A plane is flying 510 mph on a bearing of 295°. A wind of 35 mph is blowing at a bearing of 130°. Calculate the ground speed of the plane and its bearing.
I have the following picture that shows my work
I don't understand bearing properly, can anyone explain it a bit clearly to me on how I can go about and solve this problem? I don't see a problem with the diagram, an explanation along with the correction to the diagram would really help
Best Answer
It seems you define "bearing" of an aircraft's path differently than I do. I define it so that the zero bearing points toward the top of the page and 90 degrees points to the right, because that's how it is on aeronautical charts. But that isn't the reason you got the wrong answer; if you consistently measure bearing so that zero bearing is toward the right and 90 degrees is toward the top of the page, then (assuming no other errors) you would get the correct numeric results.
I found at least three errors in your calculations. First, you used $130$ rather than $35$ as the length of the wind vector. You also mishandled the angle you got from the arc tangent function; I'm not quite sure how you got the final result you got, but it looks like you subtracted the result of the arc tangent from $270$ degrees, and when you rounded it to an integer you chose $251$ when the nearest integer to $270 - 19.9$ is actually $250.$ The correct answer (if the wind speed really had been $130$) would have been $289.9$ degrees.
On top of all that, the calculations are so weirdly disorganized that even the grader apparently couldn't figure out what you were doing. The "sign error" they marked ($-215.53$ to $215.53$) was actually not one of the several errors. (At least, what you wrote was equal to the value it was derived from on the line above it.)
Part of the disorganization is having equations for one of the early steps of the problem at the bottom of the workspace, with later steps of the calculation above them, and with stray "$=$" signs to the left of some things where there's no equal thing to the left or above them. You also seem to want to put every vector inside an absolute value, that is, you write $\lvert \vec v\rvert + \lvert \vec u\rvert$ where you intended to take the sum of two vectors, that is, $\vec v + \vec u.$
But what really makes your paper hard to understand, I think, is that you decided to subtract $90$ degrees from all the bearings in the question. So instead of taking $510\cos(295^\circ)$ and $510\sin(295^\circ)$ as the coordinates of your air velocity vector, you take $510\cos(205^\circ)$ and $510\sin(205^\circ).$ If you can deal with trig functions of $205^\circ,$ you can deal with trig functions of $295^\circ,$ and you don't have the extra $90^\circ$ that you have to remember to add back at the end.
Converting a vector from magnitude and direction into rectangular components does not have to be this complicated. If the vector $v$ has magnitude $\lVert v\rVert$ and direction $\psi,$ the component of $v$ in the "zero bearing" direction is $\lVert v\rVert \cos\psi$ and the component in the "$90$ degrees bearing" direction is $\lVert v\rVert \sin\psi.$ The angle $\psi$ is just the direction angle exactly as it was given to you, in this case the bearing.
Also remember when you convert from components back to magnitude and direction, the arc tangent function only has output from $-90$ degrees to $90$ degrees. That's $180$ degrees of coverage. The other $180$ degrees' worth of directions that vectors can go will never be just given to you by the arc tangent function; instead, it will give you the opposite direction, and you have to add or subtract $180$ degrees to make it right. One way to see the correct answer is to plot the components of the vector; a quicker way is, if the "$\cos\theta$" component of the vector is negative, add or subtract $180$ degrees from the arc tangent. The way to remember that is that the cosines of the angles you can get from the arc tangent function are all positive, so if the angle should have a negative cosine, the arc tangent gives you the wrong direction.