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April 20, 2014

April 20, 2014

Posted by **Diego** on Thursday, October 11, 2012 at 9:15pm.

of 9.31 m/s. He intends to swim directly

across a river that has a downstream current

of 4.35 m/s.

a) How many degrees from straight across

the river should he head? Let upstream be a

positive angle.

Answer in units of degrees

b) What is the magnitude of the swimmer’s

velocity relative to the bank?

Answer in units of m/s

- Physics -
**Elena**, Friday, October 12, 2012 at 2:50pm(a)tanα =u/v=4.35/9.31=0.467.

α=arctan 0.467=25.04°

(b) v(rel) =sqrt(v²-u²)=sqrt(9.31²-4.35²)=8.23 m/s

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