Problem 8 - Introductory Algebra for College Students

Question

Each exercise is a problem involving motion. The joys of the Pacific Coast! You drive 90 miles along the Pacific Coast Highway and then take a 5 -mile run along a hiking trail in Point Reyes National Seashore. Your driving rate is nine times that of your running rate. If the total time for driving and running is 3 hours, find the average rate driving and the average rate running.

Step-by-Step Solution

Answer

The running rate is 5 miles per hour and the driving rate is 45 miles per hour.

Step 1: Formulate the equations

Let \( r \) represent the running rate in miles per hour, and \( d \) represent the driving rate, again in miles per hour. The total time spent driving and running is 3 hours – this can be expressed as a time = distance / rate equation, i.e., \(90/d + 5/r = 3\). The driving rate is nine times the running rate, or \(d = 9r\).

Step 2: Solve for the running rate

Substitute \( d = 9r \) into the first equation to get \( 90/9r + 5/r = 3 \). Simplifying gives \(10/r + 5/r = 3\). Multiply through by \( r \) to clear the fraction bringing the equation to \(10 + 5 = 3r\), which simplifies to \(3r = 15\), i.e. \( r = 5 \) miles per hour.

Step 3: Solve for the driving rate

Substitute \( r = 5 \) in the original equation \( d = 9r \), we get \( d = 9 × 5 = 45 \) miles per hour.

Key Concepts

Understanding the Distance-Rate-Time Relationship

One fundamental concept in solving motion problems is the relationship between distance, rate, and time. This relationship is summarized by the formula:\[\text{Distance} = \text{Rate} \times \text{Time}\]This means if you know any two of these values, you can solve for the third. For instance, if you're given the distance and rate, simply divide the distance by rate to find the time. This formula helps break down real-world problems into simpler mathematical equations that are easier to solve.
In the context of our problem, we're dealing with two different activities—driving and running—each with their own distances and rates. By representing each motion with its own time equation, like \(\frac{90}{d}\) for driving and \(\frac{5}{r}\) for running, we can combine these to find the total time spent, which is given as 3 hours. Remember, consistently keeping track of units (like hours and miles) and converting rates appropriately is key in avoiding mistakes.

Solving Equations to Find Rates

Solving equations is a critical part of assessing the conditions given in a problem and finding unknown values. When dealing with equations from motion problems, we use algebraic manipulations to make the unknown the subject of the formula. This typically involves isolation techniques that simplify the equation step-by-step.
For instance, after formulating our equations for the problem—one for each activity—we look for relationships between them. In this problem, we know that the driving rate \(d\) is nine times the running rate \(... d = 9r ...\). Such an equation allows us to express all variables in terms of one—here, in terms of \(r\) (the running rate).
After substitution, we simplify to isolate \(r\). Adding terms, clearing fractions, and then dividing through by coefficients are pivotal. This resulted in \(... r = 5 ...\) miles per hour for our running rate once this process concluded. Being meticulous in each calculation step ensures the accuracy of your solution.

Using Systems of Equations in Motion Problems

A system of equations arises when multiple equations describe different constraints of the same situation. These can be solved simultaneously to find the values of multiple variables. In this exercise, the system consists of:

• \(\frac{90}{d} + \frac{5}{r} = 3\) — This represents the total time for driving and running.
• \(d = 9r\) — This connects the rates by stating the driving rate is nine times the running rate.

To solve, we substitute one equation into another. For example, substitute \(... d = 9r ...\) into the first equation, allowing us to solve for \(r\) without needing \(d\) immediately. Once \(r\) is found, its value is plugged back to find \(d\), ensuring each equation holds true.
This method enhances your problem-solving capability, particularly in scenarios requiring the reconciliation of different physical principles. Identifying and using such systems is a skill that can handle both simple and complex real-world challenges efficiently.