Problem 28 - Algebra A Combined Function

Question

A semi-truck travels 300 miles through the flatland in the same amount of time that it travels 180 miles through mountains. The rate of the truck is 20 miles per hour slower in the mountains than in the flatland. Find both the flatland rate and mountain rate.

Step-by-Step Solution

Answer

Flatland rate: 50 mph; Mountain rate: 30 mph.

Step 1: Define Variables

First, establish the variables for the truck's rates. Let \( r \) be the rate in the flatlands (in miles per hour), and \( r - 20 \) be the rate in the mountains, since it is 20 miles per hour slower.

Step 2: Set Up Equations Based on the Problem

The time taken to travel a distance is calculated as \( \frac{\text{Distance}}{\text{Rate}} \). Given that the time taken in the flatland and mountains is the same, equate the time expressions: \[ \frac{300}{r} = \frac{180}{r - 20} \]

Step 3: Cross-Multiply to Eliminate Fractions

To solve the equation, cross-multiply to get rid of the fractions: \[ 300(r - 20) = 180r \]

Step 4: Simplify the Equation

Expand and simplify the equation: \[ 300r - 6000 = 180r \]. This simplifies to \[ 120r = 6000 \] after moving terms involving \( r \) to one side.

Step 5: Solve for the Flatland Rate

Divide both sides by 120 to solve for \( r \): \[ r = \frac{6000}{120} = 50 \]. So, the rate in the flatlands is 50 miles per hour.

Step 6: Determine the Mountain Rate

Since the mountain rate is 20 miles per hour less than the flatland rate, calculate it as: \( r - 20 = 50 - 20 = 30 \). So, the mountain rate is 30 miles per hour.

Key Concepts

Rate of Change

In algebra, the "rate of change" refers to how one quantity changes in relation to another. It is essentially a measure of how much something changes over a specified interval.This concept is critical when dealing with motion, as with our semi-truck problem. Here, the rate of change is about how fast the truck travels. In this context, it's expressed as miles per hour.

The key idea is:

• In the flatlands, the truck travels at a rate of \( r \),
• In the mountains, it's slower at a rate of \( r - 20 \).

This change in the rate is consistent with everyday experiences (like slowing down in mountainous terrain). Understanding the rate of change helps us set up equations reflecting these physical scenarios.

Equations and Expressions

In the realm of algebra, equations and expressions are fundamental. An equation is a statement that asserts the equality of two expressions. Before solving an algebraic problem, defining variables and creating equations based on the relationships outlined in the problem is important.

In the given scenario:

• The truck's rate in the flatlands is \( r \),
• The rate in the mountains is \( r - 20 \) because it's slower by 20 miles per hour.

Building our equation starts with the time formula: Time = \( \frac{\text{Distance}}{\text{Rate}} \). Since the time for both distances (300 miles in the flatlands and 180 miles in the mountains) is equal, we equate the two expressions:
\[ \frac{300}{r} = \frac{180}{r - 20} \] This equation uses the given rates and distances to express the problem mathematically.

Distance-Rate-Time Problems

Distance-rate-time problems are common in algebra and are solved using the formula \( \,Distance = Rate \, \times \, Time\). This formula is manipulated in different ways to solve for one of the three variables, depending on the problem's requirements.

In the semi-truck scenario, we have:

• Distance covered in flatland and mountains
• Rates: \( r \) for flatlands and \( r - 20 \) for mountains
• Equal times for both regions

By setting the time expressions equal, \( \frac{300}{r} = \frac{180}{r - 20} \), the problem sees us solve for the flatland rate \( r \) using algebraic manipulation (cross-multiplying and simplifying). Finally, we determine the mountain rate by subtracting 20 from the flatland rate. These steps illustrate the power of algebra in unraveling real-world problems.