91Ó°ÊÓ

In physics, change in velocity, often represented as \( \Delta \mathbf{v} \), is fundamental to understanding motion and acceleration. It represents how much an object's velocity changes over a period. In our example, at time \( t=10 \mathrm{s} \), the velocity of the particle is \( 0.1 \mathbf{i} + 2 \mathbf{j} \mathrm{m/s} \). By time \( t=10.1 \mathrm{s} \), the velocity changes to \( -0.1 \mathbf{i} + 1.8 \mathbf{j} \mathrm{m/s} \).

To find the change in velocity, we subtract the initial velocity from the final velocity:

• Initial velocity, \( \mathbf{v_1} = 0.1 \mathbf{i} + 2 \mathbf{j} \).
• Final velocity, \( \mathbf{v_2} = -0.1 \mathbf{i} + 1.8 \mathbf{j} \).
• Change, \( \Delta \mathbf{v} = \mathbf{v_2} - \mathbf{v_1} = (-0.1 - 0.1) \mathbf{i} + (1.8 - 2) \mathbf{j} \).
• So, \( \Delta \mathbf{v} = -0.2 \mathbf{i} - 0.2 \mathbf{j} \ \mathrm{m/s} \).

This change in velocity is key to determining the average acceleration.

Time interval, denoted as \( \Delta t \), is the duration over which the change in velocity occurs. It is crucial in the computation of average acceleration, as acceleration is defined as the change in velocity per unit time.

In the exercise, the particle's velocity changed from \( t=10 \) seconds to \( t=10.1 \) seconds, making the time interval \( \Delta t = t_{final} - t_{initial} \). Calculating this we have:

• \( t_{final} = 10.1 \text{ s} \)
• \( t_{initial} = 10 \text{ s} \)
• \( \Delta t = 10.1 - 10 = 0.1 \text{ s} \)

This simple calculation shows how minor our measurement of time can be when evaluating motion, especially when it comes to minute intervals of change.

Velocity components refer to the breakdown of velocity into its respective parts along the coordinate axes, typically \( x \) and \( y \) in two-dimensional motion. Understanding these components allows us to analyze motion in multiple directions separately.

For the initial velocity of our particle at \( t=10 \) seconds, the components are:

• \( v_{x_1} = 0.1 \ \mathrm{m/s} \)
• \( v_{y_1} = 2 \ \mathrm{m/s} \)

At \( t=10.1 \) seconds, the components change to:

• \( v_{x_2} = -0.1 \ \mathrm{m/s} \)
• \( v_{y_2} = 1.8 \ \mathrm{m/s} \)

The change in each component gives us the velocity components' change:

• \( \Delta v_x = v_{x_2} - v_{x_1} = -0.2 \ \mathrm{m/s} \)
• \( \Delta v_y = v_{y_2} - v_{y_1} = -0.2 \ \mathrm{m/s} \)

This breakdown helps calculate the overall change in velocity effectively.

The angle \( \theta \) that the average acceleration vector makes with the positive \( x \)-axis helps us understand the direction of motion. It can be found using the tangent function in trigonometry, specifically \( \theta = \arctan\left(\frac{a_y}{a_x}\right) \).

Here, \( a_x \) and \( a_y \) are the components of acceleration, computed from the change in velocity over the time interval. With \( \Delta \mathbf{v} = -0.2 \mathbf{i} - 0.2 \mathbf{j} \ \mathrm{m/s} \) and \( \Delta t = 0.1 \ \mathrm{s} \), the average acceleration is:

• \( a_{x}= -2 \ \mathrm{m/s^2} \)
• \( a_{y} = -2 \ \mathrm{m/s^2} \)

Substituting into the formula: \[\theta = \arctan\left(\frac{-2}{-2}\right) = \arctan(1) = 45^\circ\]However, since the vector is in the third quadrant, where both \( a_x \) and \( a_y \) are negative, the true angle from the positive \( x \)-axis is:

• \( \theta = 225^\circ \)

This angle describes how the acceleration vector is directed downward and to the left.