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A canoe has a velocity of 0.550m/s southeast relative to the earth. The canoe is on a river that is flowing at 0.540m/s east relative to the earth. enter image description here

Alright so I attempted many different things an still do not understand the nature of the question nor do I understand how to get an answer. Find the magnitude of Vc/r of the canoe relative to the river now the next question Find the direction of the velocity of the canoe relative to river expressed as an angle measured south of west. Now I figured that the angle would be equal to theta if you laid our the vectors tail to tip with -Vr/e + Vc/e however I have no earthly idea on how to calculate this other than the equation: arcsin(Ay/A with the resulting vector of -Vr/e + Vc/e being defined as A and Ay being the y component of A but im not sure how to calculate the y component of A. If this question doesn't make much sense I apologize I am struggling with the concept as a whole.

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closed as off-topic by John Rennie, Dimensio1n0, Dilaton, jinawee, Kyle Kanos Mar 2 '14 at 16:22



1 Answer 1

up vote 1 down vote accepted

The canoe has a relative velocity to the river which we simply call $\vec{v}_{c/r} =( \dot{x}, \dot{y} )$. The motion of the river relative to the earth is $\vec{v}_{r/e} = ( 0.54, 0 )$. The canoe relative to the earth is $\vec{v}_{c/e} = ( 0.55\cos(-45^\circ), 0.55\sin(-45^\circ) ) $.

All together you have

$$ \vec{v}_{c/e} = \vec{v}_{c/r} + \vec{v}_{r/e} $$

which is used the find the speed of the canoe relative to the river

$$ \vec{v}_{c/r} = \vec{v}_{c/e} - \vec{v}_{r/e} = ( 0.389, -0.389 ) - ( 0.54, 0 ) = ( -0.151, -0.389 ) $$

So the speed of the boat is $||( -0.151, -0.389 )|| = 0.417$ and the direction relative to east ${\rm atan2}(-0.389, -0.151) = -111.2^\circ$.

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Thank you for your answer and for walking e through the problem. –  user41607 Mar 2 '14 at 4:09
It would be nice if you didn't give the solution, just the key steps. –  jinawee Mar 2 '14 at 17:15
Yes, but I get a sense the OP was confused on how to use $\arctan$ and so I wanted to demonstrate it. I consider it a learning opportunity since they now understand the details of such problems. –  ja72 Mar 2 '14 at 17:26
