91Ó°ÊÓ

The concept of velocity is crucial when studying motion. In physics, velocity is not just about how fast something moves but also in which direction. In mathematical terms, it is the derivative of the position function with respect to time. This means that velocity measures how position changes as time progresses. Given the problem, the velocity function is presented as:\[ v(t) = 9.8t + 5 \]- **Derivative**: This function tells us that the rate of change of position over time (velocity) is dependent on time \( t \). Specifically, as time increases, the velocity also changes. - **Understanding Change**: The linear nature of this specific velocity function indicates a constant acceleration, making it an important concept when relating motion to physics. This means as every moment passes, the object's speed increases by 9.8 units for every second, added to an initial velocity starting at 5 units/second at \( t = 0 \). Understanding this gives us insight into how integrating this function will help us find the general position at any given time.

Integration is a fundamental concept in calculus used to calculate quantities such as areas under curves or, as in this case, to determine the original function given its derivative. Here, our task is to derive the position function \( s(t) \) from the velocity function \( v(t) \).The velocity is given by:\[ v(t) = 9.8t + 5 \]- **Integral of Velocity**: To find the position function, integrate the velocity function: \[ s(t) = \int (9.8t + 5) \, dt \]- **Computing the Integral**: This breaks down into two simpler integrals: - \( \int 9.8t \, dt = 4.9t^2 \) - \( \int 5 \, dt = 5t \)Combining these results, the position function becomes a polynomial:\[ s(t) = 4.9t^2 + 5t + C \]where \( C \), the constant of integration, represents any initial position we need to consider later.

Initial conditions are necessary in calculus to solve for constants of integration that appear after indefinite integrals. They help us tailor general solutions to specific situations. In this problem, the initial condition is given as \( s(0) = 10 \).- **Application of Initial Condition**: By substituting \( t = 0 \) into the position equation, we solve for \( C \): \[ s(0) = 4.9(0)^2 + 5(0) + C = 10 \]- **Solving for \( C \)**: Simplifying the equation, we find that: \[ C = 10 \]With the constant found, substitute it back into the position function:\[ s(t) = 4.9t^2 + 5t + 10 \]This final function describes the position of the object at any time \( t \), grounded by the fact that it was at position 10 when \( t = 0 \). Initial conditions ensure that our mathematical models accurately reflect real-world scenarios.