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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?

Ask by Schofield Love. in the United States
Feb 27,2025

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Answer

To cross the river directly opposite the dock, the boat should be aimed approximately 8.6° upstream from the perpendicular direction. It will take about 1.5 minutes to cross the 0.5 km wide river.

Solution

We want the boat’s net motion to be directly across the river despite the 3‑km/h current. Step 1. Adjusting for the Current Let θ be the angle (relative to directly across the river) at which the boat must be aimed upstream. The boat’s speed relative to the water is 20 km/h. Its component in the horizontal (across the river) direction is 20 cos θ, while its component in the upstream direction is 20 sin θ. The river’s current of 3 km/h pushes the boat downstream. To compensate, the upstream component of the boat’s velocity must cancel the downstream current. Thus, we set   20 sin θ = 3. Solving for θ gives   sin θ = 3/20  →  θ = sin⁻¹(3/20). Numerically, sin⁻¹(0.15) is approximately 8.6°. So the boat must be aimed about 8.6° upstream from the perpendicular (i.e. from the line directly across the river). Step 2. Finding the Crossing Time The effective speed directly across the river is given by the component of the boat’s velocity perpendicular to the bank,   20 cos θ. Using the fact that cos θ = √(1 – sin²θ), we have   cos θ = √(1 – (3/20)²) = √(1 – 9/400) = √(391/400) = √391/20. Thus, the effective speed is   20 cos θ = 20 (√391/20) = √391 km/h. However, it is easier to directly compute the time: The river is ½ km wide, so   time = (distance)/ (speed across)      = (½ km) / (20 cos θ). Substitute cos θ:   time = 0.5 / (20 × (√391/20)) = 0.5 / √391 hours. Calculating numerically, note that √391 is approximately 19.8, so   time ≈ 0.5/19.8 ≈ 0.0253 hours. To convert to minutes, multiply by 60:   0.0253 × 60 ≈ 1.52 minutes. Summary of Answers: • The motorboat should be headed at an angle θ where sin θ = 3/20, i.e. approximately 8.6° upstream from the straight‐across direction. • The crossing will take about 1.5 minutes.

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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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