Problem 19 - Precalculus Student Solutions Manual 5th

Question

Solve each problem. Margaret drove to a business appointment at 50 mph. Her average speed on the return trip was \(40 \mathrm{mph}\). The return trip took \(\frac{1}{4} \mathrm{hr}\) longer because of heavy traffic. How far did she travel to the appointment?

Step-by-Step Solution

Answer

The distance to the appointment is 50 miles.

Step 1 - Define Variables

Let the distance to the appointment be denoted as \( d \) miles. Denote the time taken to reach the appointment as \( t \) hours.

Step 2 - Express Distance in Terms of Speed and Time

Use the formula for distance: \( d = \text{speed} \times \text{time} \). For the trip to the appointment, \( d = 50 \times t \). For the return trip, the speed is 40 mph, so \( d = 40 \times (t + \frac{1}{4}) \).

Step 3 - Set Up the Equation

Since the distance to and from the appointment is the same, set the two expressions for distance equal to each other: \( 50t = 40(t + \frac{1}{4}) \).

Step 4 - Solve for \( t \)

Simplify the equation: \( 50t = 40t + 10 \). Subtract \( 40t \) from both sides to get \( 10t = 10 \). Solve for \( t \) by dividing both sides by 10: \( t = 1 \).

Step 5 - Find the Distance

Substitute the value of \( t \) back into the equation for the distance: \( d = 50 \times 1 = 50 \). Therefore, the distance to the appointment is 50 miles.

Key Concepts

distance formula

When we talk about the distance formula, we refer to the relationship between distance, speed, and time. The formula is:

\[ d = \text{speed} \times \text{time} \]
This simple equation tells us that distance (\(d\)) is the product of speed (\(s\)) and time (\(t\)).

In our problem, Margaret drove at a speed of 50 mph to her appointment. This means the distance to the appointment can be written as:
\[ d = 50 \times t \]
For the return trip, her speed was 40 mph, and it took her an extra quarter hour due to heavy traffic. So for the return trip, we can write:

\[ d = 40 \times \bigg(t + \frac{1}{4}\bigg) \]
Notice that both expressions equal the same distance because the distance to and from the appointment remains unchanged.

average speed

Average speed is another critical concept in our problem. In general, average speed describes the overall rate at which distance is covered. It is calculated by dividing the total distance traveled by the total time taken.

For Margaret's trip to the appointment, her average speed was 50 mph. It means she covered a certain distance (\(d\)) over a particular period of time (\(t\)).

Average speed formula can be presented as:
\[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \]
Here, because we have different speeds for each trip, we look at the average speed of each leg separately. The trip to the appointment was conducted at a steady speed of 50 mph, while the return trip was at 40 mph.

linear equations

Linear equations are algebraic expressions where each term is either a constant or the product of a constant and a single variable. They are fundamental for solving distance, rate, and time problems.

In our context, we set up a linear equation using the given information:
\[ 50t = 40 \times \bigg(t + \frac{1}{4}\bigg) \]
This equation states that the distance going to the appointment (\(50t\)) equals the distance coming back (\(40(t + \frac{1}{4})\)). This setup allows us to solve for \(t\), the time it took to get to the appointment.

Simplifying the equation step-by-step, we get:
\[ 50t = 40t + 10 \]
Subtracting 40t from both sides, we obtain:
\[ 10t = 10 \]
Dividing by 10, we determine:
\[ t = 1 \]

variables in algebra

In algebra, variables are symbols that represent unknown values. They are crucial for setting up and solving equations.

In the exercise, we defined:

• \(d\) as the distance to the appointment
• \(t\) as the time taken to reach the appointment

Using these variables in our equations, we can translate real-world scenarios into solvable mathematical problems.

By expressing the distance for each trip as a function of speed and time and setting them equal, we connected the variables to find that \(t = 1\), meaning it took Margaret 1 hour to reach her appointment.

Finally, we used this value to find the distance, substituting \(t\) back into the distance formula, giving us:

\[ d = 50 \times 1 = 50 \]
Therefore, Margaret traveled 50 miles to her appointment.