Problem 58 - Algebra and Trigonometry

Question

Distance, Speed, and Time Two cyclists, 90 mi apart, start riding toward each other at the same time. One cycles twice as fast as the other. If they meet 2 h later, at what aver- age speed is each cyclist traveling?

Step-by-Step Solution

Answer

The slower cyclist travels at 15 mph and the faster at 30 mph.

Step 1: Understanding the Problem

We are given that two cyclists are 90 miles apart and start riding towards each other at the same time. One cyclist is twice as fast as the other. They meet after 2 hours. We need to find the average speed of each cyclist.

Step 2: Define Variables

Let the speed of the slower cyclist be \( x \) miles per hour. Since the other cyclist is twice as fast, his speed is \( 2x \) miles per hour.

Step 3: Calculate Distances Traveled

Since they meet after 2 hours, the distance traveled by the slower cyclist is \( 2x \times 2 = 2x \) miles. The distance traveled by the faster cyclist is \( 2 \times 2x = 4x \) miles.

Step 4: Set Up Equation

The total distance they cover when they meet is 90 miles. So, we set up the equation: \( 2x + 4x = 90 \).

Step 5: Solve Equation

Simplify the equation: \( 6x = 90 \). Solve for \( x \) by dividing both sides by 6: \( x = 15 \).

Step 6: Find Speeds

The slower cyclist rides at \( 15 \) miles per hour and the faster cyclist rides at \( 2x = 30 \) miles per hour.

Key Concepts

Average Speed Calculation

Calculating average speed is a common task when working with problems involving distance, speed, and time. In this scenario, we have two cyclists starting 90 miles apart and meeting after 2 hours. We need the calculation of their average speeds.

To find each cyclist's speed, first define their rates. The slower cyclist rides at a speed of \( x \) miles per hour. As the exercise reveals, the faster cyclist rides at double this speed, therefore at \( 2x \) miles per hour.

Since they meet after 2 hours, the speed of each cyclist represents the distance each covers over that time. You multiply their respective speeds by 2, revealing how far they travel:

• The slower cyclist: \( 2 \times x = 2x \) miles
• The faster cyclist: \( 2 \times 2x = 4x \) miles

Both add up to the total distance of 90 miles:

• The equation becomes: \( 2x + 4x = 90 \)

By solving this equation, you determine the average speed of each cyclist, which helps in understanding relative velocities when heading towards each other.

Algebraic Problem Solving

Algebra is a powerful tool in problem-solving. In cases like this exercise, variables allow you to handle unknown quantities effectively. Here, the unknowns are the speeds of the two cyclists.

Here's how algebra clarifies the problem:

• Let \( x \) be the speed of the slower cyclist in miles per hour.
• The faster cyclist, going twice as fast, is represented by \( 2x \).

The goal is to find the values of \( x \) and consequently \( 2x \). Given they cover a combined distance of 90 miles in 2 hours, their combined traveled paths must meet.

By setting up an equation:

• \( 2x + 4x = 90 \)

The next step is simplifying:

• Combine like terms: \( 6x = 90 \)
• Divide both sides by 6 to isolate \( x \): \( x = 15 \)

Algebra helps in understanding how parts of the problem connect, leading to an accurate solution.

Cyclist Speed Comparison

Comparing speeds can illustrate the relationship between two moving objects. In this case, understanding how one cyclist rides twice as fast as the other is fundamental.

The slower cyclist's speed is defined as \( x \) miles per hour. As established, the faster cyclist travels at \( 2x \) miles per hour. This comparison not only shows a simple multiplier relationship but also explains more profound dynamics of speed.

Over the same period, observe how the faster cyclist covers more ground:

• The slower cyclist covers \( 2x = 30 \) miles in 2 hours.
• The faster cyclist covers \( 4x = 60 \) miles in that time.

The problem effectively demonstrates the practical aspect of rates of speed, offering students insights into how different speeds influence travel times and distances.

Speed comparison like this can further be applied in analyzing scenarios like vehicle races, travel timetables, and athletic performances, enriching the understanding of rates.