Problem 12 - College Algebra with Modeling and Visualization

Question

Exercises 11 and 12: Write a symbolic representation (formula) for a function \(f\) that computes the following. (a) The distance traveled by a car moving at 50 miles per hour for \(x\) hours (b) The total number of hours in day \(x\) (c) The distance in miles between a runner and home after \(x\) hours if the runner starts 1 mile from home and jogs away from home at 6 miles per hour (d) A car's speed in feet per second after \(x\) seconds if its tires are 2 feet in diameter and rotating 14 times per second

Step-by-Step Solution

Answer

(a) \(f(x) = 50x\), (b) \(f(x) = 24x\), (c) \(f(x) = 1 + 6x\), (d) \(f(x) = 28\pi\) feet per second.

Step 1: Calculate Distance Traveled by Car

The formula for distance is given by the product of speed and time. For part (a), the car travels at 50 miles per hour for \(x\) hours. Thus, the distance traveled is calculated as:\[ f(x) = 50 \times x \]

Step 2: Calculate Total Number of Hours in Day

To find the total number of hours in day \(x\), since each day has exactly 24 hours, the function is simply:\[ f(x) = 24 \times x \]

Step 3: Calculate Runner's Distance from Home

The runner starts 1 mile from home and runs at 6 miles per hour. The distance from home after \(x\) hours is the initial distance plus the distance run in that time. Therefore:\[ f(x) = 1 + 6x \]

Step 4: Convert Car's Speed from RPM to Feet per Second

The rotational speed of the car's tires must be converted to a linear speed. The tire's circumference is the distance traveled in one rotation, which is \( \pi \times \text{diameter}\). The tires rotate 14 times per second, thus:\[ \text{Circumference} = \pi \times 2 = 2\pi \text{ feet} \]\[ \text{Linear Speed} = 14 \times 2\pi = 28\pi \text{ feet per second} \]The car's speed does not depend on \(x\), but rather is a constant based on the given tire specifications.

Key Concepts

Understanding the Distance Formula

The distance formula is crucial for calculating how far an object has traveled over a period of time. It is given in the form of the product of speed and time. For example, if a car moves at a speed of 50 miles per hour, the distance covered in 'x' hours can be calculated easily with this formula:

• Distance = Speed × Time
• Use this formula to find how far the car travels in 'x' hours: \(f(x) = 50 \times x\)

Whenever you have a consistent speed and need to know how much distance an object covers in a certain time, this simple multiplication will get you the result. The beauty of the distance formula is its simplicity and its wide applicability, from average car trips to complex physics problems.

Mastering Unit Conversion

Unit conversion allows you to understand measurements in different units and is essential for solving practical problems. In cases where measurements need to adhere to a certain unit standard, like converting speed from miles per hour to feet per second, unit conversion steps in to make this possible. Here, let's see how this works specifically for the car's speed:

• Calculate the tire's circumference: Since the tire diameter is given as 2 feet, the circumference is \(\pi \times 2 = 2\pi\) feet.
• The tires rotate 14 times per second. Thus, the linear speed of the car is \(14 \times 2\pi = 28\pi\) feet per second.

This example of unit conversion helps us understand how a rotational speed can be expressed in a linear format. By following the steps of identifying what needs to be converted and applying the correct formulas, we can navigate changes between units with ease. Remember, knowing the conversion factors is often the key!

Working with Algebraic Expressions

Algebraic expressions are at the heart of deriving formulas and solving mathematical problems. They involve combining numbers, variables, and arithmetic operators. In the given exercises, you have seen these expressions used to calculate various measures, from distance to speed.

• Consider the runner's distance from home. It begins with an initial 1-mile distance and increases by 6 miles per hour with each passing hour. Representing this as an algebraic expression becomes:\[f(x) = 1 + 6x\]
• These expressions offer precise representation of the relationship between different quantities such as time, distance, and speed.

Understanding how to manipulate these expressions is crucial for problem-solving. They help transform word problems into mathematical models that can be analyzed and interpreted efficiently. By practicing constructing and simplifying algebraic expressions, you become adept at expressing complex relations in simple mathematical terms.