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Chapter 3: Problem 72

A boat that can travel at \(4.0 \mathrm{km} / \mathrm{h}\) in still water crosses a river with a current of \(1.8 \mathrm{km} / \mathrm{h}\). At what angle must the boat be pointed upstream to travel straight across the river? In other words, in what direction is the velocity of the boat relative to the water?

Short Answer

Expert verified
Answer: The boat must be pointed about \(29.97^\circ\) upstream relative to the water.

Step by step solution

01

Understand the problem

We want to find an angle such that the boat's direction in still water has a perpendicular component that exactly cancels the river's current (so that the boat moves straight across the river). The boat's velocity in still water is \(4.0\mathrm{km/h}\). The river's current runs at \(1.8\mathrm{km/h}\).
02

Apply Pythagorean theorem

Let \(v_b\) be the velocity of the boat in still water, and \(c\) be the current of the river. We can break down \(v_b\) into two components: \(v_{bx}\) — the component parallel to the current, and \(v_{by}\) — the component perpendicular to the current. The Pythagorean theorem tells us that \(v_{b}^2 = v_{bx}^2 + v_{by}^2\).
03

Determine the component of the boat's velocity that cancels the current

In order for the boat to move straight across the river, the component of the boat's velocity perpendicular to the current (\(v_{by}\)) must equal the current (\(c\)). Thus, \(v_{by} = 1.8\mathrm{km/h}\).
04

Calculate the component of the boat's velocity parallel to the current

Using the Pythagorean theorem from Step 2 with \(v_{by} = 1.8\mathrm{km/h}\), we can calculate the component \(v_{bx}\): \(v_{bx}^2 = v_b^2 - v_{by}^2 = (4.0\mathrm{km/h})^2 - (1.8\mathrm{km/h})^2 = 10.76\mathrm{km^2/h^2}\). Then, \(v_{bx} = \sqrt{10.76}\mathrm{km/h} \approx 3.28\mathrm{km/h}\).
05

Find the angle of the boat relative to the water

Now we can use trigonometric functions to find the angle \(\theta\). From Step 3 and 4, we have the values of both the adjacent side (\(v_{bx} = 3.28\mathrm{km/h}\)) and the opposite side (\(v_{by} = 1.8\mathrm{km/h}\)) of the angle. We use the tangent function: \(\tan(\theta) = \frac{v_{by}}{v_{bx}}\), and so \(\theta = \arctan(\frac{1.8}{3.28}) \approx 29.97^\circ\). The boat must be pointed about \(29.97^\circ\) upstream relative to the water in order to travel straight across the river.

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