Problem 56 - Algebra 1: Concepts and Skills

Question

Use the following information. During the hammer throw event, a hammer is swung around in a circle several times until the thrower releases it. As the hammer travels in the path of the circle, it accelerates toward the center. This acceleration is known as centripetal acceleration. The speed \(s\) that the hammer is thrown can be modeled by the formula \(s=\sqrt{1.2 a},\) where \(a\) is the centripetal acceleration of the hammer prior to being released. Find the approximate centripetal acceleration (in meters per second per second) when the ball is thrown with a speed of 24 meters per second.

Step-by-Step Solution

Answer

The approximate centripetal acceleration when the hammer is thrown with a speed of 24 m/s is \(24^2 / 1.2\) m/s². Use a calculator to get the precise value.

Step 1: Understand the formula

We have the formula \(s=\sqrt{1.2 a}\), where \(s\) is the speed of the hammer and \(a\) is the centripetal acceleration. The goal is to rearrange this formula to express acceleration \(a\) in terms of speed \(s\).

Step 2: Rearrange the formula

Start by squaring both sides of the formula to eliminate the square root:\(s^2 = 1.2*a\).Then, solve the equation for \(a\) by dividing both sides by 1.2: \(a = s^2 / 1.2 \). This innovation expresses the acceleration \(a\) in relation to speed \(s\).

Step 3: Insert given value into the formula

Now we can use the rearranged formula to find the centripetal acceleration when the hammer is thrown with a speed of 24 m/s. Substituting \(s = 24\) m/s into the equation gives: \(a = 24^2 / 1.2\).

Key Concepts

Algebraic Manipulation

Algebraic manipulation is a fundamental tool used to solve equations and express variables in the form we need. This technique involves various operations such as adding, subtracting, multiplying, dividing, and factoring among others, to rearrange algebraic equations.

In the case of finding the approximate centripetal acceleration of a hammer in the hammer throw event, we start with the given equation for speed, which involves a square root:
\(s = \sqrt{1.2 a}\).

Our objective is to isolate the variable \(a\), the centripetal acceleration, on one side of the equation. To do so, we need to perform a sequence of algebraic manipulations:

• Square both sides of the equation to eliminate the square root.
• Rearrange the result so that \(a\) is on one side of the equation and all the terms without \(a\) are on the other side.

This process is essential in transforming the equation so that we can solve for the desired variable.

Square Root Operations

Square root operations are critical when dealing with equations that involve squared terms or when we encounter square roots directly.

In the given problem, speed is represented as the square root of a function of acceleration:
\(s = \sqrt{1.2 a}\).

This square root signifies that the speed is a function of the squared value of some constant multiplied by the acceleration. To remove the square root and make the equation easier to solve for \(a\), we square both sides:
\(s^2 = (\sqrt{1.2 a})^2\).

The squaring process is an 'undo' operation for the square root, allowing us to simplify the right side to just \(1.2 a\). It's crucial to understand this operation as it helps clear the square root and creates a solvable equation for the variable of interest.

Solving for Variables

Solving for variables is the ultimate goal in most algebraic problems; it is the process of finding the value of a variable that makes an equation true. This often requires a combination of the previously mentioned methods.

In our scenario, once we have squared both sides of the equation \(s^2 = 1.2 a\), we are left with a linear equation where the variable \(a\) is multiplied by a constant. Our next step is to isolate this variable:
\(a = s^2 / 1.2\).

Here, we divide both sides of the equation by the constant value (1.2) to find \(a\) alone on one side of the equation. This is facilitated by understanding the rules governing division and how it interacts with algebraic terms.

Once the variable \(a\) is isolated, we can substitute the known value of speed \(s\) to calculate the specific centripetal acceleration needed for our problem. Using algebraic manipulation and square root operations, we’ve translated a physical phenomenon into a solvable mathematical problem.