27. Correct Direction for Crossing a River A river has a constant current of 3 kilometers per hour. At what angle to a boat dock should a motorboat, capable of maintaining a constant speed of 20 kilometers per hour, be headed in order to reach a point directly opposite the dock? If the river is \( \frac{1}{2} \) kilometer wide, how long will it take to cross?
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The Deep Dive
To cross the river directly opposite the dock, the motorboat must navigate at a specific angle to counteract the current. Let \( v_b = 20 \) km/h be the boat's speed and \( v_c = 3 \) km/h be the current's speed. Using the Pythagorean theorem, we can calculate the angle \( \theta \) as follows: 1. The boat's velocity perpendicular to the current is \( v_b \sin(\theta) \). 2. The boat's velocity upstream to counteract the current is \( v_b \cos(\theta) \). To reach the opposite bank directly, the component of the boat's speed against the current must equal the speed of the current: \[ v_b \cos(\theta) = v_c \] \[ 20 \cos(\theta) = 3 \quad \Rightarrow \quad \cos(\theta) = \frac{3}{20} \] Now, we can find \( \theta \) using: \[ \theta = \cos^{-1}\left(\frac{3}{20}\right) \approx 72.34^\circ. \] Next, to find the time \( t \) it takes to cross the river, we can determine the time taking the width of the river (\( \frac{1}{2} \) km) and the velocity component perpendicular to the current: \[ v_{perpendicular} = v_b \sin(\theta) = 20 \sin(72.34^\circ) \approx 19.46 \text{ km/h}. \] Thus, the time taken to cross is: \[ t = \frac{0.5 \text{ km}}{19.46 \text{ km/h}} \approx 0.0257 \text{ hours} \approx 1.54 \text{ minutes}. \] So, the boat should head at approximately \( 72.34^\circ \) to the dock, taking just about 1.54 minutes to cross the river!
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