91Ó°ÊÓ

Velocity is a core concept in kinematics, representing the rate of change of position with respect to time. It describes how quickly an object is moving and in which direction. In our problem, the velocity of the object is derived by differentiating the position function.
Since the position function is \( \vec{r}(t) = \hat{\mathrm{i}} + 3t^2 \hat{\mathrm{j}} + t \hat{\mathrm{k}} \), we find the first derivative with respect to time, \( t \), for each component:

• The \( \hat{\mathrm{i}} \) component, a constant, yields a derivative of 0.
• The \( \hat{\mathrm{j}} \) component, \( 3t^2 \), becomes \( 6t \).
• The \( \hat{\mathrm{k}} \) component, \( t \), becomes 1.

Hence, the velocity function results in \( \vec{v}(t) = 0\hat{\mathrm{i}} + 6t\hat{\mathrm{j}} + 1\hat{\mathrm{k}} \). This expression tells us that the object speeds up in the \( \hat{\mathrm{j}} \) direction at a rate of 6\( t \), and consistently moves in the \( \hat{\mathrm{k}} \) direction.

Acceleration is the rate of change of velocity with respect to time. In simpler terms, it tells us how the velocity of an object is changing. If we go back to our example, we start by understanding that the acceleration function comes from differentiating the velocity function.
Beginning with the previously found velocity function \( \vec{v}(t) = 0\hat{\mathrm{i}} + 6t\hat{\mathrm{j}} + 1\hat{\mathrm{k}} \), we differentiate each term again with respect to time:

• The \( \hat{\mathrm{i}} \) component remains 0, as it was previously.
• The \( \hat{\mathrm{j}} \) component, \( 6t \), simplifies to just 6.
• The \( \hat{\mathrm{k}} \) component remains 0, as it is constant.

Thus, the acceleration function becomes \( \vec{a}(t) = 0\hat{\mathrm{i}} + 6\hat{\mathrm{j}} + 0\hat{\mathrm{k}} \). This indicates that the object experiences a constant acceleration of 6 in the \( \hat{\mathrm{j}} \) direction.

Derivatives are a fundamental tool in calculus used to determine the rate of change of a quantity. When dealing with kinematics, we primarily use derivatives to find velocity and acceleration from a position function. Differentiating the position function gives us velocity, while differentiating velocity gives acceleration.
In the context of our exercise, for a function of time, a derivative explains how a function's output changes as the input (time, in this case) changes:

• This helps identify the instantaneous rate of change at any given time.
• A first derivative (such as with the position function) will often give velocity.
• A second derivative (as with the velocity function) provides acceleration.

Derivatives simplify understanding motion by converting complex scenarios into manageable functions, helping unravel the pace and progression of movement.

A position function in kinematics illustrates the location of an object at any instant. It is crucial since it serves as the foundation for determining both velocity and acceleration. The position function used in our problem is \( \vec{r}(t) = \hat{\mathrm{i}} + 3t^2 \hat{\mathrm{j}} + t \hat{\mathrm{k}} \), which breaks down into its components:

• The \( \hat{\mathrm{i}} \) term remains constant, indicating no movement in the x-direction.
• The \( 3t^2 \hat{\mathrm{j}} \) part suggests movement that accelerates in the y-direction as time progresses.
• The \( t \hat{\mathrm{k}} \) term shows a linear change over time in the z-direction.

This function is pivotal because by differentiating it, we not only grasp the change in position but also derive insights into how velocity and acceleration evolve over time. Understanding the components of each direction is key to accurately mapping an object's motion.