91Ó°ÊÓ

In the world of physics, a displacement vector is crucial because it describes the change in position of an object. Rather than just focusing on the distance traveled, it details the shortest path between the start and end point. In our exercise, the displacement vector is calculated as the difference between the final and initial position vectors. We denote it as \( \Delta \vec{r} \) and it provides both the direction and magnitude of movement over time. Calculating it involves simple subtraction of corresponding components of the initial and final vectors:

• i-component: Final position - Initial position = (-2.0 - 5.0)
• j-component: Final position - Initial position = (8.0 + 6.0)
• k-component: Final position - Initial position = (-2.0 - 2.0)

Simplifying these, we get \( \Delta \vec{r} = -7.0 \hat{\mathrm{i}} + 14.0 \hat{\mathrm{j}} - 4.0 \hat{\mathrm{k}} \). This vector tells us how the ion moved from one point to another.

Unit-vector notation helps us express vectors easily by showing components along the coordinate axes. Here, each component (i, j, k) signifies direction along the x, y, and z axes, respectively. Vectors are made up of these components, represented by unit vectors:

• \( \hat{\mathrm{i}} \) - direction along the x-axis
• \( \hat{\mathrm{j}} \) - direction along the y-axis
• \( \hat{\mathrm{k}} \) - direction along the z-axis

In our problem, we use this notation to detail both the position vectors and the displacement vector. The use of unit vectors provides clarity and conciseness, helping us to visualize and calculate spatial information easily. It simplifies the addition and subtraction of vectors by separating each component.

Position vectors establish the location of an object in space relative to a reference point, usually the origin. In the exercise, the ion's position vectors are given at two different instances. The initial position \( \vec{r}_{\text{initial}} \) is described as \( 5.0 \hat{\mathrm{i}} - 6.0 \hat{\mathrm{j}} + 2.0 \hat{\mathrm{k}} \). The final position vector \( \vec{r}_{\text{final}} \) is \( -2.0 \hat{\mathrm{i}} + 8.0 \hat{\mathrm{j}} - 2.0 \hat{\mathrm{k}} \).Each term in the vector denotes how far the point is along the respective axis:

• \( 5.0 \hat{\mathrm{i}} \) - 5 meters along the x-axis
• \( -6.0 \hat{\mathrm{j}} \) - -6 meters means 6 meters in the negative y direction
• \( 2.0 \hat{\mathrm{k}} \) - 2 meters along the z-axis

These position vectors help us compute other physical quantities such as displacement, emphasizing the importance of understanding vector properties.

The concept of a time interval is essential when dealing with motion. It represents the duration over which change occurs when an object moves. In this problem, the time interval is the difference between the two recorded times, and it is given as 10 seconds. Time interval \( \Delta t \) is used to compute the average velocity. This time span is crucial because it provides the context needed to relate displacement to motion. In essence, by dividing the displacement vector by the time interval, we're able to determine the average velocity:\[ \vec{v}_{\text{ave}} = \frac{\Delta \vec{r}}{\Delta t} = \frac{-7.0 \hat{\mathrm{i}} + 14.0 \hat{\mathrm{j}} - 4.0 \hat{\mathrm{k}}}{10}\]This results in the average velocity being \( -0.7 \hat{\mathrm{i}} + 1.4 \hat{\mathrm{j}} - 0.4 \hat{\mathrm{k}} \) meters per second. Understanding time intervals allows for accurate measurements of dynamic changes.