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A swimmer can cross a river of width d in still water in time t0, reaching the point directly opposite on the other bank. When the river has a current, the swimmer still manages to reach the point directly opposite (taking the same straight-line path), but now takes time t1 (t1 > t0). What is the speed of the river current?

1. d / sqrt(t1 * t0)
2. d / (t1 * t0) * sqrt(t1² - t0²)
3. d / sqrt(t1² - t0²)
4. d * sqrt(t1² + t0²) / (t1 * t0)

Correct answer: d / (t1 * t0) * sqrt(t1² - t0²)

Solution

Swimmer speed vₛ = d/t0. To cross directly against current v_w: sqrt(vₛ² - v_w²) = d/t1. Squaring and solving: v_w² = d²/t0² - d²/t1² = d²*(t1²-t0²)/(t0*t1)², giving v_w = d*sqrt(t1²-t0²)/(t1*t0).

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