Relative Velocity of a swimmer

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Homework Statement



In an anniversary celebration of Marilyn Bell's 1954 crossing of Lake Ontario a swimmer set out from the shores of New York and maintained a velocity of 4m/s [N]. As the swimmer approached the Ontario shore, she encountered a cross current of 2m/s [E 25deg S]. Find her velocity with respect to the crowd observing from the beach.

The Attempt at a Solution



Firstly, am I to understand that this is a right-angle triangle? When representing it graphically, it certainly does not look like a right triangle:

http://i543.photobucket.com/albums/gg464/yowatupguystill/vector.jpg [Broken]

However, when I endeavor to solve this by converting from polar to cartesian co-ordinates, it seems that I have to assume a right-triangle.

Let S be the swimmer, W be the water, and G the ground.

sVw = 4 m/s [N] = (4, 90*)
wVg = 2 m/s [E25*S] = (2, -335*)
sVg = ?
.: sVg = sVw + wVg
= [0, 4] + [1.8, 0.84]
= [1.8, 4.84]
= (5.2, 69.5*)

I am not very confident in my answer. For starters, I am not supposed to really solve this using polar-cartesian conversion, but I was at a standstill when attempting another solution. Any light shed on a solution for this would be much appreciated.
 
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Answers and Replies

  • #2
Redbelly98
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Yo. Welcome to PF :smile:

[1.8, 0.84]
Double-check the +/- signs in those numbers. Things look fine otherwise.

I am not supposed to really solve this using polar-cartesian conversion ...
That's weird, because that is by far the standard and preferred way to solve problems like this. Alternatively, but more cumbersome, is to use the law of cosines and law of sines from trig.
 

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