# Vectors and Components-Part Three

1. Jan 5, 2014

### Medgirl314

1. The problem statement, all variables and given/known data
An aircraft travels 250 km on a heading of 330 degrees. It then turns dues east and travels 420 km. How far and on what heading did it end up from its starting point?

2. Relevant equations
Sine/cosine/tangent equations
Pythagorean theorem

3. The attempt at a solution
I have converted the 330 heading into our mathematically calculated angle of 30 degrees. I know it sounds silly, but I am stuck trying to draw a diagram. What direction was the aircraft heading initially? It would seem that it was heading Northwest(I finally figured that as I typed) but does converting it to the 30 degree angle change the direction?

2. Jan 5, 2014

### Staff: Mentor

O degrees is North, 90 is East, 180 South and 270 West so you're right its North North West

It makes a 30 degree angle with the North direction (think 11 oclock on a clock with hands ie analog / traditional clock)

Last edited: Jan 5, 2014
3. Jan 5, 2014

### Medgirl314

Okay, thanks! I get that part, but I don't understand how the direction can change just because I converted the angle. Could you please help me with understanding that part? Is it just because mathematicians measure counterclockwise, while headings are measure clockwise?

4. Jan 5, 2014

### Medgirl314

Oh, you edited your post, correct? Oops, didn't see that part. So after converting to standard, would it be going North?

5. Jan 5, 2014

### Staff: Mentor

Headings were used by sailors and pilots:

Your 30 degree triangle is not changed the hypotenuse is along the 11 o'clock direction and one side is along the 12 o'clock (north) direction.

If you use 30 degrees as a heading then you're using the 1 o'clock direction.

Last edited by a moderator: May 6, 2017
6. Jan 5, 2014

### Medgirl314

Thank you for all the information! So the main point is that the direction is still WNW even though the angle changed when I converted?

Thanks again!

7. Jan 5, 2014

### Staff: Mentor

Make that NNW since NNW= 330 and NW=315 and WNW= 300 and W=270 ...

and yes to your question the direction is still NNW for 250km and then travel E for 420km.

So what answer did you get?

8. Jan 6, 2014

### Medgirl314

I haven't calculated the heading yet, but would it end up about 337 km from its starting point?

9. Jan 6, 2014

### Staff: Mentor

I get a distance a little less than 10% larger. Show your calculations.

10. Jan 6, 2014

### Medgirl314

Sorry, I was on mobile and I haven't figured it out completely yet.

250km^2+y^2=420^2
176400km-62500km=y^2
113900=y^2
y= 337 km. I rounded, the answer on the calculator is 337.4907406.

11. Jan 6, 2014

### Staff: Mentor

Okay you probably need to show your diagram because the 250^2 + y^2 = 420^2 is clearly not right.

Think about a related problem, you travel North 250km and then east 420km so the distance is 250^2 + 420^2 = dist^2

Now for your problem, you travel NNW 250km and then east 420km you have to travel some km to get to the north vertical line and then some more to get to the final destination point so your answer should be something like:

ydist^2 + (420 - xdist)^2 = (final dist)^2

where xdist and ydist come from traveling NNW 250km

Does that make sense?

12. Jan 6, 2014

### Medgirl314

Ah, I think I see it. My hypotenuse was wrong, correct? Now I'm getting about 489 km, but that seems too large. If you don't mind, I'll go ahead and post the diagram in the morning. It's getting late here, and I really must be up on time in the morning. Thanks again for helping! I hope you don't mind continuing tomorrow! :)

13. Jan 6, 2014

### Medgirl314

Oh, sorry, now I see that answer isn't right either. I will take a better look at your explanation and post my diagram when my brain is at optimal function. ;) Thanks again!

14. Jan 6, 2014

### lendav_rott

Did you also account for the fact that the surface of the Earth is not flat?

15. Jan 6, 2014

### Medgirl314

No, I don't think my physics teacher wants to get that complicated. Thanks for catching that though!

16. Jan 6, 2014

### Medgirl314

Here we are! I often find the diagram to be the hardest part of the problem, so critique to your heart's content! :)

17. Jan 6, 2014

### Staff: Mentor

This is not the diagram that I expected to see. I expected something like this:

Code (Text):

^ north
|
|        420km
nnw  -------------------->  destination  (east direction)
\    |            .
\   |        .
\  |     .
\ |  .
\|
origin

The angle between North and NNW is 30 degrees and if you drew this to scale with 250km as 2.5 inches and 420km as 4.2 inches then you'd be able to measure the origin to dest distance and its resultant heading angle too.

Last edited: Jan 6, 2014
18. Jan 8, 2014

### Medgirl314

Okay, so sorry for the delay! I had a hectic couple days and then managed to lock myself out of my account because I forgot I had to reset my password. XD

Could you please explain lendav_rott's question? I'm confused because if we accounted for the earth not being flat every time we answered a problem, we would need an entire new set of equations for everything, wouldn't we?? Anyway, I'd like to try just the distance part first if you don't mind, so we can confirm I got that before I move on.My other problems were set up slightly differently, no wonder I was stuck trying the same methods on this. Thanks for explaining, let me try using your formula.

ydist^2 + (420 - xdist)^2 = (final dist)^2
420^2+(420-250)^2=final^2
=453 km.

That still seems wrong, as that is much more than 10% larger. There's something I'm not getting, but I'm not sure what it is. The 420-250 seems to make sense, but I'm fairly confident the rest of the numbers I put in are in the wrong order. Is it that I assumed I knew the xdist and ydist, but I actually knew one of those and the final distance? I'm not sure why this seems so much more complicated, I've done several problems like it fairly easily. Thank you for being so explanatory and patient!

19. Jan 8, 2014

### haruspex

Since you are not given a location on the earth you cannot adjust for the curvature. Anyway, the distances here are too small for that to matter unless you are quite close to a pole.
I don't understand how you get that equation from jedishfru's diagram (with which I presume you agree). How about we label some points: O = origin, A = point where course changes, B = point where we are due N of origin, C = final point. What right angled triangles do you have ?

20. Jan 8, 2014

### Staff: Mentor

I don't think the curvature of the earth needs to be considered for this problem.

Do you see the right triangle with the hypotenuse of 250 km and one side is the north direction?

Its a 30/60/90 triangle and based on that you can compute the xdist and ydist:

sin(30) = xdist / 250 and cos(30) = ydist / 250

Do you now see where you went wrong?