Problem 226 - Elementary Algebra

Question

Felecia left her home to visit her daughter driving \(45 \mathrm{mph}\). Her husband waited for the dog sitter to arrive and left home twenty minutes \((1 / 3\) hour) later. He drove 55 mph to catch up to Felecia. How long before he reaches her?

Step-by-Step Solution

Answer

Felecia's husband reaches her after 1.5 hours.

Step 1: Understand the Problem

First, identify the key information: Felecia drives at 45 mph, her husband leaves 20 minutes later (which is \(\frac{1}{3}\) hour later) and drives at 55 mph.

Step 2: Calculate Felecia’s Head Start

Felecia drives for \(\frac{1}{3}\) hour before her husband starts. Calculate the distance she covers in that time: \[ \text{Distance} = \text{Speed} \times \text{Time} = 45 \, \text{mph} \times \frac{1}{3} \, \text{hour} = 15 \, \text{miles} \]

Step 3: Set Up an Equation for Time

Let \ t \ be the time in hours it takes Felecia's husband to catch up to her after he starts driving. Felecia continues driving during this time. Write the distance equations: \[ \text{Distance}_{\text{Felecia}} = 45 \, \text{mph} \times (t + \frac{1}{3}) \] \[ \text{Distance}_{\text{Husband}} = 55 \, \text{mph} \times t \] Since the husband catches up to Felecia, both distances will be equal.

Step 4: Solve the Equation

Equate the distances and solve for \ t \ : \[ 45 \, (t + \frac{1}{3}) = 55 \, t \] Distribute and simplify: \[ 45t + 15 = 55t \] \[ 15 = 10t \] Divide by 10: \[ t = 1.5 \]

Step 5: Interpret the Result

The result indicates that it took Felecia’s husband 1.5 hours after he started driving to catch up with her.

Key Concepts

distance-time relationship

Understanding the relationship between distance, time, and speed is key to solving travel problems in algebra.
The distance-time relationship can be summarized by the formula: \( \text{Distance} = \text{Speed} \times \text{Time} \). This means that how far an object travels (distance) depends on how fast it is going (speed) and how long it has been traveling (time).
Let's look at Felecia's journey to her daughter's house:

• She drives at a constant speed of 45 mph.
• The time she has been driving when her husband starts is \( \frac{1}{3} \text{ hour} \).
• The distance she covers in that time is \( 45 \text{ mph} \times \frac{1}{3} \text{ hour} = 15 \text{ miles} \).

This head start plays a significant role, as her husband needs to cover this initial 15 miles to catch up.

speed and travel time

In this exercise, we deal with two different speeds and travel times. Felecia drives at 45 mph, while her husband drives faster at 55 mph to catch up.
Here's a breakdown:

• Felecia's speed is slower, but she has a head start of 20 minutes.
• Her husband's speed is higher, and he starts 20 minutes later.

To solve how long it will take her husband to catch up with her, we need to understand how their travel times relate.
We can denote the time from the moment her husband starts driving until he catches up with her as \( t \). During this period, Felecia continues driving, so her travel time will be \( t + \frac{1}{3} \) hour.
These two pieces of the puzzle allow us to set up an equation to find \( t \).

algebraic equations

The core of this problem is setting up and solving an algebraic equation.
We have to equate the distances traveled by both Felecia and her husband: \( 45(t + \frac{1}{3}) = 55t \). Simplifying this equation step-by-step is critical:

• Distribute the 45: \(45t + 15 = 55t \).
• Move terms involving \( t \) to one side by subtracting \( 45t \) from both sides: \( 15 = 10t \).
• Divide by 10: \( t = 1.5 \) hours.

This result tells us that it takes her husband 1.5 hours to catch up after he starts driving.
Interpreting these results is also crucial. It wraps up our understanding of the problem and ensures we grasp how travel times and speeds interplay.
It emphasizes the practical application of algebraic equations in real-world scenarios, like travel problems.