Question

a river is flowing due East with a speed 2m/s.A swimmer on the south bank of river can swim in still water at a speed of 4m/s.Swimmer swims in such a way that he reaches a directly opposite point on the north bank.The resultant speed of the swimmer is​

Answer 1

Vr = Vcos30°

Vr = 4×√3/2

Vr = 2√3 (final answer)

Answer 2

Answer: Resultant speed of swimmer is  2√3 m/s.

Explanation:

The swimmer has to reach directly opposite point on north bank. Therefore, the swimmer has to move such a way that part of his velocity can counter the velocity of the river which tends to move the swimmer towards east.

The swimmer moves such that component of velocity along west is u (to counter the speed of river) and a along north.

Speed of swimmer = 4

⇒ $\sqrt{u^{2} +v^{2} } = 4$  

Net velocity of swimmer in ground frame =  velocity of swimmer + velocity of river

Net velocity of swimmer = u west + v north + 2 east

Net velocity of swimmer = v north + (u - 2) west

West component velocity should be 0

$u -2 = 0\\u = 2$

Therefore,

$\sqrt{2^{2} +v^{2} } =4\\4 + v^{2} =16\\v^{2} = 12\\v = 2\sqrt{3}$

Resultant speed of swimmer is v m/s ,i.e., 2√3 m/s.

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