Problem 16 - Algebra 2 and Trigonometry

Question

Megan traveled 165 miles to visit friends. On the return trip she was delayed by construction and had to reduce her average speed by 22 miles per hour. The return trip took 2 hours longer. What was the time and average speed for each part of the trip?

Step-by-Step Solution

Answer

Megan's average speed was 55 mph on the way there and 33 mph on the return trip. The trip took 3 hours there and 5 hours on the return.

Step 1: Define Variables

Let \( x \) be the average speed (in miles per hour) on the way to visit the friends. The time taken for this trip is \( \frac{165}{x} \) hours. On the return trip, the average speed was \( x - 22 \) miles per hour, and the time taken was \( \frac{165}{x - 22} \) hours.

Step 2: Set Up the Equation

According to the problem, the return trip took 2 hours longer than the trip to visit the friends. Therefore, we can write the equation: \[\frac{165}{x - 22} = \frac{165}{x} + 2\]

Step 3: Solve the Equation for x

To solve for \( x \), first eliminate the fractions by multiplying through by the least common denominator, which is \( x(x-22) \). This gives us \[165x = 165(x-22) + 2x(x-22)\] Simplify and solve this quadratic equation to find the value of \( x \).

Step 4: Simplify and Calculate

Expand and simplify the equation: \[165x = 165x - 3630 + 2x^2 - 44x\]This reduces to: \[2x^2 - 44x - 3630 = 0\] Divide the entire equation by 2: \[x^2 - 22x - 1815 = 0\]. Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) with \(a = 1\), \(b = -22\), \(c = -1815\).

Step 5: Calculate Quadratic Formula

The discriminant \(b^2 - 4ac = (-22)^2 - 4(1)(-1815) = 484 + 7260 = 7744\). Compute \(x = \frac{22 \pm \sqrt{7744}}{2}\). Since speed can't be negative, use the positive root: \(x = \frac{22 + 88}{2} = 55\).

Key Concepts

Average Speed

Average speed is a fundamental concept in motion problems. In this exercise, we determine the speed at which Megan traveled to her friends. The formula for average speed is the total distance traveled divided by the time taken to travel that distance. It is represented mathematically as:

• Average Speed = \( \frac{\text{Total Distance}}{\text{Total Time}} \)

For Megan's trip, we initially denote her average speed on the trip to her friends as \( x \). Thus, the average speed is critical in understanding how differences in speed affect travel time. In this scenario, a decrease in speed on the return trip due to construction highlights the sensitivity of travel time to changes in speed.

Time Calculation

Time calculation is crucial for determining how long a journey takes. In the given problem, the time for Megan's trip to her friends is calculated using the formula:

• Time = \( \frac{\text{Distance}}{\text{Speed}} \)

For the trip there, the time is \( \frac{165}{x} \), where 165 is the distance and \( x \) is the speed. For the return trip, the speed is reduced by 22 mph, leading to:

• Return Time = \( \frac{165}{x - 22} \)

Her return trip takes 2 hours longer, forming the basis of the equation needed to solve for both speed and time. Understanding time calculation allows us to explore how delays affect overall travel time.

Distance and Speed Problems

Distance and speed problems often involve finding unknown variables like speed or time based on given conditions. These problems typically involve forming equations that relate distance, speed, and time. In this exercise, Megan's return trip presents a classic case:

• The return trip's reduced speed increases the time taken due to construction.
• We use an equation to represent this relationship: \( \frac{165}{x-22} = \frac{165}{x} + 2 \).

Such problems require an understanding of how changing one element (like speed) affects others (like time). They are solved by setting up equations that mirror real-world scenarios, making them practical and useful for testing problem-solving skills.

Quadratic Formula

The quadratic formula is a must-know tool in algebra to find variable values in equations of the form \( ax^2 + bx + c = 0 \). In Megan's case, solving for the average speed involved forming and solving a quadratic equation.Substituting the values into the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we set:

• \( a = 1 \)
• \( b = -22 \)
• \( c = -1815 \)

The discriminant, \( b^2 - 4ac \), determines the nature of the solutions. Here, it equals 7744, a perfect square, giving real and unequal roots:

• \( x = \frac{22 + 88}{2} = 55 \)

The quadratic formula thus enables calculation of the average speed, showing how algebraic tools apply to real-life contexts.