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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
Vibekjyoti Sahoo
145 Points
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