Problem 60 - Algebra and Trigonometry with Analytic Geometry

Question

The distance that a car travels between the time the driver makes the decision to hit the brakes and the time the car actually stops is called the braking distance. For a certain car traveling \(v \mathrm{mi} / \mathrm{hr}\), the braking distance \(d\) (in feet) is given by \(d=v+\left(v^{2} / 20\right)\). (a) Find the braking distance when \(v\) is \(55 \mathrm{mi} / \mathrm{hr}\). (b) If a driver decides to brake 120 feet from a stop sign, how fast can the car be going and still stop by the time it reaches the sign?

Step-by-Step Solution

Answer

(a) 206.25 feet; (b) 40 mi/hr.

Step 1: Understanding the Problem

The braking distance formula is given as \( d = v + \left( \frac{v^2}{20} \right) \). We are asked to find the braking distance when the car's speed \( v \) is 55 mi/hr and also determine the maximum speed \( v \) such that the car stops within 120 feet.

Step 1: Calculating Braking Distance for 55 mi/hr

Substitute \( v = 55 \) into the formula.\[ d = 55 + \left( \frac{55^2}{20} \right) \]\[ = 55 + \left( \frac{3025}{20} \right) \]\[ = 55 + 151.25 \]\[ = 206.25 \text{ feet} \]So, when \( v \) is 55 mi/hr, the braking distance \( d \) is 206.25 feet.

Step 2: Setting Up Equation for Maximum Speed

We need to find \( v \) such that \( d = 120 \) feet. Substitute \( d = 120 \) into the formula and solve for \( v \).\[ 120 = v + \left( \frac{v^2}{20} \right) \]Multiply the entire equation by 20 to eliminate the fraction:\[ 2400 = 20v + v^2 \]Rearrange the equation into a standard quadratic equation:\[ v^2 + 20v - 2400 = 0 \]

Step 3: Solving the Quadratic Equation

Use the quadratic formula \( v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = 20 \), and \( c = -2400 \).\[ v = \frac{-20 \pm \sqrt{20^2 - 4 \cdot 1 \cdot (-2400)}}{2 \cdot 1} \]\[ = \frac{-20 \pm \sqrt{400 + 9600}}{2} \]\[ = \frac{-20 \pm \sqrt{10000}}{2} \]\[ = \frac{-20 \pm 100}{2} \]Calculate the two possible solutions for \( v \):\[ v = \frac{80}{2} = 40 \]or\[ v = \frac{-120}{2} = -60 \]Since speed cannot be negative, \( v = 40 \) mi/hr is the valid solution.

Key Concepts

Quadratic Equation

Understanding quadratic equations is key when dealing with problems like braking distance. A quadratic equation takes the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants. In our braking distance problem, the equation \( v^2 + 20v - 2400 = 0 \) is quadratic due to the \( v^2 \) term. Quadratic equations graph as parabolas, which makes them useful in physics to describe motion scenarios, often involving maximums and minimums.

The quadratic formula, \( v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), helps find the roots or solutions of the equation. Here, \( a = 1 \), \( b = 20 \), and \( c = -2400 \), making it possible to calculate the possible speeds that satisfy the equation. This step is essential, as drivers need to know how much distance they have before stopping to adjust their speeds safely.

A critical tip is to consider only the viable solutions in physics-related problems. In this scenario, we disregard negative velocities as speed cannot be negative in real life scenarios.

Algebraic Expressions

Algebraic expressions are central to solving braking distance problems, acting as the builders of the equations we use. They include terms combined using mathematical operations: addition, subtraction, multiplication, and division.

In our braking distance formula, \( d = v + \left( \frac{v^2}{20} \right) \), the expression is made up of two parts. The first part, \( v \), represents the speed linearly, and the second part, \( \left( \frac{v^2}{20} \right) \), accounts for the quadratic component of deceleration as speed increases.

Simplifying or rearranging these expressions, like transforming \( 120 = v + \frac{v^2}{20} \) to a standard quadratic form, \( v^2 + 20v - 2400 = 0 \), is crucial in algebra. Such manipulations enable us to utilize formulas, like the quadratic formula, to find unknown variables. Understanding each component's role within the expression supports accurately modeling real-world phenomena like car braking distances.

Velocity Calculation

Velocity calculation is the backbone of determining braking distances. Velocity refers to the speed of an object in a given direction and is often expressed in units like mi/hr. For assessing braking distances, it’s crucial to understand both how to calculate and interpret these values.

With the braking distance equation \( d = v + \frac{v^2}{20} \), calculating velocity starts by understanding how different speeds affect braking performance. When solving for an unknown velocity \( v \) that allows a car to stop within a certain distance (e.g., 120 feet), algebra and the quadratic formula come into play.

For instance, by expressing the equation in standard quadratic form and using techniques like factoring or the quadratic formula, one can determine feasible speeds to ensure safety. Accurately calculating velocity helps make informed decisions about how fast you can travel before needing to begin braking to stop safely, illustrating why mastering these calculations is essential in driving scenarios.