Find ground speed and direction of airplane

[Grey_Thunderhead]

Homework Statement

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"An airplane is flying on a bearing of 340 degrees at 400mph. A wind is blowing at a bearing of 320 degrees at 30mph. Find ground speed and direction of the plane."

Homework Equations

vx=vcosϕ
vy=vsinϕ

The Attempt at a Solution

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First, my teacher told us to change from bearings to standard positions, so I did that. For the plane, I have that it is flying at 110 degrees. For the wind, I have that it is blowing at 130 degrees.

Knowing that, I calculated ground speed by doing the following:
(400cos110+30cos130)^2 + (400sin110+30sin130)^2
Then I took the square root of that and rounded to get a speed of about 428.214mph.

The issue, then, is finding the direction of the plane. I calculated the components of the plane's movement to be <-136.808,375.877> and the wind to be <-19.284,22.981>.

I then added them together for a total of <-156.092,398.858>.

I took the arctan of (398.858/-156.092) to try to get the angle, and I got -68.627 degrees. Now, I know I'm supposed to do something to this because it's not supposed to be negative, but I'm not sure where to go from here.

Are my calculations correct up to this point? If so, how do I use what I have now to find direction?

Thank you very much!

 

[haruspex]

For the plane, I have that it is flying at 110 degrees. For the wind, I have that it is blowing at 130 degrees.

How are you calculating these? You seem to be subtracting from 450.

An airplane is flying on a bearing of 340 degrees

This is misleading. The question means heading, not bearing.
See http://aviation.stackexchange.com/q...aring-vs-course-vs-direction-vs-heading-vs-tr

 

[Grey_Thunderhead]

How are you calculating these? You seem to be subtracting from 450.

My teacher said bearings are based around north, meaning that the 340 and 320 degrees mentioned in the problem are 340 and 320 degrees clockwise from north on a compass, which is 90 degrees on the standard unit circle. That means I had to convert them to what she calls "standard" position in order to do this problem.

If a vector is at 340 degrees from north (moving clockwise), then it is 20 degrees away from the north position because 360-340 is 20. I then added this 20 degrees to 90 degrees (because north corresponds to 90 degrees on the unit circle) to get 110 degrees. Similar reasoning got me to the 130 degrees for the wind.

I don't know if this is the accepted way to do it, but it's what my precalc teacher has us do. Is there an easier/better way to find these?

Regarding headings vs bearings, I'm afraid I don't know much about aviation. I'm just going off what the question says

 

[haruspex]

First, my teacher told us to change from bearings to standard positions, so I did that. For the plane, I have that it is flying at 110 degrees. For the wind, I have that it is blowing at 130 degrees.

OK. Subtracting from 90, then adding 360 if necessary to make it positive is correct.

I took the arctan of (398.858/-156.092) to try to get the angle, and I got -68.627 degrees.

You need to be careful with inverse trig functions. They introduce an ambiguity. arctan cannot distinguish between an angle and that angle plus 180 degrees, so it could be 180-68.63. You need to consider the physical situation.
In view of the convention you converted to, the angle calculated will be anticlockwise from E. You should probably convert back to the navigational convention.

 

[Grey_Thunderhead]

You need to be careful with inverse trig functions. They introduce an ambiguity. arctan cannot distinguish between an angle and that angle plus 180 degrees, so it could be 180-68.63. You need to consider the physical situation.
In view of the convention you converted to, the angle calculated will be anticlockwise from E. You should probably convert back to the navigational convention.

180-68.63 is 111.37 degrees, which is 338.63 degrees for navigation. Is that the answer the problem is looking for and if so, did I need to convert to "standard" convention at all?

 

[haruspex]

which is 338.63 degrees for navigation

Sounds right.

did I need to convert to "standard" convention at all?

No. The mapping is a reflection about the line x=y, so it would have been easier simply to swap the x and y axes.

 

[Grey_Thunderhead]

Sounds right.

No. The mapping is a reflection about the line x=y, so it would have been easier simply to swap the x and y axes.

Thank you very much for your help!