Problem 17 - A Graphical Approach to College Algebra

Question

$$\begin{aligned} &\text {Solve each problem analytically, and support your solution}\\\ &\text {graphically.} \end{aligned}$$ Motion\(\quad\) A car went 372 miles in 6 hours, traveling part of the time at 55 miles per hour and part of the time at 70 miles per hour. How long did the car travel at each speed?

Step-by-Step Solution

Answer

The car traveled 3.2 hours at 55 mph and 2.8 hours at 70 mph.

Step 1: Define Variables

Let the time traveled at 55 miles per hour be \( t_1 \) hours and the time traveled at 70 miles per hour be \( t_2 \) hours. We know from the problem that the total travel time is 6 hours, so we can write the equation: \( t_1 + t_2 = 6 \).

Step 2: Set Up Equation for Distance

Define the expressions for the distances traveled with each speed. The distance at 55 mph is \( 55t_1 \), and the distance at 70 mph is \( 70t_2 \). Since the total distance is 372 miles, we write: \( 55t_1 + 70t_2 = 372 \).

Step 3: Solve the System of Equations

We have two equations: 1. \( t_1 + t_2 = 6 \) 2. \( 55t_1 + 70t_2 = 372 \)From the first equation, express \( t_1 = 6 - t_2 \), and substitute it into the second equation: \( 55(6-t_2) + 70t_2 = 372 \).

Step 4: Simplify and Solve for \( t_2 \)

Expand and simplify the equation: \( 330 - 55t_2 + 70t_2 = 372 \) Combine like terms to get: \( 15t_2 = 42 \) Solve for \( t_2 \): \( t_2 = \frac{42}{15} = 2.8 \text{ hours} \).

Step 5: Solve for \( t_1 \)

Substitute \( t_2 = 2.8 \) back into the equation \( t_1 = 6 - t_2 \): \( t_1 = 6 - 2.8 = 3.2 \text{ hours} \).

Step 6: Conclusion and Verification

The car traveled 3.2 hours at 55 miles per hour and 2.8 hours at 70 miles per hour. Verify by calculating the total distance: \( 55 \times 3.2 + 70 \times 2.8 = 176 + 196 = 372 \text{ miles} \), which matches the given total distance.

Key Concepts

Understanding Motion Problems

Motion problems are a fascinating part of algebra that involve finding unknown variables related to speed, distance, and time. They require using the mathematical relationship between these three elements to find a solution. In many cases like the one provided, you will have to determine how long a car or another vehicle was traveling at two different speeds to cover a specific distance.

• The key is to identify all the given information clearly. This includes speeds, total distance, and total time.
• You need to define variables that represent the unknown quantities. These might be the time spent traveling at each speed.

Motion problems are practical and help us understand real-world scenarios effectively.

Solving Systems of Equations

A powerful tool to solve motion problems is setting up a system of equations. Once all information is translated into equations, we can solve them simultaneously to find unknowns. The following are the general steps:

• Define the equations: Start by identifying each piece of given information and relate it in terms of equations.
• Substitution or elimination: Use one of these methods to solve the equations. In our example, substitution is used where one variable is expressed in terms of another.

Systems of equations are essential in solving not just motion problems but many other algebraic challenges. They provide the structure needed to connect various pieces of information.

Distance-Rate-Time Relationship

The relationship between distance, rate (speed), and time is foundational in algebraic problem-solving. This is often represented in the formula:

• \[ ext{Distance} = ext{Rate} imes ext{Time} \]

This simple yet powerful equation is the cornerstone of solving motion problems. In the context of the given problem:

• We use the formula twice: once for each speed the car traveled.
• By combining these expressions, we can form the system of equations that help unravel the time spent at each speed.

Understanding and applying this relationship allows for precise and clear problem-solving.