Problem 49 - Precalculus: Mathematical for Calculus

Question

A woman keeps fit by bicycling and running every day. On Monday she spends \(\frac{1}{2}\) hour at each activity, covering a total of \(12 \frac{1}{2} \mathrm{mi} .\) On Tuesday, she runs for 12 min and cycles for 45 min, covering a total of 16 mi. Assuming her running and cycling speeds don't change from day to day, find these speeds.

Step-by-Step Solution

Answer

Running speed is 5 mph, cycling speed is 20 mph.

Step 1: Define Variables

Let \( r \) be the woman's running speed in miles per hour, and \( c \) be her cycling speed in miles per hour.

Step 2: Convert Time into Hours

Convert all time measurements into hours since speed is in miles per hour. On Monday, she spends \( \frac{1}{2} \) hour running and \( \frac{1}{2} \) hour cycling. On Tuesday, 12 minutes is \( \frac{1}{5} \) hour and 45 minutes is \( \frac{3}{4} \) hour.

Step 3: Write Equations for Monday

On Monday, she covers \( 12 \frac{1}{2} = \frac{25}{2} \) miles. The equation for Monday is: \[ \frac{1}{2}r + \frac{1}{2}c = \frac{25}{2} \].

Step 4: Write Equations for Tuesday

On Tuesday, she covers 16 miles. The equation for Tuesday is: \[ \frac{1}{5}r + \frac{3}{4}c = 16 \].

Step 5: Solve the System of Equations

We have two equations: \( \frac{1}{2}r + \frac{1}{2}c = \frac{25}{2} \) and \( \frac{1}{5}r + \frac{3}{4}c = 16 \). Solve the first equation for one variable: \( r + c = 25 \). Substitute \( r = 25 - c \) in the second equation, leading to solving for \( c \): \[ \frac{1}{5}(25-c) + \frac{3}{4}c = 16 \].

Step 6: Simplify and Solve for Cycling Speed

Simplify the second equation: \( 5 - \frac{1}{5}c + \frac{3}{4}c = 16 \). Combine terms, \( \frac{11}{20}c = 11 \), leading to \( c = 20 \).

Step 7: Calculate Running Speed

Substitute \( c = 20 \) back into \( r + c = 25 \): \[ r + 20 = 25 \], thus \( r = 5 \).

Step 8: Verify the Solution

Calculate using both speeds in the original distances and times to ensure correctness. Verify Monday equation: \( \frac{1}{2}(5) + \frac{1}{2}(20) = \frac{25}{2} \), and Tuesday: \( \frac{1}{5}(5) + \frac{3}{4}(20) = 16 \). Both check out.

Key Concepts

System of Equations

A system of equations involves finding solutions for multiple equations that share common variables. When you have two or more equations, each with the same unknowns, you can use algebra to determine the values of these unknowns that satisfy all the equations simultaneously. In this exercise, we have two equations based on the woman's activities on Monday and Tuesday. Each equation relates her speeds to the distance covered in a given time.
The equations are:

• Monday: \( \frac{1}{2}r + \frac{1}{2}c = \frac{25}{2} \) (where \( r \) is her running speed and \( c \) is her cycling speed).
• Tuesday: \( \frac{1}{5}r + \frac{3}{4}c = 16 \).

To solve these, you can use substitution or elimination methods. In this example, substitution was employed by solving the first equation for one variable and substituting that value into the second equation. This technique helps to reduce complexity, isolating and focusing on one unknown at a time.

Distance-Speed-Time Relationship

Understanding the relationship between distance, speed, and time is fundamental in solving problems involving motion. The formula is simple: \( \text{Distance} = \text{Speed} \times \text{Time} \). In this case, the woman covers different distances by running and cycling at constant speeds on different days. By knowing either speed or time and the total distance, you can calculate the unknown value.
For example:

• On Monday, she spends \( \frac{1}{2} \) hour at each activity, and the equation is \( \frac{1}{2}r + \frac{1}{2}c = \frac{25}{2} \).
• On Tuesday, the times are \( \frac{1}{5} \) hour running and \( \frac{3}{4} \) hour cycling, forming \( \frac{1}{5}r + \frac{3}{4}c = 16 \).

Recognizing these relationships allows you to formulate equations that describe the situation accurately. It's crucial to convert time into hours consistently, as speed is measured in miles per hour.

Algebraic Manipulation

Algebraic manipulation is essential for solving equations like those in this exercise. Once you have the equations set up, you need to manipulate them to isolate the variable you're solving for.
Here’s a simplified procedure:

• Rewrite the first equation in terms of one variable (e.g., \( r + c = 25 \)).
• Use substitution to replace one variable in the second equation (e.g., substitute \( r = 25 - c \) into the second equation).
• Solve this simplified equation to find the value of one variable (\( c = 20 \)).
• Substitute this value back into one of the original equations to find the other unknown (\( r = 5 \)).

This step-by-step manipulation process transforms complex equations into simpler ones, revealing the solutions progressively. It’s a powerful skill that unlocks the ability to solve various algebraic equations.