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A river flows due south with a speed of 2ms^–1. A man steers a motor boat across the river. His velocity relative to water is 4ms^–1 due east. If the river is 800m wide, the velocity of man relative to earth is

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6ms^–1, at an angle tan^–1(2) with the stream

4.5ms^–1, at an angle tan^-1(1/2) to the stream

6ms^–1, at an angle 90^o to the stream

4.5ms^–1, at an angle tan^–1(2) with the stream

answer is A.

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Detailed Solution

Velocity of man w.r.t ground is the resultant velocity is

$\begin{array}{l} = \sqrt{{V_{BW}}^{2} + {V_{WG}}^{2}} \\ = \sqrt{4^{2} + 2^{2}} \\ = \sqrt{20} = 4.5 m / s \\ \tan θ = \frac{V_{BW}}{V_{WG}} = 2 \end{array}$

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