Problem 48 - Algebra 2

Question

A wooden pole swings back and forth over the cup on a miniature golf hole. One player pulls the pole to the side and lets it go. Then it follows a swing pattern of 25 centimeters, 20 centimeters, 16 centimeters, and so on until it comes to rest. What is the total distance the pole swings before coming to rest?

Step-by-Step Solution

Answer

The total distance the pole swings is 150 cm.

Step 1: Identify the Pattern

The swing pattern is 25 cm, 20 cm, 16 cm, and so forth. Observe that each swing is decreasing consistently in a specific pattern.

Step 2: Identify the Sequence

Recognize the pattern in swings is described by an arithmetic sequence. The first term, \(a_1 = 25\) cm, the second term, \(a_2 = 20\) cm, showing a common difference \(d = -5\) cm \(a_2 - a_1\).

Step 3: Express the Sequence Formula

The swing lengths form an arithmetic sequence, expressed as \(a_n = a_1 + (n - 1)d\), where \(a_n\) is the nth term, \(a_1 = 25\) cm, and \(d = -5\) cm.

Step 4: Determine the Number of Terms (n)

To find when the sequence ends, set \(a_n = 0\) (when it comes to rest) and solve for \(n\): \[0 = 25 + (n - 1)(-5)\] Solve: \[-25 = (n - 1)(-5)\] \[5n - 5 = 25\] \[5n = 30\] \[n = 6\]This shows that after 6 swings, the pole comes to resting position.

Step 5: Calculate the Total Distance

The total distance swung is twice the sum of the arithmetic sequence since each swing goes back and forth. First, find the sum of the sequence:\[S_n = \frac{n}{2} \times (a_1 + a_n)\]\[S_6 = \frac{6}{2} \times (25 + 0)\]\[S_6 = 3 \times 25 = 75 \text{ cm}\]Then double the sum because of the back and forth motion:\[Total = 2 \times S_6 = 2 \times 75 = 150 \text{ cm}\]

Key Concepts

Common Difference

An arithmetic sequence is a sequence of numbers where each term after the first is obtained by adding a constant, known as the **common difference**. In this problem, the swing of the wooden pole follows an arithmetic sequence where each swing decreases. You can identify the common difference by subtracting the first term from the second term. For this swing pattern:

• The first swing is 25 cm.
• The second swing is 20 cm.

The common difference is calculated as follows: \[ d = 20 - 25 = -5 \]This tells us that each subsequent swing is 5 cm less than the previous one. Understanding this difference is essential as it allows us to predict subsequent terms and understand the pattern of the swings.

Sequence Formula

In arithmetic sequences, you can use the **sequence formula** to find any term in the sequence. The formula is:\[ a_n = a_1 + (n - 1) imes d \]where:- \(a_n\) is the nth term you want to find,- \(a_1\) is the first term of the sequence,- \(d\) is the common difference between the terms, and- \(n\) is the term number.

In our example, the first swing length \(a_1\) is 25 cm and the common difference \(d\) is -5 cm. If we want to find the nth term when the swing comes to rest, this formula helps solve for \(n\) when \(a_n = 0\). This practical application of the sequence formula helps find the number of swings before the pole comes to a complete stop.

Sum of Arithmetic Sequence

The **sum of an arithmetic sequence** is useful when you want to know the total of the terms up to a certain point. In this problem, it helps calculate the total distance the pole swings. The sum \(S_n\) of the first \(n\) terms is found using:\[ S_n = \frac{n}{2} \times (a_1 + a_n) \]Here:- \(n\) is the number of terms,- \(a_1\) is the first term,- \(a_n\) is the last term in the sequence.

For our swinging pole:- \(n = 6\) (the number of swings until it stops),- \(a_1 = 25 \) cm,- \(a_6 = 0\) cm (the last swing as it stops).

Substitute these into the formula:\[ S_6 = \frac{6}{2} \times (25 + 0) = 3 \times 25 = 75 \text{ cm} \]This sum reflects the total distance for one way, which must be doubled because each swing travels forward and back.

Back and Forth Motion

When considering a swinging motion, such as in this scenario with the wooden pole, it's crucial to account for the **back and forth motion**. This means each swing travels twice the calculated distance because it swings forward and then back to its starting position. For example, when calculating the total swing distance:

• The sum of one complete sequence in one direction is 75 cm.
• Due to the back and forth motion, the actual total distance is doubled.

So, the total swing measured in both directions is:\[ \text{Total Distance} = 2 \times 75 = 150 \text{ cm} \]This step ensures you capture the entirety of the physical motion, accounting for every part of its journey before it comes to rest. It highlights the entire length covered across both directions of each swing.