91Ó°ÊÓ

In this exercise, distance and speed play crucial roles. Distance is simply how far an object moves, while speed is how quickly it covers that distance. When defining the variables, we use the length of the pool as distance and the speeds of the swimmers. If the faster swimmer swims at speed \(v_f\) and the slower one at \(v_s\), their speeds help us figure out how long it takes each swimmer to travel certain distances. The moments they meet at specific points (75 feet from one end and 25 feet from the other) allow us to use distance and speed concepts to establish relations between their travels. Understanding these basic principles helps us set up the ratio equations that connect the speeds with the distances traveled.

Ratios and proportions are fundamental in solving this problem. A ratio compares two quantities, showing how many times one value contains or is contained within the other. In this exercise, the ratio of the distances the swimmers travel corresponds to the ratio of their speeds. We denote this ratio as \(k\), where \(\frac{v_f}{v_s} = k\). For the first meeting, the ratio of the distances they swim, 75 feet for the faster swimmer and \(L - 75\) for the slower one, can be written as \(\frac{75}{L - 75} = k\). Similarly, for the second meeting, the distances traveled also form a ratio, allowing us to write \(\frac{2L - 25}{L + 25} = k\). By setting these equal, we establish a proportion that we can use to solve for the length of the pool.

A quadratic equation is a polynomial equation of the form \(ax^2 + bx + c = 0\). In this exercise, establishing the equality of ratios gives us a quadratic equation that we solve to find the pool's length. After setting \(\frac{75}{L - 75} = \frac{2L - 25}{L + 25}\), we cross-multiply to clear the fractions, resulting in \(\frac{75(L + 25) = (L - 75)(2L - 25)}\). Simplifying this yields a quadratic equation in the form \(2L^2 - 250L = 0\). We solve this by factoring out common terms, leading us to \(2L(L - 125) = 0\). Solving for \(L\), we find the length of the pool is 125 feet.

Algebra is the branch of mathematics dealing with symbols and the rules for manipulating those symbols. This exercise uses algebra extensively. Initially, we define the variables and establish relationships between them using ratios. Solving \(\frac{75}{L - 75} = \frac{2L - 25}{L + 25}\) involves the algebraic step of cross-multiplying and distributing terms. Simplifying gives us \75L + 1875 = 2L^2 - 175L - 1875\. Additional algebraic manipulation involves combining like terms and moving terms to one side to form the quadratic equation. Finally, we factor and solve the equation using algebraic techniques, thus determining the length of the pool.