Problem 85 - Algebra and Trigonometry

Question

Stopping Distance For a certain model of car the distance \(d\) required to stop the vehicle if it is traveling at \(v\) mi/h is given by the formula $$ d=v+\frac{v^{2}}{20} $$ where \(d\) is measured in feet. Kerry wants her stopping distance not to exceed 240 \(\mathrm{ft}\) . At what range of speeds can she travel?

Step-by-Step Solution

Answer

Kerry can travel at speeds from 0 to 60 mi/h.

Step 1: Understand the Formula

The formula given is \( d = v + \frac{v^2}{20} \), where \( d \) is the stopping distance in feet and \( v \) is the speed in miles per hour. We need the stopping distance \( d \) to be less than or equal to 240 feet.

Step 2: Set Up the Inequality

We want the stopping distance \( d \) not to exceed 240 feet. So we set up the inequality: \[ v + \frac{v^2}{20} \leq 240. \]

Step 3: Clear the Fraction

Multiply the entire inequality by 20 to eliminate the fraction: \[ 20v + v^2 \leq 4800. \]

Step 4: Rearrange Into Standard Form

Rearrange the inequality into standard quadratic form: \[ v^2 + 20v - 4800 \leq 0. \]

Step 5: Solve the Quadratic Inequality

To solve the quadratic inequality \( v^2 + 20v - 4800 \leq 0 \), first find the roots of the quadratic by setting the equation to equal zero: \( v^2 + 20v - 4800 = 0 \). Use the quadratic formula \( v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 20 \), and \( c = -4800 \).

Step 6: Calculate the Discriminant

Calculate the discriminant \( b^2 - 4ac \): \( 20^2 - 4 \times 1 \times (-4800) = 400 + 19200 = 19600 \). The discriminant is 19600, which is a perfect square.

Step 7: Solve for Roots

Using the quadratic formula, the roots are: \[ v = \frac{-20 \pm \sqrt{19600}}{2}. \] Simplifying further, \( \sqrt{19600} = 140 \), so, \( v = \frac{-20 \pm 140}{2} \). This gives the roots \( v = 60 \) and \( v = -80 \).

Step 8: Determine the Range of Speeds

Since speed cannot be negative, we consider the range for \( v > 0 \). Check the intervals \((-\infty, -80), (-80, 60), (60, \infty)\). The quadratic is negative between \(v = -80\) and \(v = 60\), but because speed cannot be negative, the range is \([0, 60]\).

Step 9: Conclude the Answer

The inequality \( v^2 + 20v - 4800 \leq 0 \) holds true for \( v \) in the interval \([0, 60]\). Therefore, the stopping distance of less than or equal to 240 feet is maintained for speeds in this range.

Key Concepts

Understanding Stopping Distance

Stopping distance is the length a vehicle takes to come to a complete stop after the brakes are applied. It combines both the reaction distance and braking distance. For the car model in the exercise, the formula for stopping distance in feet is given by \( d = v + \frac{v^2}{20} \). This equation indicates that

• \(v\) is the speed in miles per hour.
• The stopping distance \(d\) depends on both the magnitude of the speed and its square.

Understanding this formula helps in predicting how quickly a car can stop depending on various speeds.
Here, Kerry wants her stopping distance not to exceed 240 feet. This sets a practical limit on how fast she can safely travel in that vehicle.

Quadratic Formula Application

The quadratic formula is a crucial tool in algebra for solving equations of the form \( ax^2 + bx + c = 0 \). In the given stopping distance problem, once the quadratic inequality is arranged as \( v^2 + 20v - 4800 \leq 0 \), we utilized the quadratic formula to find the roots of the equation:

• \(v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
• Where \(a = 1\), \(b = 20\), and \(c = -4800\).

These roots help determine the speed range where the stopping distance condition is met. It simplifies solving inequalities by providing numerical boundaries that can be used to analyze the behavior of the quadratic.

Exploring Speed Range

Determining the speed range in this context means finding all possible values of \(v\) that satisfy the quadratic inequality \(v^2 + 20v - 4800 \leq 0\). Using the quadratic formula gives us two roots, \(v = 60\) and \(v = -80\). Since speed must be a non-negative value in realistic terms, we only consider the range between \(0\) and \(60\):

• \(0\) is the lowest possible speed (when the car is stationary).
• \(60\) is derived from the higher valid root of the quadratic equation.

Thus, Kerry can drive between 0 and 60 miles per hour while keeping her stopping distance within 240 feet.

Understanding the Discriminant

The discriminant in a quadratic equation \(ax^2 + bx + c = 0\) is given by \(b^2 - 4ac\). It helps to reveal the nature of the roots of the equation. In our stopping distance problem, the discriminant is calculated as:

• \(20^2 - 4 \times 1 \times (-4800) = 19600\).

Because the discriminant is a perfect square (19600 is \(140^2\)), we know that our quadratic equation has two distinct real roots. These roots were essential to finding the valid speed range, proving the equation solvable and visually understanding how speed affects stopping distances. Having a discriminant that's a perfect square also made the process of solving cleaner and more straightforward.