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Chapter 3: Problem 12

Describe how you would instruct someone to walk the line \(y=x\), where \(x\) is measured in seconds and \(y\) is measured in feet. Describe how to walk the line \(y=x\), where \(x\) is measured in seconds and \(y\) is measured in meters. Which line represents a faster rate? Explain.

Short Answer

Expert verified
Walking 1 meter per second is faster than walking 1 foot per second.

Step by step solution

01

Understanding the Line Equation

The line equation given is \(y = x\). This means for every unit increase in \(x\), there is an equal unit increase in \(y\). In this context, \(x\) is time measured in seconds, and \(y\) is the distance either in feet or meters.
02

Walking the Line in Feet

For the line \(y = x\) where \(y\) is measured in feet, for every second that passes \(x\), you walk one foot \(y\). This translates to a walking speed of 1 foot per second. So, for every time unit (second), you move forward one distance unit (foot).
03

Walking the Line in Meters

For the line \(y = x\) where \(y\) is measured in meters, the same principle applies. For every second that passes \(x\), you move one meter \(y\). This means you are walking at a speed of 1 meter per second, which is approximately 3.281 feet per second (since 1 meter 鈮� 3.281 feet).
04

Comparing the Rates

To determine which line represents a faster rate, compare the speeds. Walking 1 meter per second is faster than walking 1 foot per second because 1 meter is approximately 3.281 feet. Therefore, the line representing meters (1 meter per second) is faster than the line representing feet (1 foot per second).

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Key Concepts

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

Equation of a Line
In mathematics, the equation of a line gives us an analytical way to define the linear relationship between two variables. The most basic form of a line equation is expressed as \(y = mx + c\), where \(m\) is the slope and \(c\) is the y-intercept. However, in the given context, the equation of the line is \(y = x\), which is a simple form where:
  • The slope \(m\) is 1, meaning a perfect diagonal line with every increase in \(x\) mirrored by the same increase in \(y\).
  • The y-intercept \(c\) is 0, indicating that the line passes through the origin (0,0).
This means that as time progresses in seconds (\(x\)), the distance walked (\(y\)) increases by an equal measure, resulting in an identical increment in both units if they were referred to in the same measurement system like feet or meters.
Distance and Speed
Distance and speed are core elements of motion description. In this exercise, they manifest around how far you move over time. Here's what it means:
  • The distance \(y\) denotes how far you walk relative to the time passed, shown by \(y = x\); you move forward one unit (in this instance, either a foot or a meter) for each second passed.
  • Speed comes into play as it is the rate at which the distance changes with time. If you cover 1 foot each second, your walking speed is 1 foot per second.
To compare different speeds, consider the units. 1 meter per second is faster than 1 foot per second because 1 meter equals approximately 3.281 feet. Hence, in meters, the speed is faster when both distances refer to being traveled per second.
Unit Conversion
Unit conversion is the process by which you convert one type of unit into another, to have uniformity in measurements and permit easy comparability. Using the exercise example:
  • Feet and meters are both units of length, but they are not the same and differ in magnitude.
  • To convert between these units, remember that 1 meter is equal to approximately 3.281 feet.
Considering the conversion rate, when you walk 1 meter per second, it's actually covering 3.281 feet each second. Therefore, if asked to convert this speed into feet, you would simply multiply meters walked by 3.281 to find the corresponding feet walked. Understanding this relationship is crucial for comparing speeds in different units effectively.

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Most popular questions from this chapter

APPLICATION To plan a trip downtown, you compare the costs of three different parking lots. ABC Parking charges \(\$ 5\) for the first hour and \(\$ 2\) for each additional hour or fraction of an hour. Cozy Car charges \(\$ 3\) per hour or fraction of an hour, and The Corner Lot charges a \(\$ 15\) flat rate for a whole day. a. Make a table similar to the one shown. Write recursive routines to calculate the cost of parking up to 10 hours at each of the three lots. b. Make three different scatter plots on the same pair of axes showing the parking rates at the three different lots. Use a different color for each parking lot. Put the hours on the horizontal axis and the cost on the vertical axis. c. Compare the three scatter plots. Under what conditions is each parking lot the best deal for your trip? Use the graph to explain. d. Would it make sense to draw a line through each set of points? Explain why or why not.

APPLICATION A car is moving at a speed of \(68 \mathrm{mi} / \mathrm{h}\) from Dallas toward San Antonio. Dallas is about \(272 \mathrm{mi}\) from San Antonio. a. Write a recursive routine to create a table of values relating time to distance from San Antonio for 0 to \(5 \mathrm{~h}\) in \(1 \mathrm{~h}\) intervals. b. Graph the information in your table. c. What is the connection between your plot and the starting value in your recursive routine? d. What is the connection between the coordinates of any two consecutive points in your plot and the rule of your recursive routine? e. Draw a line through the points of your plot. What is the real-world meaning of this line? What does the line represent that the points alone do not? f. When is the car within \(100 \mathrm{mi}\) of San Antonio? Explain how you got your answer. g. How long does it take the car to reach San Antonio? Explain how you got your answer.

APPLICATION A long-distance telephone carrier charges \(\$ 1.38\) for international calls of 1 minute or less and \(\$ 0.36\) for each additional minute. a. Write a recursive routine using calculator lists to find the cost of a 7 -minute phone call. (a) b. Without graphing the sequence, give a verbal description of the graph showing the costs for calls that last whole numbers of minutes. Include in your description all the important values you need in order to draw the graph.

Consider the expression \(\frac{4(y-8)}{3}\). a. Find the value of the expression if \(y=5\). Make a table to show the order of operations. (a) b. Solve the equation \(\frac{4(y-8)}{3}=8\) by undoing the sequence of operations. (a)

Consider the equation \(2(x-6)=-5\). a. Solve the equation. b. Show how you can check your result by substituting it into the original equation.
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