Navigating across a river: The impact of angle and current on direct motion

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Implication of boat pointed at a constant angle φ to the perpendicular to the river bank
byu/DigitalSplendid inLinearAlgebra

Disclaimer: This article was created with the assistance of an AI language model and is intended for informational purposes only. Please verify any technical details before implementation.

 

The key to understanding the impact of the angle (\phi) is recognizing that the river’s current affects the boat’s motion. Here’s the breakdown of the situation and implications:

Given Setup

1. Boat speed through still water: The boat can travel at (s) meters per second in still water.
2. River current speed: The river flows from left to right at (v) meters per second.
3. Desired path: You want the boat to move directly across the river from point (A) to point (B), which is directly opposite (A).
4. Angle: The boat is pointed at an angle (\phi) relative to the perpendicular to the riverbank, aiming slightly upstream to counteract the river’s flow.

Implication of Pointing at Angle (\phi)

If the boat were pointed directly at (B) (without adjusting for the current), it would be pushed downstream by the current and would miss point (B), arriving somewhere downstream. Therefore, to reach (B) directly, the boat needs to be aimed at an angle (\phi) upstream. This angle allows the boat to compensate for the river’s current while maintaining a net movement directly across the river.

How the Angle (\phi) Works

To analyze this, think of the boat’s velocity vector ( \vec{s} ) in the water and the river’s current velocity vector ( \vec{v} ):

1. Boat’s effective velocity: When the boat is pointed at an angle (\phi) relative to the perpendicular, its velocity component perpendicular to the bank becomes (s \cos(\phi)).
2. Horizontal component: The boat’s horizontal (upstream) velocity component (s \sin(\phi)) partially counters the river current’s speed (v).
3. Resultant velocity: To move straight across (from (A) to (B)), the horizontal component of the boat’s velocity (s \sin(\phi)) must exactly cancel the river current (v), so:
   [
   s \sin(\phi) = v
   ]

This gives:
[
\sin(\phi) = \frac{v}{s}
]

Thus, (\phi = \arcsin\left(\frac{v}{s}\right)), meaning the boat must be aimed at an angle (\phi) upstream to reach (B) directly.

Summary of the Angle’s Implication
The angle (\phi) is crucial because it compensates for the river’s current, allowing the boat to travel in a straight line directly across the river. Without adjusting the boat’s angle, the current would push it downstream, and it would not reach point (B) directly. This example illustrates the concept of vector addition in physics, where the resultant velocity is the combination of the boat’s velocity in still water and the river’s flow.

---

To solve this, let’s go through the conditions given in the problem.

Given Information

1. Boat speed through still water = ( s ) meters per second.
2. River width = ( d ) meters.
3. River current speed = ( v ) meters per second.
4. The boat angle: The boat is pointed at an angle (\phi) to counteract the current and arrive directly across from point (A) at point (B).

Approach to Solve

To move directly across the river from (A) to (B) without drifting downstream:
– The horizontal component of the boat’s velocity must counterbalance the river’s current.
– The boat’s speed, ( x ), can be represented as the hypotenuse of a right triangle formed by the components of the boat’s velocity in still water ((s)) and the current’s speed ((v)).

Analyzing the Expressions

1. Expression for ( x ) (the effective speed of the boat)

To travel straight across, we need to combine the boat’s speed through still water (s) and the river current (v) using the Pythagorean theorem. The effective speed ( x ) is given by:

[
x = \sqrt{s^2 + v^2}
]

This is because ( x ) is the resultant speed when the boat’s own speed ( s ) and the current ( v ) are at right angles to each other.

2. Expression for ( \phi ) (the angle to counteract the current)

The angle ( \phi ) can be calculated using the sine function, as we know:
[
\sin(\phi) = \frac{v}{x}
]

Solving for (\phi), we get:
[
\phi = \sin^{-1} \left( \frac{v}{x} \right) = \sin^{-1} \left( \frac{v}{\sqrt{s^2 + v^2}} \right)
]

Correct Answer

From these derivations, the correct answer is:

[
x = \sqrt{s^2 + v^2} \quad \text{and} \quad \phi = \sin^{-1} \left( \frac{v}{s} \right)
]

So, the answer that matches these expressions is:

Option 2:
[
x = \sqrt{s^2 + v^2} \quad \text{and} \quad \phi = \sin^{-1} \left( \frac{v}{s} \right)
]

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