- #1
EaGlE
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A canoe has a velocity of 0.420 m/s southeast relative to the earth. The canoe is on a river that is flowing 0.550 m/s east relative to the earth.
A.) Find the magnitude of the velocity of the canoe relative to the river.
B.) Find the direction of the velocity of the canoe relative to the river. Express your answer as an angle measured south of west.
ok first thing i did was drew a picture, cause i don't get the question.
to find the unknown side(vector Y) i did
0.550^2 + Y^2 = 0.420^2
Y = 0.355m/s
i thought that vector Y was unknown because it gave the southeast side, which is the longest side right? and it also gave the east side, which is vectory X.
and for B.) arctan(0.355/0.550) = 32.84 degrees
well that's all i did so far, i don't even know if either of them is correct. i really doubt that any of them are correct, because it seemed to easy.
A.) Find the magnitude of the velocity of the canoe relative to the river.
B.) Find the direction of the velocity of the canoe relative to the river. Express your answer as an angle measured south of west.
ok first thing i did was drew a picture, cause i don't get the question.
to find the unknown side(vector Y) i did
0.550^2 + Y^2 = 0.420^2
Y = 0.355m/s
i thought that vector Y was unknown because it gave the southeast side, which is the longest side right? and it also gave the east side, which is vectory X.
and for B.) arctan(0.355/0.550) = 32.84 degrees
well that's all i did so far, i don't even know if either of them is correct. i really doubt that any of them are correct, because it seemed to easy.