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April 18, 2014

April 18, 2014

Posted by **melissa** on Tuesday, January 1, 2008 at 8:12pm.

- physics -
**bobpursley**, Tuesday, January 1, 2008 at 9:05pmIf it is to go directly across,then it must be headed upstream.

Theta=arcsin(.5/6.75)

Then velocity across= 6.75*arccosTheta

If you need more assistance, ask a followup question.

- physics -
**Damon**, Tuesday, January 1, 2008 at 9:12pmsay the boat travels at angle T from straight across. Then the upstream component of the boat velocity relative to water must be .5 m/s to counteract current

so 6.75 sin T = .5 m/s

sin T = .5/6.75 = .0741

so T = sin^-1 (.0741) = 4.25 degrees from straight across toward upstream both ways.

The component of velocity across the river is then

6.75 cos 4.25 deg = 6.73 m/s

You did not say how wide the river is so I will call it D

time = 2 D/6.73

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