Problem 22 - Algebra 2 and Trigonometry

Question

Anthony rode his bicycle to his friend's house, a distance of 1 mile. Then his friend's mother drove them to school, a distance of 12 miles. His friend's mother drove at a rate that is 25 miles per hour faster than Anthony rides his bike. If it took Anthony \(\frac{3}{5}\) of an hour to get to school, at what average rate does he ride his bicycle? (Use distance \(=\) time for each part of the trip to school.)

Step-by-Step Solution

Answer

Anthony rides his bicycle at 5 miles per hour.

Step 1: Understand the Problem

Anthony rides his bicycle for a distance of 1 mile to his friend's house and then his friend’s mother drives them 12 miles to school. We know it took a total of \( \frac{3}{5} \) of an hour to get from Anthony’s house to school at combined rates. We need to determine Anthony's bicycle riding rate.

Step 2: Define Variables and Equations

Let \( r \) be the rate at which Anthony rides his bicycle (in miles per hour). Therefore, the time it takes him to ride to his friend's house is \( \frac{1}{r} \) hours. His friend’s mother drives them 12 miles at a rate of \( r + 25 \) mph, so the time for that part is \( \frac{12}{r + 25} \) hours.

Step 3: Set Up the Equation

The total time for the trip is given as \( \frac{3}{5} \) hours. So, we set up the equation: \[ \frac{1}{r} + \frac{12}{r + 25} = \frac{3}{5} \]

Step 4: Solve for r

To solve the equation above, we need a common denominator, which is \( r(r+25) \):1. Multiply every term by \( 5r(r+25) \) to clear the fractions: \[ 5(r+25) + 60r = 3r(r+25) \]2. Simplify the equation: \[ 5r + 125 + 60r = 3r^2 + 75r \]3. Combine like terms: \[ 65r + 125 = 3r^2 + 75r \]4. Rearrange to form a quadratic equation: \[ 3r^2 + 75r - 65r - 125 = 0 \] \[ 3r^2 + 10r - 125 = 0 \]5. Use the quadratic formula, \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 3 \), \( b = 10 \), and \( c = -125 \). 1. Calculate the discriminant: \( b^2 - 4ac = 10^2 - 4 \times 3 \times (-125) = 100 + 1500 = 1600 \) 2. Find \( r \): \[ r = \frac{-10 \pm \sqrt{1600}}{6} \] \[ r = \frac{-10 \pm 40}{6} \] - Take the positive root: \( r = \frac{30}{6} = 5 \) mph

Step 5: Verify the Solution

Check if setting \( r = 5 \) mph satisfies the original total time:1. Time on a bike: \( \frac{1}{5} = 0.2 \) hours2. Time by car: \( \frac{12}{30} = 0.4 \) hours3. Verify total time: \( 0.2 + 0.4 = 0.6 \) hours, which is \( \frac{3}{5} \) hours.This confirms the solution is correct.

Key Concepts

Distance-Rate-Time Problems

Distance-rate-time problems are a classic theme in algebra that deals with how distance, rate (speed), and time are interrelated. These problems can be solved using the basic formula:

• Distance (d) = Rate (r) × Time (t)

This formula allows us to find the missing component if we know the other two. For example, if we know the distance and speed, we can find the time it takes to travel that distance by rearranging the formula to suggest: time = distance ÷ rate.

In the solution exercise involving Anthony, this relationship is applied twice: once for the biking segment and once for the car segment. Anthony's biking distance is 1 mile, and the driving distance is 12 miles.

• The time for biking: Since the distance is 1 mile, the time Anthony takes can be written as \( \frac{1}{r} \) where \( r \) is the biking rate in mph.
• The time for commuting by car: Relating to 12 miles, the equation becomes \( \frac{12}{r+25} \), where \( r + 25 \) mph is the driving speed of his friend’s mother.

Together, both times add up to the given total time of \( \frac{3}{5} \) hours, demonstrating how each segment contributes to the total journey time.

Rational Equations

Rational equations are equations that contain at least one fraction whose numerator and/or denominator is a variable expression. Solving these involves finding a common denominator and then eliminating the fractions by multiplying all terms by this common denominator.

In Anthony's case, the equation arose from combining time equations from two legs of the trip:

• \( \frac{1}{r} + \frac{12}{r+25} = \frac{3}{5} \)

The primary task was to find a common denominator, which was the product of the individual denominators: \( r(r+25) \). By multiplying both sides of the equation by this common denominator, we eliminate the fractions and simplify the equation to solve for \( r \). This method ensures the problem is handled accurately while considering both segments' contributions to travel time.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to solve for an unknown. Steps include combining like terms, factoring, and using formulas, like the quadratic formula, when necessary. In problems like Anthony's, algebraic manipulation is key to simplifying and solving the final quadratic equation.

• After multiplying and simplifying, the resulting equation was \( 3r^2 + 10r - 125 = 0 \).

To solve for \( r \), the quadratic formula was used: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 3 \), \( b = 10 \), and \( c = -125 \).
Understanding the quadratic formula is crucial for finding roots of the equation when factoring might be challenging. In this exercise, the positive solution, \( r = 5 \) mph, was applicable because it made sense in the real-world context of speed. Algebraic manipulation thus connects theoretical math with practical scenarios, making it a fundamental skill in solving equations like rational equations encountered in distance-rate-time problems.