Problem 302 - Elementary Algebra

Question

In the following exercises, solve. Aaron left at 9:15 to drive to his mountain cabin 108 miles away. He drove on the freeway until \(10: 45,\) and then he drove on the mountain road. He arrived at 11:05. His speed on the freeway was three times his speed on the mountain road. Find Aaron's speed on the freeway and on the mountain road.

Step-by-Step Solution

Answer

Mountain road: 22.36 mph. Freeway: 67.08 mph.

Step 1 - Define Variables

Let Aaron's speed on the mountain road be denoted as \(s\) (in miles per hour). Then, his speed on the freeway is \(3s\).

Step 2 - Calculate Time Spent on Freeway

Aaron drove from 9:15 to 10:45 on the freeway, which is 1.5 hours (1 hour and 30 minutes).

Step 3 - Calculate Time Spent on Mountain Road

Aaron drove from 10:45 to 11:05 on the mountain road, which is 20 minutes or \(\frac{1}{3}\) hours.

Step 4 - Set Up Distance Equation

The total distance to the cabin is 108 miles. The distance Aaron traveled on the freeway is \(3s \times 1.5\), and the distance on the mountain road is \(s \times \frac{1}{3}\). The equation is: \[ 3s \times 1.5 + s \times \frac{1}{3} = 108 \]

Step 5 - Solve the Equation

First, simplify the equation: \[ 4.5s + \frac{s}{3} = 108 \] Combine like terms by finding a common denominator: \[ 4.5s + \frac{1s}{3} = 4.5s + 0.333s = 4.833s = 108 \] Solve for \(s\): \[ s = \frac{108}{4.833} \approx 22.36 \text{ miles per hour} \]

Step 6 - Calculate Freeway Speed

Since the speed on the freeway is three times the speed on the mountain road, \[ 3s \approx 3 \times 22.36 \approx 67.08 \text{ miles per hour} \]

Key Concepts

Solving Equations

Solving equations involves finding the value of the variable that makes the equation true.
To solve an equation, you need to perform operations that will isolate the variable on one side of the equation.
Here is how it's done step by step:
1. Simplify both sides of the equation if needed by combining like terms.
2. Use addition or subtraction to move terms that don’t contain the variable to the other side.
3. Use multiplication or division to solve for the variable.
In this problem, we started with the equation \[ 3s \times 1.5 + s \times \frac{1}{3} = 108 \] and simplified it to \[ 4.833s = 108 \] This allowed us to solve for the variable: \[ s = \frac{108}{4.833} \ s \approx 22.36 \] Breaking down each step helps in understanding how the operations transform the equation and why the solution is valid.

Distance-Rate-Time Relationship

The distance-rate-time relationship is a fundamental concept in algebra and helps to solve word problems involving motion.
The formula is:

\[ Distance = Rate \times Time \] In simpler words, the distance traveled is equal to the speed (rate) multiplied by the time spent traveling.

In this exercise, Aaron's trip to his cabin required calculating how far he traveled at different speeds.
His speed on the freeway is given as three times his speed on the mountain road.

By breaking down the total travel time, we know:

• He drove 1.5 hours on the freeway
• He drove \( \frac{1}{3} \) hours on the mountain road

This allows us to use the distance formula for each part of his journey:

• Freeway: \( 3s \times 1.5 \)
• Mountain road: \( s \times \frac{1}{3} \)

Summing these distances and setting them equal to the total distance (108 miles) led us to our distance equation.

Variable Definition

Defining variables is the first step in solving word problems in algebra.
Variables represent unknown quantities and help in setting up equations to solve problems.
For instance, in this problem:

• Let \( s \) be the speed on the mountain road in miles per hour.
• Then his speed on the freeway is \( 3s \) because it is stated that it is three times the speed on the mountain road.

By clearly defining the variables, we created a basis for our equations.
The defined variables were then used to express other related quantities in the problem
(like distances traveled on different parts of the road).
This systematic assignment of variables simplifies the process of building and solving equations.