Vector Addition Help (given heading and distance)

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tcco94
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Homework Statement


To avoid restricted airspace, a pilot on a VFR cross country flight plans the following 3 legs immediately after takeoff.

Leg A (280 heading 40nm distance)
Leg B (240 heading 30nm distance)
Leg C (200 heading 25nm)

What one heading and distance from takeoff would have taken him to the same position of these 3 legs?

Homework Equations


Pythagoreum Theorem and sin theta, cos theta, tan theta

The Attempt at a Solution


I have been struggling drastically to even figure out how to draw these triangles out. I have a quiz on it tomorrow for an introductory college level physics.

The answer comes out to heading 247 degrees and 80 nm but I cannot figure out how to get that answer at all.

Thank you guys so much for all the help. New here and looking forward to it.
 
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tcco94 said:

Homework Statement


To avoid restricted airspace, a pilot on a VFR cross country flight plans the following 3 legs immediately after takeoff.

Leg A (280 heading 40nm distance)
Leg B (240 heading 30nm distance)
Leg C (200 heading 25nm)

What one heading and distance from takeoff would have taken him to the same position of these 3 legs?

Homework Equations


Pythagoreum Theorem and sin theta, cos theta, tan theta

The Attempt at a Solution


I have been struggling drastically to even figure out how to draw these triangles out. I have a quiz on it tomorrow for an introductory college level physics.

The answer comes out to heading 247 degrees and 80 nm but I cannot figure out how to get that answer at all.

Thank you guys so much for all the help. New here and looking forward to it.

It's the Pythagorean Theorem.

Regardless of the method of solution, making a quick sketch is a good way to examine the heading/distance data all at once.

BTW: If you plot Legs A, B, and C, you may find you are not dealing with one triangle, or even any triangles. :wink:
 
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I'm just supposed to solve it by algebra. Professor isn't picky about the graphs unless it helps but at this point I'd just like to know how to set up the algebra to get the answer. I wasn't sure if it was triangles or not because some vectors are added by doing triangles and some aren't? I don't know what algebra equation to use other than Pythagorean Theorem...(thank you)...

If I try to draw it would I have three different lines or do the lines stay connected to each other? As a pilot, it sounds like he is plotting the courses to avoid the TFR so there should just be three different lines all in the negative X and Y axis with distances 40, 30, 25 but I am struggling to even figure out where to start from here? That's why I wasn't able to post an attempt to finishing the equation (which I'm terribly sorry that I haven't) because I am lost as to how to start it to get that answer...

Thank you guys very much for the help.
 
tcco94 said:
I'm just supposed to solve it by algebra. Professor isn't picky about the graphs unless it helps but at this point I'd just like to know how to set up the algebra to get the answer. I wasn't sure if it was triangles or not because some vectors are added by doing triangles and some aren't? I don't know what algebra equation to use other than Pythagorean Theorem...(thank you)...

If I try to draw it would I have three different lines or do the lines stay connected to each other? As a pilot, it sounds like he is plotting the courses to avoid the TFR so there should just be three different lines all in the negative X and Y axis with distances 40, 30, 25 but I am struggling to even figure out where to start from here? That's why I wasn't able to post an attempt to finishing the equation (which I'm terribly sorry that I haven't) because I am lost as to how to start it to get that answer...

Thank you guys very much for the help.
Adding vectors algebraically when you know the direction and length of each is quite straightforward.
Pick a Cartesian co-ordinate system, East on the positive X axis and North on the positive Y axis, say.
For each vector, resolve it into its X component (with sign) and Y component. Do you know how to do that?
Add up the X components, add up the Y components.
 
tcco94 said:
I'm just supposed to solve it by algebra. Professor isn't picky about the graphs unless it helps but at this point I'd just like to know how to set up the algebra to get the answer. I wasn't sure if it was triangles or not because some vectors are added by doing triangles and some aren't? I don't know what algebra equation to use other than Pythagorean Theorem...(thank you)...
It's perfectly acceptable to draw a sketch to help you organize your thoughts and put together the right algebraic equations to solve. You don't have to show your professor the sketch if you don't want to. :smile:

If I try to draw it would I have three different lines or do the lines stay connected to each other? As a pilot, it sounds like he is plotting the courses to avoid the TFR so there should just be three different lines all in the negative X and Y axis with distances 40, 30, 25 but I am struggling to even figure out where to start from here? That's why I wasn't able to post an attempt to finishing the equation (which I'm terribly sorry that I haven't) because I am lost as to how to start it to get that answer...

Thank you guys very much for the help.

Keep things simple. On your sketch, draw an x-y plane. Assume that the Origin of the x-y plane is the point from which the pilot takes off. Now, he's flying cross country, so Leg A starts at the take-off point. Once you have drawn Leg A, draw Leg B starting at the end of Leg A opposite the starting point, draw Leg C from the end of Leg B, and so on. The pilot must fly each leg in turn as he tries to navigate around the restricted air space.

Sometimes, the most important factor in obtaining the solution to a problem is not to get in a hurry: Relax and think about the situation being discussed, drawing sketches or whatever you need to help you visualize what is being described. Once you have your thoughts organized, then you can start throwing math at the problem to develop a solution.