Problem 91 - Algebra and Trigonometry

Question

Distance, Speed, and Time 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 \(\mathrm{min}\) more time than the first leg, how fast was he driving between Ajax and Barrington?

Step-by-Step Solution

Answer

The salesman was driving about 40 mph from Ajax to Barrington.

Step 1: Define Variables

Let \( x \) be the speed (in miles per hour) the salesman drove from Ajax to Barrington. The time taken for this journey can be expressed as \( \frac{120}{x} \) hours.

Step 2: Express Second Leg Variables

For the second leg of the trip, the salesman increased his speed by 10 miles per hour, making his speed \( x + 10 \) miles per hour. The time for this leg is \( \frac{150}{x + 10} \) hours.

Step 3: Set Up Time Equation

According to the problem, the second leg took 6 more minutes than the first leg. Convert minutes to hours (6 minutes = 0.1 hours) and set up the equation: \( \frac{150}{x + 10} = \frac{120}{x} + 0.1 \).

Step 4: Solve the Equation

Multiply through by \( x(x + 10) \) to clear the denominators: \[ 150x = 120(x + 10) + 0.1x(x + 10) \] Expand and simplify: \[ 150x = 120x + 1200 + 0.1x^2 + x \]Rearrange into a standard quadratic form: \[ 0.1x^2 - 29x - 1200 = 0 \].

Step 5: Solve the Quadratic Equation

First, multiply the entire equation by 10 to eliminate the decimal: \[ x^2 - 290x - 12000 = 0 \]. Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -290 \), and \( c = -12000 \). Calculate the discriminant: \[ b^2 - 4ac = (-290)^2 - 4(1)(-12000) = 84100 + 48000 = 132100 \]. Find \( x \): \[ x = \frac{290 \pm \sqrt{132100}}{2} \].

Step 6: Compute the Quadratic Solutions

Calculate \( \sqrt{132100} \approx 363.55 \). Thus, \[ x = \frac{290 + 363.55}{2} \approx 326.775 \]. Since this is not practical for driving speed, compute for the negative root: \( x = \frac{290 - 363.55}{2} \), resulting in a valid speed calculation.

Step 7: Find the Practical Solution

Compute \( x = \frac{290 - 363.55}{2} \approx 36.775 \). The practical driving speed initially was approximately 40 mph.

Key Concepts

Quadratic Equation

The problem at hand involves solving a quadratic equation to find the speed. A quadratic equation has the general form of \( ax^2 + bx + c = 0 \). In this exercise, the equation stems from a real-world context involving distance, speed, and time. To derive the equation, we first set up a relationship between the time for two different segments of a journey. After converting the narrative into an equation, the result is a quadratic expression. This form appears when multiplying and rearranging terms that include both a term squared in speed, and linear terms. Quadratic equations typically have two solutions because their graphs are parabolas, and they potentially intersect the x-axis at two points.

Time Conversion

In many real-world problems like this one, time needs to be converted between units to ensure consistency in calculations. Specifically, this exercise requires converting minutes to hours. We know there are 60 minutes in an hour, so to convert minutes to hours, you divide by 60. Here, the salesman finds that one segment of his trip takes 6 more minutes than another. To use this information in the equation, convert 6 minutes into hours: \( 6 \text{ minutes} = 0.1 \text{ hours} \). This conversion is crucial, as it allows us to form a unified equation with consistent time units, making calculations straightforward.

Discriminant Calculation

Calculating the discriminant is a key step in solving quadratic equations. The discriminant, denoted as \( b^2 - 4ac \), provides valuable information about the number and type of solutions of a quadratic equation. In our problem, after arranging the equation in standard form \( x^2 - 290x - 12000 = 0 \), we identify \( a = 1 \), \( b = -290 \), and \( c = -12000 \). Substituting these values into the discriminant formula gives us

• \( b^2 = 84100 \)
• \( -4ac = -48000 \)

Adding these results in a discriminant of \( 132100 \). A positive discriminant indicates two distinct real roots, which is what we need to determine the possible speeds.

Solving Equations

With the determined discriminant, solving the quadratic equation involves using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula gives the roots of any quadratic equation, assuming the discriminant condition is met. In this scenario,

• \( x = \frac{290 \pm \sqrt{132100}}{2} \)

This results in two potential solutions for speed. These calculations yield two values for \( x \), representing the speeds at which the salesman might be traveling. However, not all solutions may be practical, especially when considering real-world context like speed limits. Calculations must be checked against practical constraints to ensure they are reasonable, which is why the valid solution here is approximately 40 mph.