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6 February, 11:59

A swimmer heading directly across a river 200 m wide reaches the opposite bank in 6 min 40 s, during which time she is swept downstream 480m. how fast can she swim in still water

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  1. 6 February, 12:20
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    Short method:

    Assuming the direction that she is swimming is directly across the river (y-direction), and she is swept downstream (x-direction) by the current in the river. We can directly calculate her y-velocity:

    y = v*t

    v = y/t = 200 / (6*60 + 40) = 0.5 m/s

    Long method:

    The river’s speed does not contribute to her progress across the river. It only provides an x-direction component. If we want to calculate her component of velocity over the resultant path, we would calculate the resultant distance and velocity.

    v_x = 480m/400s = 1.2 m/s

    v_x is the speed of the river current

    The distance covered is calculated using hypotenuse formula:

    d = sqrt (200^2 + 480^2) = 520 m

    The resultant velocity is:

    v = d/t = 520/400 = 1.3 m/s

    However this is still the resultant speed of the swimmer and the river current together. Therefore using the hypotenuse formula again to find for v_x:

    v^2 = v_x) ^2 + (v_y) ^2

    v^2 - (v_x) ^2 = (v_x) ^2

    v_y = sqrt [ v^2 - (v_x) ^2 ] = sqrt (1.3^2 - 1.2^2) = 0.5 m/s

    This is the swimmers velocity without the current.

    Answer:

    0.5 m/s
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