Problem 71 - Precalculus Mathematics for Calculus

Question

John and Mary leave their house at the same time and drive in opposite directions. John drives at \(60 \mathrm{mi} / \mathrm{h}\) and travels \(35 \mathrm{mi}\) farther than Mary, who drives at \(40 \mathrm{mi} / \mathrm{h}\). Mary's trip takes 15 min longer than John's. For what length of time does each of them drive?

Step-by-Step Solution

Answer

John drives for 2.25 hours, and Mary drives for 2.5 hours.

Step 1: Identify the Variables

Let's call the time John spends driving as \( t \) hours. Then, the distance John travels is \( 60t \) miles since his speed is 60 mi/h. For Mary, her time of driving is \( t + \frac{15}{60} = t + \frac{1}{4} \) hours, as her trip is 15 minutes longer. The distance Mary travels is \( 40 \left(t + \frac{1}{4} \right) \) miles.

Step 2: Set Up the Distance Equation

Since John travels 35 miles farther than Mary, we write the equation for their distances: \[ 60t = 40 \left(t + \frac{1}{4} \right) + 35 \].

Step 3: Simplify and Solve the Equation

First, expand the equation: \( 60t = 40t + 10 + 35 \). Simplify this to get \( 60t = 40t + 45 \). Then, subtract \( 40t \) from both sides to isolate \( t \): \( 20t = 45 \). Dividing both sides by 20 gives \( t = \frac{45}{20} = \frac{9}{4} \). Thus, John drives for \( \frac{9}{4} \) hours or 2.25 hours.

Step 4: Determine Mary's Driving Time

Substitute \( t = \frac{9}{4} \) back into Mary's time equation: Mary drives for \( \frac{9}{4} + \frac{1}{4} = \frac{10}{4} = 2.5 \) hours.

Key Concepts

Linear Equations

Linear equations are the backbone of solving problems involving relationships between different quantities. In this exercise, we used a linear equation to relate the distances traveled by John and Mary. Let's break down how linear equations work.

A linear equation is any equation that can be written in the form \( ax + b = c \), where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable we want to solve for. In this problem:

• We identified \( t \) as the variable representing John's driving time.
• The relation between John's and Mary's distances was expressed as \( 60t = 40(t + \frac{1}{4}) + 35 \).
• We simplified and solved this equation to find \( t \).

Understanding the structure of linear equations allows us to efficiently solve for unknowns, like the driving times in real-world distance problems.

Distance Formula

The distance formula is crucial when calculating how far someone or something travels over a period of time. This problem utilizes the simple form of the distance formula, which states:

• Distance \( = \, \text{Speed} \times \, \text{Time} \)

The problem involves determining distances based on known speeds and unknown times. For John and Mary:

• John drives with a speed of 60 mi/h, so his distance is \( 60t \) miles.
• Mary drives at 40 mi/h, and considering she drives for \( t + \frac{1}{4} \) hours, her distance becomes \( 40(t + \frac{1}{4}) \) miles.

This formula is a foundational tool in physics and algebra, enabling problem solvers to find missing values when two of the three variables—distance, speed, or time—are known.

Time Calculation

Calculating time involves understanding how we convert units and adjust calculations based on different scenarios. In our exercise, time calculation was key in solving for both John’s and Mary’s driving times and involved the following steps:

• Identifying John's driving time as \( t \) hours.
• Recognizing that Mary drove 15 minutes longer than John, leading to her driving time being \( t + \frac{1}{4} \) hours. Note that we converted 15 minutes into a quarter of an hour because there are 60 minutes in an hour.
• Solve the problem by substituting these expressions into the distance formula, leading us to determine actual times: 2.25 hours for John and 2.5 hours for Mary.

Time calculations often involve conversions between minutes to hours to facilitate use with speed and distance, ensuring consistent units across all calculations.