91Ó°ÊÓ

An acceleration function describes how a particle's velocity changes over time. In our problem, the acceleration function is given by \( a(t) = 3e^{-0.2t} \). This tells us that the acceleration of the particle changes exponentially with time, specifically decreasing as \( t \) increases. The negative exponent indicates a decaying factor, meaning the rate of acceleration reduces as time passes. Understanding this function helps us predict how the particle's motion will evolve, setting the foundation for finding other related quantities like velocity and position.
Key points about acceleration functions:

• They are typically given as functions of time \( t \).
• They can tell us how quickly or slowly the velocity of the particle is changing.
• Knowing the function helps us derive velocity, through integration, since acceleration is the derivative of velocity.

Integration is a crucial mathematical tool used to find quantities like velocity and position when given an acceleration function. It essentially reverses differentiation.
When dealing with acceleration \( a(t) \), integrating it yields the velocity function \( v(t) \). Similarly, integrating the velocity function gives us the position function \( x(t) \). For our specific problem, integrating the acceleration \( a(t) = 3e^{-0.2t} \) with respect to \( t \) helped us find the velocity:
\[ v(t) = \int 3e^{-0.2t} \, dt = -15e^{-0.2t} + C \]
We integrate once more to find the particle's position:
\[ x(t) = \int (-15e^{-0.2t} + 15) \, dt = 75e^{-0.2t} + 15t + C_1 \]
Integrating allows us to build functions for quantities that are built on the changes described by acceleration or velocity functions.

• Integration is used to reverse differentiation.
• Enables the calculation of velocity from acceleration and position from velocity.
• Always results in an integration constant that needs to be determined with initial conditions.

Initial conditions are known values at a specific time that help solve for constants after integrating. They provide the details necessary to tailor general solutions to particular scenarios.
For our problem, initial conditions are given at \( t = 0 \) with \( v(0) = 0 \) and \( x(0) = 0 \). After integrating, these initial conditions are crucial to find the integration constants \( C \) and \( C_1 \). Applying the initial condition for velocity, we substitute into the velocity equation:
\[ 0 = -15e^0 + C \rightarrow C = 15 \]
Similarly, for the position function:
\[ 0 = 75e^0 + C_1 \rightarrow C_1 = -75 \]
Initial conditions ensure the solution fits the particular motion scenario of the problem, whether it's a starting point, velocity, or another state variable's initial value.

• Provide specific starting points or known values for velocity and position.
• Used to determine integration constants.
• Make the solution unique to the given problem situation.

Velocity and position functions describe the motion details of a particle after considering its acceleration and initial conditions. The velocity function is obtained first by integrating the acceleration function.
In this case, the velocity at any time \( t \) is represented by \( v(t) = -15e^{-0.2t} + 15 \). Calculating for a specific time, \( t = 0.5 \), the velocity becomes \( v(0.5) \approx 1.428 \text{ ft/s} \).
Similarly, a second integration gives the position function \( x(t) \). In the exercise, at \( t = 0.5 \), the position is \( x(0.5) \approx 0.36 \text{ ft} \).
Together, velocity and position functions give complete information about a particle's motion relative to time.

• Velocity is derived from acceleration and integration.
• Position is derived from velocity through a second integration.
• Provide insight into a particle's trajectory and timing.
• Critical for determining exact states of motion at any time \( t \).