Problem 71 - Algebra and Trigonometry

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 \(\mathrm{min}\) 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 and Relationships

Let \( t_j \) be the time John drives and \( t_m \) be the time Mary drives, with \( t_m = t_j + \frac{15}{60} \) converting 15 minutes into hours. The distance John travels is \( 60t_j \) and the distance Mary travels is \( 40t_m \). It's given that John's distance is 35 miles more than Mary’s distance, so \( 60t_j = 40t_m + 35 \).

Step 2: Substitute Relationship of Times

Substitute the expression for \( t_m \) from the time relationship: \( 60t_j = 40(t_j + \frac{15}{60}) + 35 \). This becomes \( 60t_j = 40t_j + 10 + 35 \).

Step 3: Simplify the Equation

Simplify the equation to find \( t_j \): \( 60t_j = 40t_j + 45 \). Rearranging terms gives \( 20t_j = 45 \).

Step 4: Solve for John's Time

Solve for \( t_j \): \( t_j = \frac{45}{20} = 2.25 \) hours.

Step 5: Solve for Mary's Time

Substitute \( t_j \) back to find \( t_m \): \( t_m = 2.25 + \frac{1}{4} = 2.5 \) hours.

Step 6: Verify the Solution

Calculate the distances: John travels \( 60 \times 2.25 = 135 \) miles and Mary travels \( 40 \times 2.5 = 100 \) miles. John indeed travels 35 miles more, confirming the solution is correct.

Key Concepts

Variable Definition

When tackling distance, speed, and time problems, defining your variables clearly is crucial. Variables represent the unknown elements of the problem that you will solve for.
In this exercise, two main variables are defined to represent the time each person drives:

• Let \( t_j \) represent the time John drives.
• Let \( t_m \) represent the time Mary drives.

To connect these times, use given relationships in the problem:
Since Mary's trip takes 15 minutes longer than John's, we convert 15 minutes to hours \( \left( \frac{15}{60} \right) \) and set up the relation: \( t_m = t_j + \frac{15}{60} \).
This conversion ensures that all the time units in the problem are consistent, which is essential for solving the problem accurately.

Equation Solving

With variables defined, you can set up equations based on the relationships described in the problem statement. A key perspective in this scenario is how the distances relate to each other.

The equation comes from two main points:

• The relationship between distances: John drives a distance of \( 60t_j \), and Mary drives \( 40t_m \), with John's distance being 35 miles more than Mary's, forming the equation \( 60t_j = 40t_m + 35 \).
• The substitution of \( t_m \): Replace \( t_m \) using \( t_m = t_j + \frac{15}{60} \), giving you a single equation \( 60t_j = 40(t_j + \frac{15}{60}) + 35 \).

Simplifying, you'll balance and solve for \( t_j \). Rearranging terms leads to the equation \( 20t_j = 45 \), which simplifies further to \( t_j = 2.25 \) hours. This step-by-step substitution and equation solving help unravel the mystery of each participant's driving time.

Problem Verification

Once you've reached a solution, verifying your answer ensures the logic holds throughout the process and aligns with the problem's requirements.
Verification involves checking if the distances calculated meet the original conditions.

• Calculate John's distance: \( 60 \times 2.25 = 135 \) miles.
• Calculate Mary's distance: \( 40 \times 2.5 = 100 \) miles.

Now, compare the two distances: Is John's distance 35 miles more than Mary's? Yes, since \( 135 - 100 = 35 \), confirming the initial condition is met.
This check provides reassurance that the calculated times \( t_j = 2.25 \) hours and \( t_m = 2.5 \) hours are indeed correct. It is a testament to the reliability of algebraic methods when solving real-world problems.