Calculating Distance Traveled A Step By Step Guide
In the realm of mathematics and physics, understanding the relationship between speed, distance, and time is fundamental. This article delves into a specific scenario involving Margarita's travel, where her speed is expressed as an algebraic fraction, and we aim to determine the distance she covers in a given time. This problem not only reinforces basic physics principles but also hones algebraic simplification skills, crucial for various mathematical applications. Our primary focus will be on meticulously breaking down each step, ensuring a comprehensive understanding of the solution. We will start by understanding the problem statement, identifying key variables, and then proceed to apply the distance formula. The challenge here lies in simplifying the algebraic expressions involved, which requires factoring polynomials and canceling common factors. This process highlights the interconnectedness of algebra and physics, demonstrating how algebraic techniques are essential tools for solving real-world problems. This exploration is beneficial for students learning algebra, physics enthusiasts, and anyone interested in practical mathematical applications. Through this article, readers will gain a clearer understanding of how to approach similar problems and enhance their problem-solving abilities in related areas. Let’s embark on this mathematical journey to unravel the distance Margarita covers.
Understanding the Problem
To accurately calculate the distance Margarita travels, we must first dissect the given information. Margarita's speed is defined as $\frac{s-4}{s^2-9s+20}$ miles per hour, and she travels for $ rac{s-5}{4}$ hours. The fundamental principle connecting these quantities is the formula: Distance = Speed × Time. This formula forms the bedrock of our calculation. The challenge, however, lies in the algebraic expressions representing speed and time. These expressions involve polynomials, necessitating simplification before direct multiplication. Specifically, the denominator of the speed expression, $s^2 - 9s + 20$, is a quadratic polynomial, which we need to factor. Factoring this polynomial will reveal common factors with the numerator and the time expression, allowing us to simplify the overall calculation. This process underscores the importance of algebraic manipulation in solving problems involving variable quantities. By simplifying the expressions, we not only make the calculation easier but also gain deeper insights into the relationships between the variables. Understanding the problem thoroughly sets the stage for a methodical approach to the solution, ensuring accuracy and clarity in each step. This groundwork is essential for anyone tackling mathematical problems involving algebraic expressions and physical quantities.
Applying the Distance Formula
With a clear understanding of the problem, we now apply the distance formula: Distance = Speed × Time. Substituting the given values, we have:
Distance = $\frac{s-4}{s^2-9s+20} \times \frac{s-5}{4}$
This equation represents the core of our problem. The next crucial step is to simplify this expression. The complexity arises from the quadratic polynomial in the denominator. To simplify, we must factor the quadratic expression $s^2 - 9s + 20$. Factoring involves finding two binomials that, when multiplied, yield the quadratic polynomial. This process is a fundamental skill in algebra and is essential for simplifying rational expressions. Once we factor the quadratic, we can look for common factors in the numerator and denominator, which can be canceled out. This cancellation is the key to simplifying the expression and arriving at the final answer. The application of the distance formula is straightforward, but the algebraic manipulation required to simplify the expression is where the challenge lies. Successfully navigating this algebraic terrain will lead us to the solution, showcasing the power of algebra in solving practical problems.
Factoring the Quadratic Expression
The heart of simplifying the distance expression lies in factoring the quadratic polynomial $s^2 - 9s + 20$. Factoring a quadratic expression involves rewriting it as a product of two binomials. To factor $s^2 - 9s + 20$, we look for two numbers that multiply to 20 (the constant term) and add up to -9 (the coefficient of the linear term). These numbers are -4 and -5. Therefore, we can write the quadratic expression as:
This factorization is a crucial step because it reveals common factors with the other parts of the distance expression. Specifically, we see the factors $(s - 4)$ and $(s - 5)$, which also appear in the numerator of the speed expression and the time expression, respectively. This allows us to simplify the overall expression by canceling out these common factors. The ability to factor quadratic expressions is a fundamental skill in algebra, with applications in various areas of mathematics and beyond. It allows us to rewrite expressions in a more manageable form, making it easier to solve equations, simplify expressions, and understand the underlying relationships between variables. In this case, factoring the quadratic is the key to unlocking the simplification of the distance expression and arriving at the final answer.
Simplifying the Expression
With the quadratic expression factored, we can now substitute it back into the distance equation and simplify:
Distance = $\frac{s-4}{(s-4)(s-5)} \times \frac{s-5}{4}$
Now we can cancel the common factors. We have $(s - 4)$ in the numerator and denominator of the first fraction, and $(s - 5)$ in the numerator of the second fraction and the denominator of the first fraction. Canceling these common factors, we get:
Distance = $\frac{\cancel{(s-4)}}{\cancel{(s-4)}\cancel{(s-5)}} \times \frac{\cancel{(s-5)}}{4}$
This simplifies to:
Distance = $\frac{1}{4}$
The simplified expression reveals that the distance Margarita covers is $ rac{1}{4}$ miles, regardless of the value of s (as long as s is not 4 or 5, which would make the original expression undefined). This result is a testament to the power of algebraic simplification. By factoring and canceling common factors, we were able to transform a complex expression into a simple, constant value. This highlights the elegance and efficiency of algebraic techniques in solving mathematical problems. The simplification process not only provides the answer but also offers insights into the underlying relationships between the variables. In this case, it shows that the distance is independent of s, a perhaps surprising result given the initial complexity of the expression.
Expressing the Answer
After simplifying the expression, we arrive at the answer: Margarita covers a distance of $\frac{1}{4}$ miles. This answer is already in its simplest form, as 1 and 4 have no common factors other than 1. Expressing the answer in its simplest form is a crucial step in mathematical problem-solving. It ensures clarity and avoids any ambiguity. In this case, the simplified fraction $ rac{1}{4}$ is easily understood and represents a quarter of a mile. The journey from the initial complex expression to this simple answer underscores the importance of algebraic manipulation. Factoring, canceling common factors, and simplifying are essential skills that enable us to solve problems efficiently and accurately. This problem exemplifies how algebraic techniques can transform seemingly complex scenarios into straightforward solutions. The final answer, $\frac{1}{4}$ miles, provides a clear and concise representation of the distance Margarita covers, highlighting the power and elegance of mathematics in solving real-world problems. This result not only answers the specific question but also reinforces the broader principles of problem-solving and algebraic simplification.
In conclusion, this exercise demonstrates the application of algebraic principles to solve a practical problem involving distance, speed, and time. We began with a complex expression for speed and a given time, and through careful algebraic manipulation, we determined that Margarita covers a distance of $\frac{1}{4}$ miles. The key steps in this process included understanding the distance formula, factoring the quadratic expression, canceling common factors, and simplifying the resulting fraction. Each of these steps highlights the importance of algebraic skills in solving mathematical problems. The ability to factor polynomials, simplify rational expressions, and apply fundamental formulas is crucial for success in mathematics and related fields. This problem serves as a valuable example for students learning algebra and physics, illustrating how these subjects are interconnected and how algebraic techniques can be used to solve real-world problems. The process of simplifying the expression not only leads to the answer but also provides insights into the relationships between the variables. The final, simplified answer of $\frac{1}{4}$ miles showcases the power and elegance of mathematics in transforming complex scenarios into clear and concise solutions. By mastering these algebraic techniques, students can enhance their problem-solving abilities and gain a deeper understanding of the mathematical principles that govern the world around us. This journey through the problem-solving process underscores the value of persistence, attention to detail, and a methodical approach in tackling mathematical challenges.
Keywords: Algebraic simplification, distance calculation, factoring quadratics, speed and time, mathematical problem-solving.
Key phrases: distance formula, simplifying expressions, canceling common factors, simplest form, quadratic polynomial.
