Problem 20 - Algebra 1

Question

Effects connection Different masses \(M\) (in kilograms) are hung from a spring. The distances \(d\) (in centimeters) that the spring stretches are shown in the table. Test different models to see which type of model best fits the data. Write a model that accurately represents the data. (IMAGE CAN'T COPY)$$\begin{array}{|l|c|c|c|c|c|c|c|}\hline \text { Mass, M } & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\\\hline \text { Distance, } d & 2.6 & 5.2 & 7.8 & 10.4 & 13 & 15.6 & 18.2 \\\\\hline\end{array}$$

Step-by-Step Solution

Answer

The model that accurately represents the data is \(d = 2.6M\). This model was found by observing a constant rate of change in distance for every 1Kg increase in mass which prompted a linear model. The slope of the linear model was determined to be 2.6 by calculating the rate of change between two data points and assuming no initial stretch on the spring gave a y-intercept of 0.

Step 1: Understand the Data

The first step is to understand the relationship between the mass \(M\) and distance \(d\). We can do this by plotting the values on a graph. When the mass is 1Kg, the distance is 2.6cm and as we increase the mass, the distance also increases. By observing the data, it seems like for each increase in mass by 1Kg, the distance increases by approximately 2.6cm.

Step 2: Analyze Possible Models

Next, start analyzing different mathematical models to see which one fits the data best. For instance, we can try linear, polynomial, exponential, and logarithmic models. However, looking at the pattern of the data, a linear model (y = mx + c) seems to best fit since the distance increases at a constant rate as the mass increases. The slope (m) of this line will be the constant rate of change or the amount the distance increases for every 1Kg increase in mass, and the intercept (c) will be the initial distance when the mass is zero.

Step 3: Write the Model

The model that best fits this data is the linear model. In this model, the slope \(m\) is calculated by taking any two points from the data and finding the change in distance divided by the change in mass. For instance, taking the first two readings, we find that \(m\) = (5.2 - 2.6) / (2 - 1) = 2.6. The intercept \(c\) ideally should represent the distance when the mass is zero. However, from the data given this is not possible. Thus, we can assume that the spring has no initial stretch so that \(c = 0\).Therefore, the accurate model that represents this data is \(d = 2.6M\).

Step 4: Test the Model

The last step is to test the model against the data in the table. If we plug in the values of mass from the table into the equation, it should give us the corresponding distances. For instance, when \(M = 6Kg\) the model predicts \(d = 2.6 * 6 = 15.6cm\), which is exactly what's in the table. So, it confirms that the model is accurate.

Key Concepts

Linear Model

When analyzing a set of data, one of the simplest models to consider is the linear model. This model is often used when there is a constant rate of change between two variables. In the context of our exercise, we have masses and distances stretched by a spring. A linear model can help illustrate how the distance changes as we alter the mass.
A linear model is represented by the equation:

• \( y = mx + c \)

Here, \( y \) is the dependent variable (distance), \( x \) is the independent variable (mass), \( m \) is the slope, and \( c \) is the y-intercept. The slope \( m \) captures the constant rate of change, indicating how much the distance increases for each kilogram increase in mass. The intercept \( c \) is the starting value, representing the initial distance when no mass is added.
For a linear relationship, plotting the points should give a straight line. In our exercise, the distances increased consistently with the masses, supporting a linear relationship where distance is directly proportional to mass, represented by the equation \( d = 2.6M \). Here, \( c = 0 \) because the problem assumes no initial stretch of the spring with zero mass.

Slope Calculation

The slope is a crucial component of the linear model, as it defines the rate at which one variable changes relative to another. In mathematical terms, the slope \( m \) in the linear equation is determined by the ratio of the 'rise' (change in distance) to the 'run' (change in mass).
To find it, we can use any two data points from our mass and distance pairs and apply the formula:

• \( m = \frac{{ ext{{change in distance}}}}{{ ext{{change in mass}}}} \)

For example, using the points (1, 2.6) and (2, 5.2) from our dataset, we calculate the slope as:

• \( m = \frac{{5.2 - 2.6}}{{2 - 1}} = 2.6 \)

This consistent slope across all intervals confirms the linear relationship, demonstrating that for each additional kilogram of mass, the spring stretches an additional 2.6 centimeters. This slope calculation is critical because it directly influences the accuracy of the linear model.

Data Analysis

Data analysis involves interpreting and extracting meaningful insights from our data. In this exercise, our focus is to determine how the stretch of a spring varies with different masses. By looking closely at the information provided, we sought a pattern or trend.
The first step is visualizing the data, typically by plotting it on a graph. This makes it easier to notice if the relationship between the variables is consistent with a straight line, supporting a linear model. Every calculated data point supports the hypothesis of a linear relationship, where the increase in distance is consistent with the increasing mass.
Once we've established the pattern, it becomes essential to determine which model fits best. While numerous models can be tested, a linear pattern was evident in this data. This choice simplifies our approach and provides a useful, easy-to-understand representation of the relationship. Testing our linear model further by substituting back into the original data and comparing predicted distances to actual values, confirms the accuracy and reliability of our model.
Thus, data analysis not only guides us in identifying the right model but also validates its applicability to represent real-world phenomena accurately.