A river is $90 \mathrm{~m}$ wide and flows East $\left(90^{\circ}\right)$ at $1.6 \mathrm{~m} / \mathrm{s}$.
A ferry, which travels at $2.6 \mathrm{~m} / \mathrm{s}$ in still water, crosses the river.
The ferny needs to head upstream to cross to a point directly opposite.
Calculate the bearing in degrees ${ }^{\circ}$ at which it must head.
Answer
Final Answer: The bearing at which the ferry must head is approximately \(\boxed{148.39°}\).
Step 1 ：The river is 90 m wide and flows East (90°) at 1.6 m/s. A ferry, which travels at 2.6 m/s in still water, crosses the river. The ferry needs to head upstream to cross to a point directly opposite.
Step 2 ：The ferry needs to head upstream to counteract the flow of the river. This means that the ferry's velocity relative to the river is not directly across, but at an angle to the flow of the river. We can use trigonometry to calculate this angle.
Step 3 ：The ferry's velocity relative to the river is the vector sum of its velocity in still water and the river's velocity. We can represent these velocities as vectors, with the river's velocity pointing east and the ferry's velocity in still water pointing north. The resultant vector, representing the ferry's velocity relative to the river, will point in the direction the ferry needs to head.
Step 4 ：We can use the Pythagorean theorem to calculate the magnitude of the resultant vector, and then use the inverse tangent function (also known as arctan or atan) to calculate the angle of the resultant vector. This angle is the bearing at which the ferry must head.
Step 5 ：Given that the velocity of the river is 1.6 m/s and the velocity of the ferry is 2.6 m/s, we can calculate the magnitude of the resultant vector as \(\sqrt{(1.6)^2 + (2.6)^2} = 3.0528675044947495\) m/s.
Step 6 ：We can then calculate the angle of the resultant vector as \(\arctan\left(\frac{2.6}{1.6}\right) = 31.607502246248906°\).
Step 7 ：However, since the ferry needs to head upstream, the bearing at which the ferry must head is \(180° - 31.607502246248906° = 148.3924977537511°\).
Step 8 ：Final Answer: The bearing at which the ferry must head is approximately \(\boxed{148.39°}\).
