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A swimmer heads directly across a river, swimming at 1.10 m/s relative to the water. She arrives at a point 60.0 m downstream from the point directly across the river, 80.0 m wide. What is the speed of the river current?

What is the swimmer's speed relative to the shore?
In what direction (as an angle relative to a direct line across the river) should the swimmer aim instead, so that she arrives at the point directly opposite her starting point?

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2 answers
  1. (a)
    s=sqrt(80²+60²) = 100 m.

    sinα=60/100=v/1.1.
    v=60•1.1/100 = 0.66 m/s.
    (b)
    sin φ= 0.66/1.1 = 0.06
    φ=arcsin 0.06 = 3.44º

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  2. Elena, i have a little problem with your calculation. I was thinking the velocity of the swimmer should tally with one of the adjacent sides of the triangle and not the hypothenus. the hypothenus should be in line with the resultant velocity of the swimmer.

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