91Ó°ÊÓ

Understanding how to progress from acceleration to velocity in kinematics is crucial for analyzing motion. Acceleration, represented by \( \mathbf{a}(t) \), is the rate at which an object's velocity changes with time. To find velocity, we integrate the acceleration function with respect to time. This process essentially sums up all the infinitesimal changes in velocity that occur as time advances.

For instance, if we have an acceleration function \( \mathbf{a}(t) = t \mathbf{j} + t \mathbf{k} \), the integration process will involve finding the antiderivative of each component of the acceleration vector. The result of integrating \( t \mathbf{j} \) and \( t \mathbf{k} \) with respect to time will provide us with velocity components in the \( \mathbf{j} \) and \( \mathbf{k} \) directions, but we will also need to add a constant vector \( \mathbf{C} \) to account for initial conditions. This constant is determined by inserting known values of velocity at a particular time.

Once we have integrated acceleration to get a general velocity formula, we need to apply initial conditions to find the exact velocity vector \( \mathbf{v}(t) \). The given initial condition, \( \mathbf{v}(1) = 5 \mathbf{j} \), is plugged into the integrated velocity equation to solve for the constant of integration.

In the provided exercise, the integration yielded \( \frac{1}{2}t^2 \mathbf{j} + \frac{1}{2}t^2 \mathbf{k} + \mathbf{C} \). By inserting \( t = 1 \) and the known velocity value, we found that \( \mathbf{C} = 4.5 \mathbf{j} \), concluding that the velocity at any time \( t \) can be expressed as \( \mathbf{v}(t) = \frac{1}{2}t^2 \mathbf{j} + \frac{1}{2}t^2 \mathbf{k} + 4.5 \mathbf{j} \).

The process of determining the position vector \( \mathbf{r}(t) \) is analogous to calculating velocity from acceleration. Here, we take the velocity function and integrate it with respect to time to obtain the position function. The integration accounts for the cumulative distance traveled as time passes.

Utilizing the velocity vector we found earlier and integrating each component with respect to time, we add another constant of integration \( \mathbf{D} \) to include initial position information. For example, after integrating \( \frac{1}{2}t^2 \mathbf{j} + \frac{1}{2}t^2 \mathbf{k} + 4.5 \mathbf{j} \), we add \( \mathbf{D} \) to represent the initial position. By using the provided condition \( \mathbf{r}(1) = \mathbf{0} \) and substituting \( t = 1 \) into our general position equation, we can solve for \( \mathbf{D} \) to find the definitive position vector for any time \( t \).

Initial conditions are the cornerstone for solving kinematics problems involving integration. They anchor our general equations to a particular scenario by specifying values at a known point in time, usually \( t = 0 \) or another specified time. These values typically include an object's initial velocity and position.

In our exercise, the initial conditions given were \( \mathbf{v}(1) = 5 \mathbf{j} \) and \( \mathbf{r}(1) = \mathbf{0} \). They provided the necessary information to solve for the constants of integration \( \mathbf{C} \) and \( \mathbf{D} \) in the integrated forms of the velocity and position functions. Without these initial conditions, we would not have been able to determine the exact motion of the object. Initial conditions ensure our answers are not just mathematically correct but also physically meaningful.