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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.  Search ResultsWeb results Rituraj Tiwari
11 months ago 16 days ago

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