Problem 32 - Algebra Form and Function

Question

Let \(d(v)\) be the braking distance in feet of a car traveling at \(v\) miles per hour. Explain the meaning of the following statements. (a) \(d(30)=111\) (b) \(\quad d(a)=10\) (c) \(d(10)=b\) (d) \(s=d(v)\)

Step-by-Step Solution

Answer

Answer: The statement d(30) = 111 represents that the braking distance for a car traveling at a speed of 30 miles per hour is 111 feet. This means it takes the car 111 feet to come to a complete stop when applying brakes at this speed.

Step 1: (a) d(30) = 111

In this statement, we have the function \(d(v)\) evaluated at \(v = 30\). This means that the car is traveling at a speed of 30 miles per hour. The result given is 111 feet, which indicates that the braking distance for a car traveling at 30 miles per hour is 111 feet. In other words, it takes the car 111 feet to come to a complete stop when applying brakes at this speed.

Step 2: (b) d(a) = 10

In this statement, the function \(d(v)\) is evaluated at \(v = a\), where \(a\) represents an unknown speed in miles per hour. The result given is 10 feet, which means that the braking distance for a car traveling at a speed of \(a\) miles per hour is 10 feet. To find the actual speed \(a\), one would need more information about the relationship between braking distance and speed.

Step 3: (c) d(10) = b

In this statement, the function \(d(v)\) is evaluated at \(v = 10\). This means that the car is traveling at a speed of 10 miles per hour. The result given is \(b\) feet, which indicates that the braking distance for a car traveling at 10 miles per hour is \(b\) feet. Here, \(b\) represents an unknown braking distance, and one would need more information about the relationship between braking distance and speed to determine the actual value of \(b\).

Step 4: (d) s = d(v)

In this statement, we are given that \(s\) is equal to the function \(d(v)\). Here, \(s\) represents the braking distance of a car in feet, and \(v\) represents the speed of the car in miles per hour. Thus, this statement is simply expressing the overall relationship between braking distance and speed, which is what the function \(d(v)\) represents. In other words, for any given speed \(v\), the braking distance \(s\) can be found by evaluating the function \(d(v)\).

Key Concepts

Algebra

Whenever you're dealing with mathematics, especially word problems, algebra often sneaks in to help make sense of things. Algebra allows us to take real-world situations, like braking distances, and translate them into mathematical expressions and equations. This process helps us make predictions and solve problems. For example, when we say that the braking distance is expressed as a function of speed, we're using algebra to express a relationship:

• Variables: In algebra, variables like \(v\) and \(a\) in our problem hold place for values we might not know yet, such as speed or braking distance.
• Equations: An equation like \(d(30)=111\) shows that algebra is describing a specific condition - it's telling us the exact braking distance for the car at 30 mph.
• Evaluation: When we evaluate equations, we're simply plugging in values (like in \(d(a)=10\)) to find unknowns or verify information.

Algebra isn't just about crunching numbers, it's about finding relationships and making predictions by representing complex realities with simple equations. This makes it easier to understand real-world mechanics, such as how quickly a car stops at different speeds.

Functions

Functions in mathematics serve as special rules that assign each input to exactly one output. Think of a function like a magical box:

• Input: You put a specific value into the box (like the speed \(v\) of a car).
• Magical Processing: The function processes this input using some internal rules.
• Output: Out comes the resulting value (like the braking distance \(d(v)\)).

This ability to create a predictable outcome based on a given input is one of the reasons functions are so crucial in math. They allow us to form consistent relationships between different variables, like speed and braking distance. In our problem, the function \(d(v)\) represents the unique link between a car's speed \(v\) and its braking distance \(d\). It helps us determine how fast a car needs to be going to achieve a certain braking distance or, conversely, how far it will travel before stopping.

Evaluation of Functions

Evaluating a function means calculating the output for a given input value. This process involves using the rules defined by the function to turn a specific input into an output. Let's break down how we evaluate functions in the context of braking distance:

• Substitute the Input: Take your input value (like \(v = 30\)) and plug it into the function \(d(v)\).
• Calculate the Output: Use the function's rules to determine the corresponding output (braking distance). For \(d(30) = 111\), this means the car needs 111 feet to stop at 30 mph.
• Understanding Unknowns: Sometimes, you may have variables or unknowns as outputs, like in \(d(10) = b\). Here, \(b\) is a placeholder for the braking distance that the function's evaluation hasn't fully defined yet.

By evaluating functions, you can better understand how changing your input affects the outcome, aiding in decision-making and predicting future results. This is especially important in day-to-day applications like managing vehicle speeds and safety.