Problem 40 - Algebra A Combined Function

Question

Solve the following. Suppose two cars leave Brinkley, Arkansas, at the same time, traveling in opposite directions. One car travels 8 mph faster than the other car. In 2.5 hours, the cars are 280 miles apart. Find the speed of each car.

Step-by-Step Solution

Answer

The slower car's speed is 52 mph, and the faster car's speed is 60 mph.

Step 1: Define Variables

Let the speed of the slower car be \( x \) mph. Then, the speed of the faster car would be \( x + 8 \) mph because it travels 8 mph faster.

Step 2: Establish Equation

The distance traveled by each car after 2.5 hours can be expressed as their rate multiplied by time. The slower car travels \( x \times 2.5 \) miles, and the faster car travels \( (x + 8) \times 2.5 \) miles. The sum of these distances is given as 280 miles. So, the equation becomes:\[ 2.5x + 2.5(x + 8) = 280 \]

Step 3: Simplify the Equation

Distribute and combine like terms in the equation:\[ 2.5x + 2.5x + 20 = 280 \]\[ 5x + 20 = 280 \]

Step 4: Solve for x

Subtract 20 from both sides to solve for \( x \):\[ 5x = 260 \]Now, divide both sides by 5:\[ x = 52 \]

Step 5: Find Speeds of Both Cars

The speed of the slower car is \( x = 52 \) mph. The speed of the faster car, which is 8 mph faster, is \( 52 + 8 = 60 \) mph.

Key Concepts

Distance-rate-time problems

Understanding distance-rate-time problems is a fundamental part of algebra and involves scenarios where you'll need to find one of these three elements when the other two are known.
In this context, when two objects move apart from each other, the overall distance is a sum of each individual path they cover.
For our exercise, imagine two cars moving in opposite directions.
The key here is knowing how each car contributes to the total distance over a given period.

• The slower car moves at a speed of \( x \) miles per hour,
• while the faster car travels at \( x + 8 \) miles per hour, given it is 8 miles per hour faster.

After 2.5 hours, these cars together cover 280 miles.
This distance-rate-time relationship can be expressed in the equation:\[ (2.5x) + (2.5(x + 8)) = 280 \]By understanding how time and speed relate to distance, you can solve for the unknown speeds.

Linear equations

Linear equations are mathematical statements that show the equality of two expressions involving variables and constants.
In the context of distance-rate-time problems, linear equations are a powerful tool for finding unknown values, such as the speed in this exercise.
Let's examine our equation:

• The equation \(2.5x + 2.5(x + 8) = 280 \) arises from combining distance elements of both cars.
• By simplifying, we collect like terms: \(5x + 20 = 280 \).

These steps turn a word problem into an algebraic expression that is easier to manage.
The focus of solving a linear equation is to isolate the variable, which gives insights into the scenario – here, it tells us the speeds.
Subtract 20 from each side and then divide by 5 to find \( x = 52 \), revealing the speed of the slower car, while \( x + 8 = 60 \) is the speed of the faster car.

Problem-solving strategies

Effective problem-solving strategies are crucial in tackling algebra word problems. Here’s how you can approach them efficiently:
1. **Understand the Problem:** Comprehend what the question is asking by identifying given values and unknowns. For our problem, you recognize the kinds of speeds and the total distance.2. **Define Variables:** Assign variables to unknowns, such as \( x \) for the slower car's speed.3. **Translate Words into Equations:** Convert textual information into mathematical equations. Terms like '8 mph faster' guide us how to setup equations.4. **Solve Step by Step:** Break down the equation solving process into manageable steps, ensuring you handle operations in sequence. For instance, combine terms first before isolating the variable.5. **Verify Your Solution:** Substitute back to check that solutions satisfy the original conditions, i.e., do the speeds lead to 280 miles in 2.5 hours?
Using these methods helps you systematically approach the problems, leading to a clearer understanding and more confident solutions.