Correct option is (b) 3km/h
Time to cross the river = 5 min
The displacement perpendicular to flow = 200 m
∴ Velocity of swimmer
\(v_{sy} = \frac{200}{5 \times 60} = \frac 23 m/s\)
The displacement covered in the direction of flow = 300 m in 10 min.
∴ The velocity of river flow
\(v_r = \frac{300}{10 \times 60} = \frac 12 m/s\)
The velocity of swimmer in the direction of flow
\(\vec {v_{sx}} = \vec v_r + \vec v_s\)
or \(v_{sx} = \frac 12 + 0 =\frac 12 m/s\)
His velocity with respect to bank
\(v_S = \sqrt{v_{sx}^2 + v_{sy}^2}\)
\(= \sqrt{(\frac23)^2 + (\frac12)^2}\)
\(=\frac 56 m/s\)
\(= 3 km/h\)
The velocity \(\vec{v_s}\) makes an angle θ with the bank, then
\(\tan \theta = \frac{v_{sy}}{v_{sx}} =\frac{2/3}{1/2}\)
\(= \frac 43\)
or \(\theta = \tan^{-1}(4/3)\)
