Problem 77 - Precalculus: Functions and Graphs

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 mph.

Step 1: Substitute the Speed into the Formula

For part (a), we need to find the braking distance when the car's speed is 55 mph. Substitute \( v = 55 \) into the formula \( d = v + \frac{v^2}{20} \).

Step 2: Calculate the Braking Distance

Using the formula with \( v = 55 \), we have: \( d = 55 + \frac{55^2}{20} \). Calculate \( 55^2 \) then divide by 20. Add this result to 55.

Step 3: Solve the Equation

Calculate \( 55^2 \) which is 3025. Then divide by 20 to get 151.25. Add 55 to 151.25 to get a total braking distance of 206.25 feet.

Step 4: Set Up the Equation for Given Distance

For part (b), set \( d = 120 \) and solve the equation \( 120 = v + \frac{v^2}{20} \) for \( v \).

Step 5: Rearrange the Equation

Multiply the entire equation by 20 to eliminate the fraction: \( 20 \times 120 = 20v + v^2 \), leading to \( v^2 + 20v - 2400 = 0 \).

Step 6: Solve the Quadratic Equation

This equation \( v^2 + 20v - 2400 = 0 \) can be solved using the quadratic formula \( v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 20, c = -2400 \).

Step 7: Apply the Quadratic Formula

Here, calculate \( b^2 - 4ac = 20^2 - 4(1)(-2400) = 400 + 9600 = 10000 \), and then solve for \( v = \frac{-20 \pm 100}{2} \).

Step 8: Compute the Valid Solution for Speed

Solving gives \( v = \frac{80}{2} = 40 \) and \( v = \frac{-120}{2} = -60 \). Since speed cannot be negative, the valid solution is \( v = 40 \).

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation that take the form \( ax^2 + bx + c = 0 \). In these equations, \( x \) is the variable and \( a \), \( b \), and \( c \) are constants. The solutions to quadratic equations are the values of \( x \) that make the equation true. These solutions can be found using different methods such as factoring, completing the square, or using the quadratic formula.

• The quadratic formula is derived from the quadratic equation and is given by: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• This formula provides two values of \( x \), which correspond to the two possible solutions of the quadratic equation.

Quadratic equations appear frequently in real-world problems, especially in physics and engineering. They help us model the behavior of objects under certain conditions. For example, in the given exercise, we rearranged the braking distance formula into a quadratic equation to determine the speed of a car that can safely stop within a specified distance.

Distance Formula

The braking distance of a vehicle is calculated using a specific distance formula that accounts for both the velocity and the deceleration of the vehicle. This formula is given by \( d = v + \frac{v^2}{20} \), where \( d \) is the braking distance in feet and \( v \) is the speed in miles per hour.

• The first part of the formula, \( v \), represents the initial distance covered by the vehicle.
• The second part, \( \frac{v^2}{20} \), accounts for the distance needed to decelerate the vehicle to a complete stop.

This is not a linear relationship due to the \( v^2 \) term, indicating that braking distance increases with the square of the speed. This aspect explains why faster speeds result in significantly longer stopping distances. Understanding how this formula works is crucial for safe driving practices, especially in calculating safe stopping distances when approaching obstacles like stop signs.

Physics of Motion

The physics of motion involves the study of how objects move through space and time. This involves several concepts such as speed, velocity, and acceleration. In the context of braking distance, these factors play a crucial role.

• Speed refers to how fast an object is moving along its path. In our exercise, we specified the car's speed to calculate its braking distance.
• Velocity is similar to speed but includes the direction of movement. For simplicity, we often treat speed and velocity interchangeably in linear motion along a straight path.
• Acceleration occurs when there is a change in the velocity of an object. When braking, acceleration is actually negative, often called deceleration.

The reason the braking distance is quadratic in \( v \) is due to the physics of acceleration and deceleration. An object moving with constant acceleration will cover more distance with time, which explains the quadratic term in our distance formula. Thus, understanding the physics of motion is crucial not only in academic pursuits but also in practical applications like driving safety.