91Ó°ÊÓ

The concept of acceleration is central to understanding motion along a straight line. Acceleration, often denoted as \( a(t) \), describes how the velocity of an object changes over time. In the given exercise, the acceleration function is \( a(t) = \frac{2t}{(t^2 + 1)^2} \). This function tells us that the acceleration of the object is dependent on time \( t \), and it incorporates terms that can change the rate of acceleration as \( t \) increases. To work with this function, it’s crucial to understand how the components interact:

• \( 2t \): This linear term indicates that acceleration is directly proportional to time.
• \( (t^2 + 1)^2 \): This part of the function moderates the growth of the acceleration, creating a more complex relationship by making the denominator larger as \( t \) grows.

By gaining insight into what the acceleration function looks like and how it's structured, you build a foundation for using it effectively to find other kinematic quantities.

Integration is a key technique used in calculus to find functions from their derivatives. When dealing with motion, integrating the acceleration function gives us the velocity function. In this case, we start with \( a(t) = \frac{2t}{(t^2 + 1)^2} \) and seek a way to integrate it.A common strategy for integration is a substitution method. Here, you can simplify the integral by letting \( u = t^2 + 1 \), which turns \( du = 2t \, dt \). This transforms the original integral into a simpler form: \( \int \frac{1}{u^2} \, du \). When you integrate with respect to \( u \), you obtain \( -\frac{1}{u} \), and by substituting back the value of \( u \), you find: \[ v(t) = -\frac{1}{t^2 + 1} + C \] Integration also helps to determine the position function from the velocity function. By integrating the velocity function \( v(t) \), you ascertain the position function \( s(t) \). Recognize that each integration step introduces a constant of integration (\( C \) or \( D \)), which reflects the initial conditions necessary to pinpoint the specific solution to the problem.

Initial conditions are essential to uniquely determining a solution in physics problems dealing with motion. They act as boundary values that allow the complete specification of functions describing motion.In this exercise, the initial velocity \( v(0) = 0 \) and the initial position \( s(0) = 0 \) are used to solve for the integration constants.

• When finding the velocity function, \( v(t) = -\frac{1}{t^2 + 1} + C \), apply the initial condition \( v(0) = 0 \) to find \( C \). Setting \( 0 = -\frac{1}{1} + C \) gives \( C = 1 \).
• For the position function, \( s(t) = -\arctan(t) + t + D \), use the initial condition \( s(0) = 0 \) to solve for \( D \). This results in \( D = 0 \) when substituting \( t = 0 \).

These initial conditions allow you to find specific solutions for both the velocity and position functions, tailoring them to the particular physical context of this motion.