91Ó°ÊÓ

The position function describes the location of a particle along a path at any given time. In this exercise, the position function is given by \( s(t) = \sqrt{t+1} \). This function indicates the directed distance of the particle from a reference point, labeled as point O, in feet (ft). It is essential to understand the position function as it serves as the foundation for finding other key quantities like velocity and acceleration. When we know the position function, we can use calculus to derive these other quantities.
To grasp the concept of the position function fully, remember:

• The function must be continuous to describe a realistic movement of a particle.
• It can take various forms: linear, quadratic, or even more complex functions as we see here with the square root.
• The domain (time, t) should be considered to ensure that it aligns with the physical context (e.g., time cannot be negative).

The derivative is a fundamental tool in calculus that helps us find the rate of change of a function. In this problem, we need to find the derivative of the position function to determine the velocity function.
Derivatives have a wide range of applications including physics, engineering, and economics. For a position function \( s(t) \), its first derivative with respect to time \( t \) gives us the velocity function. Specifically, when we take the derivative of \( s(t) = \sqrt{t+1} \), we apply the chain rule:
\[ v(t) = \frac{ds}{dt} = \frac{d}{dt} (t+1)^{1/2} = \frac{1}{2} (t+1)^{-1/2} \left( \frac{d}{dt} (t+1) \right) \] Since \( \frac{d}{dt} (t+1) = 1 \, \) the expression simplifies to:
\[ v(t) = \frac{1}{2} \frac{1}{\sqrt{t+1}} \]
This derivative tells us how fast the position of the particle changes with respect to time.

The velocity function describes the rate at which the position of the particle changes with time. It is obtained by taking the first derivative of the position function.
In our example, the position function is \( s(t) = \sqrt{t+1} \, \) and its derivative gives us the velocity function:
\[ v(t) = \frac{1}{2} \frac{1}{\sqrt{t+1}} \]
The velocity function \( v(t) \frac}{1}{2} \frac{1}{\sqrt{t+1}} \) tells us how quickly or slowly the particle is moving at any time \( t \). This helps us find the speed of the particle at specific instances. For example, to find the velocity at \( t = 3 \, \),we substitute \( t \) with 3 in the velocity function:
\[ v(3) = \frac{1}{2} \frac{1}{\sqrt{3+1}} = \frac{1}{2} \frac{1}{\sqrt{4}} = \frac{1}{2} \frac{1}{2} = \frac{1}{4} \text{ ft/sec} \]
This calculation reveals that the particle's instantaneous velocity at \( t = 3 \) seconds is \( \frac{1}{4} \text{ ft/sec} \).

Instantaneous velocity refers to the speed and direction of a particle at a specific moment in time. Unlike average velocity, which considers the total displacement over a time interval, instantaneous velocity is concerned with a single point in time.
To determine instantaneous velocity, we need the derivative of the position function. For the position function \( s(t) = \sqrt{t+1} \, \), we have already found that:
\[ v(t) = \frac{1}{2} \frac{1}{\sqrt{t+1}} \]
When we want the instantaneous velocity at \( t_1 = 3 \, \), we simply plug this value into our velocity function:
\[ v(3) = \frac{1}{2} \frac{1}{\sqrt{3+1}} = \frac{1}{2} \frac{1}{\sqrt{4}} = \frac{1}{2} \frac{1}{2} = \frac{1}{4} \text{ ft/sec} \]
This result shows that at exactly 3 seconds, the particle is moving at a speed of \( \frac{1}{4} \text{ ft/sec} \, \) giving us a precise understanding of its motion at that moment.