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. Search Results Web results

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Rituraj Tiwari · Answer

Vibekjyoti Sahoo · Answer

AB = d / 2

V R = √ (V m )^2 +V^2 + 2Vm Vcos120

V R = √V m^ 2 +V^2 −V m V .......(1)

Let θ be the angle made by the resultant velocity vector.
From parallelogram law of vector addition.

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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