91Ó°ÊÓ

Partial fraction decomposition is a method used to break down complex rational expressions into simpler fractions, which are easier to work with in calculus, especially during integration. In this exercise, we are given a velocity function in the form of a fraction, \[ v(t) = \frac{t^2 + 14t + 27}{(2t+1)(t+5)^2}. \]This fraction is decomposed into several simpler fractions, known as partial fractions. The goal of decomposition is to express the given function as a sum of simpler fractions, which makes it possible to integrate each component more easily. The general form that the decomposition takes is:\[ \frac{At + B}{2t+1} + \frac{Ct + D}{t+5} + \frac{E}{(t+5)^2}. \]Here’s a quick way to approach it:

• Identify and factor the denominator completely, if possible.
• Assign unknown coefficients to potential numerators corresponding to each distinct linear and quadratic term in the denominator.
• Equate the original fraction to the sum of these components and solve for the unknowns by substituting strategic values for the variable or comparing coefficients.

By successfully finding the constants (\(A, B, C, D, E\)), the decomposed function is ready for integration.

The velocity function for an object gives us crucial information about its motion. In this exercise, the velocity function is defined as a fraction of polynomials:\[ v(t) = \frac{t^2 + 14t + 27}{(2t+1)(t+5)^2}. \]The function describes how the velocity of the object varies with respect to time. As you analyze the behavior of such a function, you can determine periods of acceleration and deceleration by examining how the function changes:

• Velocity is positive, indicating forward motion.
• Velocity is negative, suggesting backward motion.
• Velocity function slopes upwards, indicating acceleration.
• Velocity function slopes downwards, indicating deceleration.

For this exercise, integrating the velocity function over a specific interval will give you the total distance traveled by the object. This is because the integral of velocity with respect to time provides the displacement, which, for this context, is considered as the distance traveled, skipping any backward motion verification.

Distance traveled refers to the total path an object covers over a specific period. In this problem, we are interested in calculating how far an object has traveled along a straight line over the first 2 seconds. This is done using the integral of the velocity function from \( t = 0 \) to \( t = 2 \).When we integrate the velocity function,\[ s = \int_{0}^{2} \frac{t^2 + 14t + 27}{(2t+1)(t+5)^2} \, dt, \]the result is the total distance the object has covered during this time frame. Integration literally sums up all those little bits of motion into one comprehensive measure of distance:

• The lower limit of the integral is \( t = 0 \), the start time.
• The upper limit of the integral is \( t = 2 \), the end time.
• Each infinitesimal segment under the velocity curve represents a tiny contribution to the total distance.

Through this integration, we acquire the full picture on how far the object has traveled.

A definite integral is used to calculate the net area under a curve within given bounds, capturing the total accumulation of a quantity. In this exercise, we use the definite integral to determine the distance traveled by the object over a time interval.For the velocity function \( v(t) = \frac{t^2 + 14t + 27}{(2t+1)(t+5)^2} \), we set up the definite integral as:\[ \int_{0}^{2} \frac{t^2 + 14t + 27}{(2t+1)(t+5)^2} \, dt. \]The process involves:

• Setting the limits of integration from \( t = 0 \) to \( t = 2 \).
• Calculating the integral, which may involve several techniques, including partial fraction decomposition.
• Evaluating the resulting expression at these two bounds and subtracting the results.

The outcome of this integral gives the total distance traveled, integrating the effect of all instantaneous velocities across the interval. It provides concrete numerical insight into physics and applications emphasizing how integration helps in solving real-world problems.