Question

An accepted relationship between stopping distance \(d\) (in feet), and the speed \(v\) of a car (in \(\mathrm{mph}\) ), is \(d=1.1 v+0.06 v^{2}\) on dry, level concrete. (a) How many feet will it take a car traveling \(45 \mathrm{mph}\) to stop on dry, level concrete? (b) If an accident occurs 200 feet ahead of you, what is the maximum speed you can be traveling to avoid being involved?

Short answer

(a) 171 feet (b) approximately 49.25 mph

Step-by-step solution

Understand the Equation

The stopping distance formula is given by \(d = 1.1 v + 0.06 v^2\), where \(d\) is in feet and \(v\) is in mph.

Plug in Speed for Part (a)

For part (a), the speed \(v\) is given as 45 mph. Substitute \(v = 45\) into the formula: \(d = 1.1(45) + 0.06(45)^2\).

Calculate the Stopping Distance

Calculate the terms separately: \(1.1(45) = 49.5\) and \(0.06(45)^2 = 0.06(2025) = 121.5\). Add these results: \(49.5 + 121.5 = 171\). So, the stopping distance is 171 feet.

Set Up the Equation for Part (b)

For part (b), you're given the stopping distance \(d = 200\) feet. Set the equation equal to 200: \(200 = 1.1 v + 0.06 v^2\).

Rearrange the Equation

Rearrange the equation to \(0.06 v^2 + 1.1 v - 200 = 0\). This is a quadratic equation in the standard form \(ax^2 + bx + c = 0\), where \(a = 0.06\), \(b = 1.1\), and \(c = -200\).

Apply the Quadratic Formula

Use the quadratic formula to solve for \(v\): \(v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute the values: \(v = \frac{-1.1 \pm \sqrt{(1.1)^2 - 4(0.06)(-200)}}{2(0.06)}\).

Simplify Under the Square Root

Calculate the discriminant: \((1.1)^2 - 4(0.06)(-200) = 1.21 + 48 = 49.21\), then take the square root: \(\sqrt{49.21} \approx 7.01\).

Solve for \(v\)

Plug the square root value back into the formula: \(v = \frac{-1.1 \pm 7.01}{0.12}\). This gives two solutions: \(v = \frac{5.91}{0.12} \approx 49.25\) and \(v = \frac{-8.11}{0.12} \approx -67.58\). Discard the negative value as speed cannot be negative. So, the maximum speed is approximately 49.25 mph.

Key concepts

Quadratic Equations

A quadratic equation is a type of polynomial equation of the form \( ax^2 + bx + c = 0 \). In our problem, the quadratic equation is used to find the maximum speed at which a car can travel to stop within a given distance. The general formula to solve quadratic equations is known as the quadratic formula: \[ v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Here, \( a \), \( b \), and \( c \) are coefficients obtained from rearranging our stopping distance equation. This method helps us find the possible values of \( v \) (in our case, speeds) that satisfy the given distance.

Speed and Distance Relationship

The stopping distance \( d \) of a car is related to its speed \( v \) through the equation \( d = 1.1 v + 0.06 v^2 \). This formula has two parts:

• A linear term \( 1.1v \) which denotes the proportion of speed contributing directly to the stopping distance.
• A quadratic term \( 0.06v^2 \) accounting for how higher speeds increase the stopping distance at an accelerating rate because reaction times and braking efficiency decrease.

Understanding this relationship is essential for estimating safe stopping distances and preventing accidents.

Solving Equations

Solving equations involves rearranging and substituting values to find unknown variables. For the exercise:

1. We first substituted the given speed into the stopping distance equation to find the distance. For example, to calculate the stopping distance for a car traveling at 45 mph, we plugged in \( v = 45 \) into the equation and solved for \( d \).

2. For the second part, we set the stopping distance to 200 feet and rearranged the formula to form a quadratic equation, \( 0.06v^2 + 1.1v - 200 = 0 \). Using the quadratic formula, the speeds giving a stopping distance of 200 feet were calculated, keeping only the positive value as speed must be a non-negative number.

These steps are crucial for solving real-world problems involving physics and motion, providing accurate and practical results.