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anonymous
 3 years ago
PLEASE, SOMEONE HELP ME! THIS IS DUE TOMORROW, AND I'VE BEEN STUCK FOR SEVERAL DAYS!!!! The first stage of a trip has a displacement vector in the negative x direction. The second stage has a displacement vector at an angle of 30 above the positive x axis. The final stage has a displacement 40 degrees below the +x axis at a magnitude of 15km. If after all three stages you end up where you started, what was the magnitudes of the displacement of the first two stages?
anonymous
 3 years ago
PLEASE, SOMEONE HELP ME! THIS IS DUE TOMORROW, AND I'VE BEEN STUCK FOR SEVERAL DAYS!!!! The first stage of a trip has a displacement vector in the negative x direction. The second stage has a displacement vector at an angle of 30 above the positive x axis. The final stage has a displacement 40 degrees below the +x axis at a magnitude of 15km. If after all three stages you end up where you started, what was the magnitudes of the displacement of the first two stages?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This is what I'm thinking: dw:1381891283053:dw I was thinking that d1=a+b and d2 is the sum of vectors b and c. If I could find x/y comp of a and b, then I could add up the comp to find displacement... I need HELP though, IM STUCK!!!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Is there a simpler way???????

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yah! I think this may be late, but use the ordering of the stages to your advantage. You know part 2 has to happen at the end of part 1, etc.. so dw:1381915710776:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0θ = 180 9030 = 60 ψ = 40 φ = 1809040 = 50 (θ+φ) = 110 So you have triangle ABC and then use law of sines to find side lengths (for magnitudes only) dw:1381916712888:dw
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