Question

A river is flowing due east with a speed of 3m/s. A swimmer can swim in still water at a speed of 4m/s. If he starts swimming due north, what will be his resultant velocity (Magnitude and Direction) ?

Answer 1

$$\begin{gathered} v_r=velocityoftheriver=3m/s \\ {v}_s=velocityoftheswimmer. \\ Resul\tan tvelocity=\sqrt{{v_r}^2+{v_s}^2} \\ V=\sqrt{9+16} \\ v=5m/s \\ Direction \\ \tan \theta =\frac{v_s}{v_r}=\frac{4}{3} \\ \theta ={\tan }^{-1}\left(\frac{4}{3}\right) \\ where\theta istheanglebetweenresul\tan tvelocityandvelocityoftheriver. \end{gathered}$$