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A swimmer crosses a river of wifth d flowing at velocity v. While swimming he keeps himself always at an angle of 120degree with river and on reaching the other end he finds a drift of d/2 in the direction of flow of river .Find the speed of the swimmer with respect to the river.
AB = d / 2
V R = √ (V m )^2 +V^2 + 2Vm Vcos120
V R = √V m^ 2 +V^2 −V m V .......(1)
θ = V mcos(120) / V+ V m sin(120) .......(2)
In ΔABC, ∠AOB = (90−θ)
tan(90−θ) = d/2 ÷ d = 1/2
cotθ = 1/2 so θ = tan^−1 2 .......(3)
tan^−1 2 = V mcos(120) / V+ V m sin(120)
From equations (2) and (3)
03.43 = Vm x 0.86 / V - 0.5 v
Vm = 1.95 v
Putting this in equation (1) we get.
V R = √ (1.91v)^2 + v^2 + - 1.95 v^2
Hence the speed of the swimmer with respect to river is V R = √ (1.91v)^2 + v^2 + - 1.95 v^2
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