91Ó°ÊÓ

The velocity function represents how fast an object moves at a particular moment in time. In this exercise, we are given the velocity function as \( v(t) = t^2 + 2 \).
This tells us that the object's speed depends on time, \( t \), and specifically that the speed is determined by the square of time plus an additional constant term.

• Time-dependent: The square term \( t^2 \) indicates that the velocity increases with time in a non-linear way.
• Constant term: The +2 indicates a baseline speed or starting speed component that is always present, adding to the speed irrespective of time.

Knowing the velocity function is essential as it allows us to find the position function, which will tell us where the object is in space at any given time. The velocity function is essentially the derivative of the position function, linking these two concepts directly in calculus.

The position function describes an object's location along a path over time. To find this function, we integrate the velocity function. In this case, the velocity function \( v(t) = t^2 + 2 \) is integrated to find the position function \( s(t) \).
The integral of the velocity function is performed as follows:

• Integrate term by term: For \( t^2 \), we get \( \frac{t^3}{3} \).
• The integration of the constant \( 2 \) results in \( 2t \).

The integral yields the expression \( s(t) = \frac{t^3}{3} + 2t + C \), where \( C \) is an unknown constant, called the constant of integration. The position function provides complete information about the object's movement, indicating exactly where the object is at any particular time \( t \).

When you integrate a function, a constant of integration \( C \) appears. This represents an unknown value that could have been any constant number, since the differentiation of a constant is zero. Without additional information, we would not know the exact value of \( C \).
In practice, we determine the value of \( C \) using an initial condition. For our exercise, we know that at \( t = 0 \), the position is 5. Substituting into the position function:

• Set \( t = 0 \) and \( s(t) = 5 \), which gives \( 5 = \frac{0^3}{3} + 2(0) + C \).
• Solve to find \( C = 5 \).

Substituting \( C = 5 \) into the position function gives us the complete solution for the object's position at any time \( t \). This step is crucial for making the position function specific to the scenario we are examining.

The initial condition is a crucial piece of information required to solve problems in calculus involving integration. It provides a specific snapshot of the system or scenario at a specific moment in time, allowing us to uniquely determine constants like \( C \).
In our situation, we used the fact that at \( t = 0 \), the object's position is 5. This condition connects the abstract formula we derived from integration with real-world specifics.

• This enables us to calculate the value of the constant of integration \( C \) based on the position function.
• Making the function \( s(t) \) = \( \frac{t^3}{3} + 2t + 5 \) precisely reflect the object's motion in the given context.

The initial condition effectively customizes our position function so it fits the particular characteristics of the problem we're analyzing. Without it, the position function would be too general, failing to provide useful predictive power for specific points in time.