Problem 42 - Algebra A Combined Function

Question

When a wind blows perpendicularly against a flat surface, its force is jointly proportional to the surface area and the speed of the wind. A sail whose surface area is 12 square feet experiences a 20 -pound force when the wind speed is 10 miles per hour. Find the force on an 8-square-foot sail if the wind speed is 12 miles per hour.

Step-by-Step Solution

Answer

The force on the 8-square-foot sail is 16 pounds.

Step 1: Understand the joint proportionality

Since the force is jointly proportional to area and wind speed, we can express it as \( F = k imes A imes v \), where \( F \) is the force, \( A \) is the area, \( v \) is the wind speed, and \( k \) is the proportionality constant.

Step 2: Determine the Proportionality Constant

Using the information given for the first sail, where \( F = 20 \) pounds, \( A = 12 \) square feet, and \( v = 10 \) mph, substitute into the equation to find \( k \): \( 20 = k \times 12 \times 10 \). Solving for \( k \), we have \( k = \frac{20}{120} = \frac{1}{6} \).

Step 3: Calculate the Force on the 8-square-foot sail

Now, use the value of \( k \) to find the force on the new sail. The area \( A \) is 8 square feet and the wind speed \( v \) is 12 mph. Plug these values into the formula: \( F = \frac{1}{6} \times 8 \times 12 \).

Step 4: Simplify the Expression

Perform the calculations: \( 8 \times 12 = 96 \), and \( \frac{1}{6} \times 96 = 16 \) pounds.

Key Concepts

Proportionality Constant

The concept of a proportionality constant is fundamental when dealing with joint proportionality, as it helps link different variables in a simplified manner. In this scenario, when we say the force of the wind is jointly proportional to the surface area of a sail and the speed of the wind, it means the force varies directly as a product of the area and the wind speed. To express this mathematically, we use a constant term, known as the proportionality constant, denoted by the letter \( k \).

• The formula used is \( F = k \times A \times v \), where:
• \( F \) is the force exerted by the wind,
• \( A \) is the surface area, and
• \( v \) is the wind speed.

The process to find \( k \) involves isolating it in the equation by substituting known values of \( F \), \( A \), and \( v \). For example, with an area of 12 square feet, wind speed of 10 miles per hour, and a force of 20 pounds, we rearrange the formula to solve for \( k \):

• \( 20 = k \times 12 \times 10 \)
• Simplifying gives \( k = \frac{20}{120} = \frac{1}{6} \).

Understanding \( k \) helps us predict how other combinations of area and speed will affect the force.

Force Calculation

Once the proportionality constant \( k \) is determined, calculating the force for different scenarios becomes straightforward. To find the force on a different sail, you substitute the respective area and wind speed into the joint proportionality formula, \( F = k \times A \times v \). This direct relationship guides how changes in area or speed influence the force.
Taking the new sail with an area of 8 square feet and a wind speed of 12 miles per hour, we use the previously calculated \( k \) value. The calculation proceeds with:

• \( F = \frac{1}{6} \times 8 \times 12 \).
• First, calculate \( 8 \times 12 = 96 \),
• then multiply by \( \frac{1}{6} \) to find \( F \).\( \frac{1}{6} \times 96 = 16 \) pounds.

This straightforward method allows you to easily determine the impact of different variable adjustments on the force.

Algebraic Expressions

Algebraic expressions play a vital role in solving problems involving joint proportionality and constant calculations. Understanding how to manipulate these expressions is crucial for effective problem-solving. At the core, you are working with equations that express the relationship between multiple variables.

• Expressions such as \( F = k \times A \times v \) can initially seem complex.
• Key operations include substituting known values and isolating variables to find unknown quantities.

In the exercise, we applied algebra to isolate the proportionality constant by rearranging the formula:

• \( 20 = k \times 12 \times 10 \) simplified to solve for \( k \).
• Through algebraic manipulation, \( k \) was isolated: \( k = \frac{20}{120} \).

This formula rearrangement and understanding of expressions are pivotal when calculating further forces or making predictions based on changes in the incorporated variables.