Problem 31 - Beginning and Intermediate Algebra

Question

Solve each problem. Round answers to the nearest tenth as needed. The distance from Jackson to Lodi is about \(40 \mathrm{mi}\), as is the distance from Lodi to Manteca. Adrian drove from Jackson to Lodi, stopped in Lodi for a highenergy drink, and then drove on to Manteca at 10 mph faster. Driving time for the entire trip was 88 min. Find her rate from Jackson to Lodi. (Data from State Farm Road Atlas.)

Step-by-Step Solution

Answer

Adrian's rate from Jackson to Lodi is approximately 50 mph.

Step 1: Define Variables

Let Adrian's rate from Jackson to Lodi be represented by \( r_1 \) mph. Then, her rate from Lodi to Manteca is \( r_1 + 10 \) mph.

Step 2: Convert Time to Hours

The total driving time is given as 88 minutes. Convert minutes to hours: \( \frac{88}{60} = 1.467 \) hours (rounded to three decimal places).

Step 3: Set Up the Distance and Rate Equation

Use the formula \( \text{distance} = \text{rate} \times \text{time} \). The distance from Jackson to Lodi (40 miles) at the rate of \( r_1 \) and the distance from Lodi to Manteca (40 miles) at the rate of \( r_1 + 10 \) will combinedly take 1.467 hours. The equation is: \[ \frac{40}{r_1} + \frac{40}{r_1 + 10} = 1.467 \]

Step 4: Solve for \( r_1 \)

Solve the equation step-by-step. Start by multiplying both sides by \( r_1(r_1 + 10) \) to clear the denominators: \[ 40r_1 + 400 + 40r_1 = 1.467r_1(r_1 + 10) \] Combine like terms and simplify: \[ 80r_1 + 400 = 1.467r_1^2 + 14.67r_1 \] Move all terms to one side to set up a quadratic equation: \[ 1.467r_1^2 + 14.67r_1 - 80r_1 - 400 = 0 \] Simplify the terms: \[ 1.467r_1^2 - 65.33r_1 - 400 = 0 \] Use the quadratic formula \( r_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1.467 \), \( b = -65.33 \), and \( c = -400 \): \[ r_1 = \frac{-(-65.33) \pm \sqrt{(-65.33)^2 - 4(1.467)(-400)}}{2(1.467)} \] Calculate the discriminant and solve for \( r_1 \): \[ r_1 = \frac{65.33 \pm \sqrt{4268.1089 + 2347.36}}{2.934} \] \[ r_1 = \frac{65.33 \pm \sqrt{6615.4689}}{2.934} \] \[ r_1 = \frac{65.33 \pm 81.33}{2.934} \] \[ r_1 = \frac{146.66}{2.934} \text{ or } r_1 = \frac{-16}{2.934} \] The negative result is not physically meaningful, so the correct rate is: \[ r_1 \approx 50 \text{ mph} \]

Key Concepts

distance-rate-time relationship

In algebra, the distance-rate-time relationship is fundamental for solving many word problems. The standard formula is \[ \text{distance} = \text{rate} \times \text{time} \]. This means that the distance traveled is equal to the speed (or rate) multiplied by the time spent traveling.

For example, if you drive 60 miles per hour for 2 hours, the distance you cover is \[ 60 \text{ mph} \times 2 \text{ hours} = 120 \text{ miles} \].

In problems where you know the distance and the time, but not the rate, you can rearrange the formula: \[ \text{rate} = \frac{\text{distance}}{\text{time}} \]. Similarly, if the rate and distance are known, you can find the time by: \[ \text{time} = \frac{\text{distance}}{\text{rate}} \].

Understanding this relationship allows you to set up equations based on given information, especially in scenarios where the total travel time involves different rates or split intervals.

quadratic formula

The quadratic formula is essential for solving quadratic equations, which are equations of the second degree (the highest exponent of the variable is 2). The general form of a quadratic equation is: \[ ax^2 + bx + c = 0 \]

The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \[ a \], \[ b \], and \[ c \] are coefficients from the quadratic equation. The term inside the square root, \[ b^2 - 4ac \], is called the discriminant and determines the nature of the roots.

• If \[ b^2 - 4ac \] is positive, there are two real roots.
• If it is zero, there is one real root.
• If it is negative, the roots are complex (not real) numbers.

Using this formula, you can find the solutions to a quadratic equation even when factoring it directly is challenging or impossible. Practicing with the quadratic formula can help you solve many different types of problems, including those related to distance-rate-time scenarios.

converting time units

Converting time units is often necessary in algebra, especially in word problems that involve speed, distance, and time. Common conversions include minutes to hours and vice versa.

For instance, 1 hour has 60 minutes. To convert minutes to hours, use the conversion factor \[ \frac{1 \text{ hour}}{60 \text{ minutes}} \].

Suppose you have 88 minutes and need to convert it to hours: \[ 88 \text{ minutes} \times \frac{1 \text{ hour}}{60 \text{ minutes}} = 1.467 \text{ hours} \]. This conversion helps in problems where you need to use a consistent unit of time.

• For converting hours to minutes, you multiply by 60.
• If you have fractional hours, like 1.5 hours, multiply by 60 to get 90 minutes.

Mastering these conversions ensures accuracy in setting up and solving algebraic equations related to real-world problems, such as calculating travel rates or determining travel durations.

word problems in algebra

Word problems in algebra involve translating a real-world scenario into an algebraic equation and then solving it. These problems require careful reading, identifying relevant information, and setting up equations correctly.

Let's take a common structure. The problem often states the distances, times, or rates and asks you to find the missing value. In our example, Adrian's travel scenario involves two parts, each with a different rate, and a total travel time.

Here's a strategy to follow:

• Define variables for the unknowns (e.g., let \[ r_1 \] be Adrian's rate).
• Set up equations based on given information (e.g., \[ \frac{40}{r_1} + \frac{40}{r_1 + 10} = 1.467 \] hours).
• Solve the equations using appropriate algebraic methods (e.g., quadratic formula).

These steps simplify the problem and make them manageable. Practice and familiarity with these kinds of problems develop your problem-solving skills and make it easier to apply algebra to daily scenarios.