Problem 56 - Precalculus: Mathematical for Calculus

Question

Distance, Speed, and Time 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: Define Variables

Let John's driving time be \( t \) hours. Because Mary drives 15 minutes longer, her time will be \( t + \frac{1}{4} \) hours, as 15 minutes equals a quarter of an hour.

Step 2: John's Distance Equation

John's distance can be calculated using the formula: \( \text{Distance} = \text{Speed} \times \text{Time} \). Therefore, John's distance is \( 60t \) miles.

Step 3: Mary's Distance Equation

Mary's distance can also be calculated using the formula: \( \text{Distance} = \text{Speed} \times \text{Time} \). So, Mary's distance is \( 40(t + \frac{1}{4}) \) miles.

Step 4: Set Up the Relationship Between Distances

Since John travels 35 miles farther than Mary, the equation \( 60t = 40(t + \frac{1}{4}) + 35 \) represents the relationship between their distances.

Step 5: Solve the Distance Equation for \( t \)

Scale it up:\[ 60t = 40t + 10 + 35 \]Simplify and solve for \( t \):\[ 60t = 40t + 45 \]Subtract \( 40t \) from both sides:\[ 20t = 45 \]Divide each side by 20:\[ t = \frac{45}{20} = 2.25 \] hours.

Step 6: Determine Mary's Driving Time

Mary drives for \( t + \frac{1}{4} \) hours. Since \( t = 2.25 \) hours, Mary's time is:\[ 2.25 + 0.25 = 2.5 \] hours.

Key Concepts

Algebraic Equations

Algebraic equations are a fundamental part of solving word problems involving distance, speed, and time. These equations help us translate a verbal description into a mathematical form that can be solved. In these types of problems, you are usually given some information and asked to find something unknown, which can be outlined with equations.

For our exercise about John and Mary, we defined equations based on their distances. First, we set up an equation for each person's travel distance, using known variables such as speed and time.

When solving these equations, always remember that algebraic equations are like scales; what you do to one side, you must do to the other. This principle helps maintain balance and allows for accurate problem-solving. The goal is to isolate the variable, usually time in these problems, to find out more about the situation given.

Relationship Between Distance and Time

The relationship between distance, speed, and time is vital in these types of word problems. Understanding this helps in forming the correct equations to find unknown values.

The basic formula to remember is:

• Distance = Speed × Time

This means that how far you travel depends on how fast you move and for how long. For John and Mary, the problem clarifies this relationship by giving their speeds and asking us to find out how long they traveled under different conditions.

The time difference in their journeys influenced their distance, showing how interconnected these elements are. The time equation for John and Mary allowed us to find their respective distances and ensure accuracy by considering additional factors like the 15-minute difference affecting Mary.

Problem-solving in Precalculus

Problem-solving in precalculus involves using mathematical strategies to find solutions to complex real-world scenarios. It's about breaking down a situation into parts, defining variables, and using equations to express these relationships.

In the scenario with John and Mary, a simple situation of driving in opposite directions turned into multiple equations that needed solving. The main challenge in precalculus problem-solving is setting up the situation mathematically and solving for unknowns.

Students should practice the following steps:

• Identify what you know and what you need to find.
• Define the variables clearly.
• Set up equations based on these variables.
• Use algebraic manipulation to solve for the unknowns.

This exercise highlights how seemingly simple daily activities translate into mathematical problems that require thorough analysis and strategic approaches in precalculus.