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L
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Joined: 12 Sep 2008
Posts: 29
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 Posted: Mon, 7 Sep 2009 02:37:08 UTC Post subject: Vector Displacement/Direction Problem |
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<i>In the morning you travel due north for 17 km and rest for lunch. After lunch, you travel 15 km in a direction 37* south of east and rest for dinner. After dinner, you take a short walk 4 km due west and set up camp. Find your total displacement (magnitude and direction) for the day.</i>
I'm having a lot of trouble understanding this type of problem. I feel like I'm just "winging it" even when I get the correct answer. Could someone lend me a hand here and let me know if I'm doing this right?
Here's what I got:
x (east/west)= 12, -4
y (north/south) = 17, -9
I was told to first list out what I know under x/y (17 and 4 in this case) and then solve for what I don't know like so: x = 15km(cos37*) =12 km y = 15km(sin37*) = 9 km. This is where I start to get confused - how do I know that for x(east/west) I use cos and for y(north/south) I use sin? I've just been going on x=cos/y=sin but is that even correct?
Also here; how can I tell what's negative and what's positive? I made 9 and 4 negative because they point in the opposite direction from the first two vectors (north/west).
In following that reasoning I did 12 + (-4) = 8 and 17 + (-9) = 8 so 8^2 + 8^2 = 11 km for the total displacement.
Now for the angle and again I get more confused. How do I know which is opposite side and which is the adjacent side? In this case it doesn't really matter since I just did 8/8 = 1 tan^-1 = 45* south of east (and again, this direction is pretty much a guess - how do I know which way it points?!)
Ugh. Sorry if that was a little hard to read there, but I feel rather lost. I've tried sketching the directions but it hasn't helped. My textbook only gives one example and it sucks. Help please? |
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Mage
Member of the 'S.O.S. Math' Hall of Fame
Joined: 06 Nov 2005
Posts: 380
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| Posted: Thu, 5 Nov 2009 05:10:17 UTC Post subject: |
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Hi!
Let me see if I can help with this:
| Quote: | | This is where I start to get confused - how do I know that for x(east/west) I use cos and for y(north/south) I use sin? I've just been going on x=cos/y=sin but is that even correct? |
This comes straight from trigonometry.
Consider the triangle:
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You are given the length of the hypotenuse of the triangle (that is your vector's length). You also know an angle, x. With that, you should be able to derive a relationship to find the horizontal and vertical component of the vector.
If you're still not sure about this, let me know, and I'll explain a bit more.
| Quote: | | Also here; how can I tell what's negative and what's positive? I made 9 and 4 negative because they point in the opposite direction from the first two vectors (north/west). |
You have the right idea here. In essence: Pick a direction that you set as positive, and stick with it.
In general, upwards and rightwards (is that even a word?) is normally set as positive (as it follows a standard cartestian axis).
| Quote: | | Now for the angle and again I get more confused. How do I know which is opposite side and which is the adjacent side? |
Standard trig here again. You are given an angle, and that is your 'reference point' in some sense. The side that touches this angle that is not the hyptenuse is the adjacent. The one opposite that angle is the opposite.
_________________
Double Major: Chemical Engineering and Chemistry
Minor:Mathematics
@Queens University in Kingston, Ontario |
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