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A swimmer heads directly across a river, swimming at 1.9 m/s relative to the water. She arrives at a point 50 m downstream from the point directly across the river, which is 79 m wide.
(a) What is the speed of the river current?

(b) What is the swimmer's speed relative to the shore?

(c) In what direction should the swimmer head so as to arrive at the point directly opposite her starting point?
° upstream from straight ahead

  • Physics -

    tan A = 50m/79m = 0.63291
    A = 32.3o East of North = 57.7o CCW.

    b. Vs = 1.9m/sin57.7 = 2.25 m.[57.7o] = Velocity of swimmer.

    Vc = 2.25m/s[57.7o] - 1.9[90o]

    X = 2.25*cos57.7 - 1.9*cos90=1.20 m/s.
    Y=2.25*sin57.7 - 1.9*sin90 = 0 m/s.

    b. Vc = 1.20 + i0 = 1.20 m/s =Velocity of the current.

    c. 32.3oWest of North.

    Vc = X/cosB = 1.20/cos0.0878=13.67m/s.

  • Physics -

    A man can swim directly across a stream of width 300 m in 4 minutes, when there is no current and in 5 minutes when there is current. Find velocity of the current.

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