Relative Motion in Two Dimension: Practice Problem & Solution
A swimmer can swim with speed of \(8\text{ m/s}\) in still water. \(800\text{ m}\) wide river is flowing with speed of \(4\text{ m/s}\). Swimmer wants to cross the river in minimum time. Velocity of swimmer with respect to ground is (approximately)
Solution Explained:
To solve this problem, we apply the core principles of Relative Motion in Two Dimension. Understanding the underlying formula is key to arriving at the correct answer below:
For minimum crossing time, the swimmer must head perpendicular to the river bank. The net ground velocity is the vector sum of swimmer's velocity and river velocity: \(v_g = \sqrt{v_{\text{sw}}^2 + v_r^2} = \sqrt{8^2 + 4^2} = \sqrt{80} \approx 8.94\text{ m/s} \approx 9\text{ m/s}\).
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