A river is flowing from west to east direction with speed of \(9\) km/hr. If a boat capable of moving at a maximum speed of \(27\) km/hr in still water, crosses the river in half a minute, while moving with maximum speed at an angle of \(150^\circ\) to direction of river flow, then the width of the river is:
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When solving river crossing problems, decompose the boat's velocity into components: one parallel to the river flow and one perpendicular to it. The perpendicular component gives the speed for crossing the river.
The boat's speed relative to the river is \( 27 \, \text{km/hr} \). The boat travels across the river at an angle of \( 150^\circ \) to the river's flow. The component of the boat's velocity perpendicular to the river flow is calculated as: \[ V_{L} = 27 \, \text{km/hr} \times \cos 60^\circ = \frac{27}{2} = 13.5 \, \text{km/hr} \] The time to cross the river is \( 30 \, \text{seconds} \), which is \( \frac{1}{2} \) minute. The distance covered (river width) is calculated using the formula: \[ S = V_t \times t = 13.5 \, \text{km/hr} \times \frac{30}{60} \, \text{hr} = 13.5 \times \frac{1}{2} = 112.5 \, \text{m} \] Therefore, the river's width is \( 112.5 \, \text{m} \).
