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Numerical · Q21

Q.A man can swim at 4 km/h in still water; the river flows at 3 km/h. He wishes to cross the river of width 400 m by the shortest path (arriving directly opposite his starting point). At what angle to the straight-across direction should he head, and how long will he take to cross?

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For the swimmer to land directly opposite his starting point, the component of his swim velocity along the upstream direction must exactly cancel the river's downstream current. If θ\theta is the angle between his heading and the straight-across direction, this requires vswimsin⁡θ=vriverv_{\text{swim}}\sin\theta = v_{\text{river}}, so sin⁡θ=34=0.75\sin\theta = \dfrac{3}{4} = 0.75, giving θ=sin⁡−1(0.75)≈48.6∘\theta = \sin^{-1}(0.75) \approx 48.6^\circ upstream of straight-across. His resultant speed directly across the river is then the remaining (across-river) component, vacross=vswim2−vriver2=42−32=7≈2.646v_{\text{across}} = \sqrt{v_{\text{swim}}^2-v_{\text{river}}^2} = \sqrt{4^2-3^2} = \sqrt7 \approx 2.646 …

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