Problem 93 - Algebra and Trigonometry

Question

A salesman drives from Ajax to Barrington, a distance of 120 \(\mathrm{mi}\) , at a steady speed. He then increases his speed by 10 \(\mathrm{mi} / \mathrm{h}\) to drive the 150 \(\mathrm{mi}\) from Barrington to Collins. If the second leg of his trip took 6 min more time than the first leg, how fast was he driving between Ajax and Barrington?

Step-by-Step Solution

Answer

The speed from Ajax to Barrington was 50 mi/h.

Step 1: Define Variables

Let's define the variables to simplify our understanding of the problem. Let \( v \) be the speed (in miles per hour) at which the salesman drives from Ajax to Barrington. The distance between Ajax and Barrington is \( 120 \, \text{mi} \). The distance between Barrington to Collins is \( 150 \, \text{mi} \), and on this leg, he drives at a speed of \( v + 10 \, \text{mi/h} \).

Step 2: Express Time in Terms of Speed

The time taken for a journey is given by the formula \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \). For the first leg (Ajax to Barrington), the time taken \( t_1 \) is \( \frac{120}{v} \) hours. For the second leg (Barrington to Collins), the time taken \( t_2 \) is \( \frac{150}{v + 10} \) hours.

Step 3: Set Up the Time Equation

According to the problem, the second leg took 6 minutes more than the first leg. Convert 6 minutes to hours because our speed is in miles per hour. Therefore, \( 6 \, \text{min} = \frac{6}{60} \, \text{hours} = 0.1 \, \text{hours} \). Set up the equation: \( \frac{150}{v + 10} = \frac{120}{v} + 0.1 \).

Step 4: Solve the Equation

To find \( v \), first eliminate the fractions by multiplying through by \( v(v + 10) \): \[ 150v = 120(v + 10) + 0.1v(v + 10) \]. Simplifying this equation, we have: \( 150v = 120v + 1200 + 0.1v^2 + v \). Collecting terms gives: \( 0.1v^2 + 29v - 1200 = 0 \). Simplify by multiplying through by 10 to get: \( v^2 + 290v - 12000 = 0 \).

Step 5: Use the Quadratic Formula

The quadratic formula to solve \( ax^2 + bx + c = 0 \) is given by \( v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = 290 \), \( c = -12000 \). Substitute into the formula: \( v = \frac{-290 \pm \sqrt{290^2 - 4 \times 1 \times (-12000)}}{2 \times 1} \). Simplify to find values of \( v \).

Step 6: Calculate the Discriminant

Calculate the discriminant: \( 290^2 + 48000 = 84100 \). Thus, \( \sqrt{84100} = 290 \). Substitute back: \( v = \frac{-290 \pm 290}{2} \).

Step 7: Solve for Positive Speed

Choose the positive solution as speed must be positive: \( v = \frac{-290 + 290}{2} = \frac{290}{2} = 50 \).

Step 8: Verify the Solution

Substitute \( v = 50 \) back into the time expressions: \( t_1 = \frac{120}{50} = 2.4 \) hours and \( t_2 = \frac{150}{60} = 2.5 \) hours. Difference is \( 0.1 \) hours, which matches the problem statement.

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of degree two. They take the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( x \) represents an unknown variable. Quadratic equations are recognizable by the presence of the \( x^2 \) term, which defines their characteristic parabolic shape when graphed.

In this exercise, we derived a quadratic equation from the given problem to determine the salesman's speed. The quadratic equation we formed was \( v^2 + 290v - 12000 = 0 \). Solving this involves using the quadratic formula:

• Identify \( a \), \( b \), and \( c \) from the equation. In our problem, \( a = 1 \), \( b = 290 \), and \( c = -12000 \).
• Use the formula \( v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find possible solutions for \( v \).
• Due to the "\( \pm \)" symbol in the quadratic formula, quadratic equations can have two solutions.

In real-world applications, like in this exercise, often only the positive solution is meaningful, especially when dealing with speeds or distances.

Distance-Speed-Time Relationship

Understanding the Distance-Speed-Time relationship is crucial for solving travel problems. The foundational formula used is: \( \text{Distance} = \text{Speed} \times \text{Time} \). By rearranging, we can find that \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \).

This principle allows us to calculate the time taken for travel when the speed and distance are known. In the provided exercise, we used this relationship to determine the time taken for two legs of a trip:

• The first leg from Ajax to Barrington was \( \frac{120}{v} \) hours.
• The second leg from Barrington to Collins was \( \frac{150}{v + 10} \) hours.

Understanding the way these variables interact allows us to create equations that model real-life situations, which is exactly what we did by expressing times and setting them equal with a condition (extra 6 minutes).

Problem Solving Steps

Approaching algebra word problems systematically can significantly simplify the process. It's essential to break down the problem into manageable parts. Here are the steps used in this exercise:

1. **Define Variables**: Start by identifying and defining all unknowns and known quantities. For this problem, \( v \) was the initial speed from Ajax to Barrington.

2. **Express Relationships Mathematically**: Translate words into algebraic expressions or equations using known formulas. We used \( \frac{\text{Distance}}{\text{Speed}} \) to express time.

3. **Set Up Equations**: Use the relationships and conditions provided in the word problem to form equations. Here, we set the time relationships equal plus the additional 6 minutes.

4. **Solve the Equations**: Employ algebraic techniques, like the quadratic formula, to solve for unknowns. After forming our quadratic, we solved for \( v \).

5. **Verify the Solution**: Check calculated values against the conditions given in the problem to ensure they satisfy all aspects. We verified both times matched given times.

Equation Solving Techniques

Solving equations efficiently involves various algebraic techniques. For this problem, we utilized a mix of approaches:

- **Fraction Elimination**: To avoid fractions when dealing with equations, a common strategy is to multiply throughout by the least common denominator. This helps simplify the equation. We multiplied by \( v(v + 10) \) to clear fractions.

- **Simplification**: Combine like terms and simplify expressions to reduce complexity. This makes the equation easier to manage and solve.

- **Quadratic Formula**: This formula solves any quadratic equation, providing both possible solutions. It is an essential tool when \( x \) appears squared. In this problem, using the formula directly led us to the required speed.

These techniques are fundamental in algebra and are useful in a wide range of applications, from simple linear equations to more complex quadratic scenarios.