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Exercises 46 to 48 refer to the following setting. Some high school physics students dropped a ball and measured the distance fallen (in centimeters) at various times (in seconds) after its release. If you have studied physics, then you probably know that the theoretical relationship between the variables is distance \(=490(\text { time })^{2}\) . A scatterplot of the students data showed a clear curved pattern. Which of the following single transformations should linearize the relationship? I. time^ \(^{2}\) II. distance \(^{2}\) III. \(\sqrt{\text { distance }}\) (a) I only (c) III only (e) I and III only (b) II only (d) I and II only

Short Answer

Expert verified
I and III only (e)

Step by step solution

01

Understand the Problem

We need to determine which transformation will make the relationship between time and distance linear, given that the equation is \( \text{distance} = 490(\text{time})^{2} \). A scatterplot shows a curved pattern, indicating a non-linear relationship.
02

Apply Transformation I

Transformation I suggests using \( \text{time}^{2} \). Since the equation is already \( \text{distance} = 490(\text{time})^{2} \), using \( \text{time}^{2} \) will turn the relationship into \( \text{distance} = 490 \times \text{time term} \). This is a linear relationship.
03

Apply Transformation II

Transformation II suggests using \( \text{distance}^{2} \). Substituting and squaring both sides gives \( \text{distance}^2 = (490(\text{time})^{2})^2 \), which results in a higher degree non-linear relationship (degree 4). This does not linearize the relationship.
04

Apply Transformation III

Transformation III suggests using \( \sqrt{\text{distance}} \). Substituting gives \( \sqrt{\text{distance}} = \sqrt{490} \times \text{time} \), which simplifies to \( \sqrt{490} \cdot \text{time} \). This is also a linear relationship.
05

Evaluate Transformations

From the above transformations, both \( \text{time}^{2} \) (I) and \( \sqrt{\text{distance}} \) (III) transform the non-linear relation into a linear one, whereas \( \text{distance}^{2} \) (II) does not.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Scatterplot Analysis
To understand scatterplot analysis, think of a scatterplot as a type of graph used to study relationships between two variables. In our physics experiment, the students created a scatterplot to show the relationship between time and distance. This scatterplot revealed a curved pattern, highlighting the non-linear relationship between the variables.
A scatterplot helps us quickly identify whether a relationship is linear or non-linear, and guides us to choose the appropriate transformation to linearize the data.
  • Look for patterns: A straight line indicates a linear relationship, while a curve indicates non-linearity.
  • Determine the correlation: The tighter the points cluster around a line or curve, the stronger the relationship between the variables.
  • Use scatterplots to predict: By analyzing patterns, students can make informed predictions.
Scatterplots thus serve as a visual tool that is crucial for tasks like identifying the right transformation in data analysis.
Non-linear Transformation
Non-linear transformation refers to changing a variable to potentially linearize a relationship that is initially non-linear. In our exercise, the relationship between time and distance was not linear, which was evident from the curved pattern in the scatterplot.
Transformations are applied to change the shape of data. Here, the theoretical equation given was \( \text{distance} = 490(\text{time})^2 \). Our task was to figure out if any of the suggested transformations (I. \( \text{time}^2 \), II. \( \text{distance}^2 \), or III. \( \sqrt{\text{distance}} \)) would linearize the data.
  • Understanding that applying \( \text{time}^2 \) simplifies our data to a linear form because \( \text{distance} \) is already a function of \( \text{time}^2 \).
  • Squaring distance would increase the non-linearity, failing to linearize the data.
  • Taking the square root of distance simplifies to a formulation involving just time, making it linear.
Non-linear transformations are powerful tools for mathematicians and scientists, making seemingly complex data easier to analyze.
Linear Relationships
Linear relationships are at the heart of many scientific studies, including our physics experiment. A linear relationship between two variables means that changes in one variable cause proportional changes in the other.
Initially, our scatterplot showed a non-linear relationship between time and distance. This was because the relationship theoretically follows \( \text{distance} = 490(\text{time})^2 \), which on its own is quadratic in nature. By identifying the appropriate transformations, the aim was to express this as a linear function.
  • The implementation of \( \text{time}^2 \) immediately transforms our equation into a linear form \( \text{distance} = 490 \cdot \text{transformed term} \).
  • Alternatively, \( \sqrt{\text{distance}} \) transformation also results in a relationship directly proportional to time.
Understanding linear relationships allows students to predict outcomes and understand the regular patterns in various datasets, making analysis and interpretation much more efficient.
Physics Experiments in Education
Physics experiments in education are crucial in helping students understand and apply theoretical concepts in a practical setting. The exercise in focus involved dropping a ball to measure the distance over time, aligning with the well-known physics equation \( \text{distance} = 490(\text{time})^2 \).
Hands-on experiments like this are valuable teaching tools for several reasons:
  • They reinforce theoretical learning: By practically applying formulas, students can see how equations work in real-life circumstances.
  • They enhance problem-solving skills: Students learn to hypothesize, test, and deduce conclusions based on experimental data.
  • They improve conceptual understanding: Observing phenomena firsthand deepens comprehension of abstract topics like the effects of gravity.
  • They connect physics with mathematics: Experiments often require data analysis, providing an opportunity to apply statistical knowledge.
Incorporating experiments into physics education not only enriches learning but also fosters a deeper appreciation for science and its applications in daily life.

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Most popular questions from this chapter

Exercises 46 to 48 refer to the following setting. Some high school physics students dropped a ball and measured the distance fallen (in centimeters) at various times (in seconds) after its release. If you have studied physics, then you probably know that the theoretical relationship between the variables is distance \(=490(\text { time })^{2}\) . A scatterplot of the students data showed a clear curved pattern. At 0.68 seconds after release, the ball had fallen 220.4 centimeters. How much more or less did the ball fall than the theoretical model predicts? (a) More by 226.576 centimeters (b) More by 6.176 centimeters (c) No more and no less (d) Less by 226.576 centimeters (e) Less by 6.176 centimeters

Beavers and beetles Do beavers benefit beetles? Researchers laid out 23 circular plots, each four meters in diameter, at random in an area where beavers were cutting down cottonwood trees. In each plot, they counted the number of stumps from trees cut by beavers and the number of clusters of beetle larvae. Ecologists think that the new sprouts from stumps are more tender than other cottonwood growth, so that beetles prefer them. If so, more stumps should produce more beetle larvae.\(^{7}\) Minitab output for a regression analysis on these data is shown below. Construct and interpret a 99% confidence interval for the slope of the population regression line. Assume that the conditions for performing inference are met. Regression Analysis: Beetle larvae versus Stumps $$ \begin{array}{lcccc} \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & -1.286 & 2.853 & -0.45 & 0.657 \\ \text { Stumps } & 11.894 & 1.136 & 10.47 & 0.000 \\ \mathrm{~S}=6.41939 & \mathrm{R}-\mathrm{Sq} & =83.98\% &\mathrm{R}-\mathrm{Sq}(\mathrm{adj}) & =83.18\% \end{array} $$

Determining tree biomass It is easy to measure the 鈥渄iameter at breast height鈥 of a tree. It鈥檚 hard to measure the total 鈥渁boveground biomass鈥 of a tree, because to do this you must cut and weigh the tree. The biomass is important for studies of ecology, so ecologists commonly estimate it using a power model. Combining data on 378 trees in tropical rain forests gives this relationship between biomass y measured in kilograms and diameter x measured in centimeters: \(^{20}\) $$\widehat{\ln y}=-2.00+2.42 \ln x$$ Use this model to estimate the biomass of a tropical tree 30 centimeters in diameter. Show your work.

Shower time (1.3, 2.2, 6.3, 7.3) Marcella takes a shower every morning when she gets up. Her time in the shower varies according to a Normal distribution with mean 4.5 minutes and standard deviation 0.9 minutes. (a) If Marcella took a 7-minute shower, would it be classified as an outlier? Justify your answer. (b) Suppose we choose 10 days at random and record the length of Marcella鈥檚 shower each day. What鈥檚 the probability that her shower time is 7 minutes or higher on at least 2 of the days? Show your work. (c) Find the probability that the mean length of her shower times on these 10 days exceeds 5 minutes. Show your work.

Multiple choice: Select the best answer for Exercises 21 to 26. What is the correlation between selling price and appraised value? (a) 0.1126 (c) -0.861 (e) -0.928 (b) 0.861 (d) 0.928

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