departure point (origin) is at the center, and the airport he wants to reach is located 2875 km away in the direction of 18 degrees east of north. We can label this direction as the desired course or true heading (TH).

2. Next, we need to take into account the 130 km/hr wind coming from due south. This means that the wind is blowing in the direction of 180 degrees, or south. We can label this direction as the wind direction or wind heading (WH).

3. To determine the pilot's ground speed (GS), we need to find the resultant of his air speed (AS) and the wind speed (WS). We can use the parallelogram method to find the resultant, which will give us the magnitude and direction of the GS. We can label the magnitude of the resultant as GR and the direction as ground heading (GH).

4. Now, we can use the cosine law to find the magnitude of the resultant (GR). The cosine law states that c^2 = a^2 + b^2 - 2ab cosC, where c is the side opposite to angle C. In our case, GR is the side opposite to the angle between the AS and WS, which we can label as angle A. So, our equation becomes GR^2 = AS^2 + WS^2 - 2(AS)(WS)cosA. We know the values of AS and WS, so we can solve for GR.

5. To find the direction of the GH, we can use the sine law. The sine law states that a/sinA = b/sinB = c/sinC, where a, b, and c are the sides of a triangle and A, B, and C are the opposite angles. In our case, we know the values of GR and the angle opposite to GH, which we can label as angle B. So, our equation becomes GR/sinB = WS/sinA. We can solve for angle B, which will give us the direction of the GH.

6. Now, we can use the triangle formed by the AS, WS, and GH to find the pilot's air heading (AH).