Problem 75 - College Physics

Question

A swimmer swims north at \(0.15 \mathrm{~m} / \mathrm{s}\) relative to still water across a river that flows at a rate of \(0.20 \mathrm{~m} / \mathrm{s}\) from west to east. (a) The general direction of the swimmer's velocity, relative to the riverbank, is (1) north of east, (2) south of west, (3) north of west, (4) south of east. (b) Calculate the swimmer's velocity relative to the riverbank.

Step-by-Step Solution

Answer

(a) North of east, (b) 0.25 m/s.

Step 1: Understand the Vector Components

The swimmer's velocity relative to still water is given as \(0.15 \, \text{m/s north}\), and the river flows from west to east at \(0.20 \, \text{m/s}\). Assign the north direction as the positive y-axis and the east direction as the positive x-axis.

Step 2: Define the Velocity Vectors

Represent the swimmer's velocity vector \(\mathbf{v}_s = 0.15 \, \text{m/s (north)}\) as \((0, 0.15)\) and the river's velocity vector \(\mathbf{v}_r = 0.20 \, \text{m/s (east)}\) as \((0.20, 0)\).

Step 3: Add the Velocity Vectors

The swimmer's velocity relative to the riverbank is the vector sum of the swimmer's and river's velocities: \(\mathbf{v}_{\text{relative}} = \mathbf{v}_s + \mathbf{v}_r = (0.20, 0) + (0, 0.15) = (0.20, 0.15)\).

Step 4: Determine the Direction of the Relative Velocity

The vector \(\mathbf{v}_{\text{relative}} = (0.20, 0.15)\) has a positive x-component and a positive y-component, indicating the general direction is north of east (choice 1).

Step 5: Calculate the Magnitude of the Relative Velocity

Use the Pythagorean theorem to calculate the magnitude: \( \|\mathbf{v}_{\text{relative}}\| = \sqrt{0.20^2 + 0.15^2} = \sqrt{0.04 + 0.0225} = \sqrt{0.0625} = 0.25 \, \text{m/s} \).

Step 6: Summarize Findings for Direction and Magnitude

We determined the swimmer's direction relative to the riverbank as north of east and calculated the magnitude of the swimmer's relative velocity to be \(0.25 \, \text{m/s}\).

Key Concepts

Vector Addition

Vector addition is crucial when dealing with movements in multiple directions, like in this swimmer exercise. Here, the swimmer swims north and the river flows east. Each movement can be represented as a vector. These vectors have both magnitude (speed) and direction. For example, the swimmer's speed of \(0.15 \, \text{m/s}\) north becomes the vector \((0, 0.15)\). Likewise, the river's flow of \(0.20 \, \text{m/s}\) east is represented as the vector \((0.20, 0)\).

Vector addition involves summing these vectors to determine the resultant vector, which gives the direction and speed relative to still surroundings. By adding the components for each direction, we get:

• East component: \(0.20\) from river, \(0.00\) from swimmer, totaling \(0.20\).
• North component: \(0.15\) from swimmer, \(0.00\) from river, totaling \(0.15\).

With the vector sum \((0.20, 0.15)\), the swimmer's movement relative to the riverbank is found.

Pythagorean Theorem

The Pythagorean theorem helps us calculate the magnitude (or length) of the resultant velocity vector from vector addition. This theorem applies to right-angled triangles, where the square of the hypotenuse (the longest side) equals the sum of the squares of the other two sides. In our problem, the swimmer's combined movement forms a right triangle with the x and y components: east (\(0.20\,\text{m/s}\)) and north (\(0.15\,\text{m/s}\)).

Using the Pythagorean theorem, we calculate:

• East component: \(0.20\)
• North component: \(0.15\)

Formula: \[\|\mathbf{v}_{\text{relative}}\| = \sqrt{0.20^2 + 0.15^2} = \sqrt{0.04 + 0.0225} = \sqrt{0.0625} = 0.25 \, \text{m/s}\]Thus, the swimmer's effective speed relative to the bank is \(0.25 \, \text{m/s}\), verifying our calculations.

Velocity Components

Velocity components simplify understanding of motion by breaking down a vector into parts aligned with coordinate axes. In the swimmer scenario, we assign the north direction as the y-axis and east as the x-axis for easier calculation.

The swimmer's velocity is \(0.15 \, \text{m/s}\) north, giving it a y-component of \(0.15\) and an x-component of \(0.00\).

• Y-component (northward): \( \text{Velocity} = 0.15 \), hence the vector \((0, 0.15)\).
• X-component (eastward): \( \text{Velocity} = 0.20 \), hence the vector \((0.20, 0)\).

By separating the velocities into components, we can perform vector addition smoothly. This approach results in a complete picture of the motion, making it easier to analyze and understand the overall velocity relative to another reference frame, such as a riverbank.