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In a river of 20 m width.Half part of river flows with speed of 10m/s and remaining half part flows with 20m/s.as shown in figure.A man starts to swim from A and reaches to point B in 10 sec .Man swims with speed vm with respect to river at an angle thita with line AB. find thita angle --ans :arctan(15/2) |
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Answer» Here the component of velocity of swimmer in AB direction is always Vmcosθ. As the swimmer crosses the river of breadth 20m in 10sec we can write 10* Vmcosθ.=20 => Vmcosθ.=2........[1] Again the net displacement along the stream is zero and each 10m breadth is crossed by the swimmer in 5sec. So we can write (Vmsinθ.-10)*5 +(Vmsinθ.-20)*5=0 => 2Vmsinθ.= 30 => Vmsinθ.= 15.....[2] Dividing [2] by [1] we get => tanθ= 15/2 => θ.= tan-1(15/2)
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