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Trigonometry or components to find the displacement

by ScullyX51
Tags: components, vectors
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ScullyX51
#1
Oct4-08, 10:56 PM
P: 36
1. The problem statement, all variables and given/known data:
Bob walks 360 m south, then jogs 490 m southwest, then walks 360 m in a direction 35 degrees east of north.
Use either trigonometry or components to find the displacement that will return Bob to his starting point by the most direct route. Give your answer as a distance.
Express your answer using two significant figures.


3. The attempt at a solution:
I drew a diagram and got the components of each.
360(cos35)(sin35)
490(cos35)(sin35)

For the components I got:
(0,360)
(401.38,281.05)
(294.89,206.49)
Then I added the x's and y's and used pythagoreum theorem, and I got d=1096.87, which isn't correct. What am I doing wrong??
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Rake-MC
#2
Oct4-08, 11:09 PM
P: 328
y components = -360m + -490cos(45) + 360cos(35)
x components = 0 + -490sin(45) + 360sin(35)

Then use pythagoras.
Make sure you get positives and negatives correct depending on which way he's travelling.

Why did you do 360(cos35)(sin35) ? and 490(cos35)(sin35) ?
You need to imagine each trajectory as a right angled triangle with the two short sides along the North/South and East/West planes and the hypotenuse being his travel path.
You know all of the angles ie. 35 degrees and SouthWest = 45 degrees west of south.
ScullyX51
#3
Oct5-08, 07:56 AM
P: 36
Thank you that worked! I was confused about what to use for the components of each vector. I did that because I thought to get the components I would multiply the length times the x and y components of each. (the sin and cosine). I don't know, I am really lost with this chapter.


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