91Ó°ÊÓ

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.

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 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.