Problem 23 - Algebra 1: Concepts and Skills

Question

In 1866 Texas cowhands used the Chisholm Trail to drive cattle north to the railroads in Kansas. The average rate r that the cattle could be moved along the trail was 11 miles per day. a. Write an equation, where \(d\) is distance and \(t\) is time in days. b. Make a table of input \(t\) and output \(d\) for \(t=7,14,\) and 28 Then graph the data. c. The distance \(d\) from San Antonio to Abilene was about 1100 miles. How long did it take to drive cattle the entire length of the trail?

Step-by-Step Solution

Answer

a. The equation is \(d = 11t\). b. The table is: \(t = 7, d = 77\), \(t = 14, d = 154\), \(t = 28, d = 308\). The graph of the equation \(d=11t\) is a straight line passing through the origin with a slope of 11. c. It takes approximately 100 days to drive cattle the entire length of the trail.

Step 1: Formulate the equation

The average rate \(r\) that the cattle could be moved along the trail is 11 miles per day given by \(r = d / t\) with \(d\) as distance and \(t\) as time in days. To get the equation in terms of \(d\) we multiply both sides by \(t\) resulting in \(d = r \cdot t\)

Step 2: Create a table and graph the data

For the table of input \(t\) and output \(d\), substitute \(t = 7, 14, 28\) into the equation \(d = 11t\) respectively to get the corresponding \(d\). Graphing these, the x-axis will represent the time \(t\) and the y-axis will represent the distance \(d\). Plot the points and join them.

Step 3: Calculate the time to drive the entire length of the trail

To find the time it took to drive the cattle the entire length of the trail, we know the distance \(d = 1100\) miles. We substitute the values into the equation representing time: \(t = d / r\) to find the value of \(t\).

Key Concepts

Rate of Change

Understanding the rate of change is fundamental in algebra, especially when dealing with real-world scenarios such as calculating speed or growth over time. In the context of our Chisholm Trail example, the rate of change represents the speed at which cattle are driven along the trail — specifically, 11 miles per day.

Mathematically, the rate of change is the ratio that shows the relationship between two variables. If we consider the variable 'distance traveled (\(d\))' and 'time (\(t\/\)) in days', their rate of change can be calculated using the formula:
\[\begin{equation}rate\ of\ change = \frac{\text{change in distance (}\Delta d\text{)}}{\text{change in time (}\Delta t\text{)}}\end{equation}\]
In a linear relationship, this rate of change is constant, just like the 11 miles per day for the cattle drive. This concept is also referred to as the slope in coordinate geometry.

Slope-Intercept Form

The slope-intercept form of a line is an important concept in algebra and is expressed as \[\begin{equation} y = mx + b\end{equation}\], where \[\begin{equation} m\end{equation}\] is the slope or rate of change, and \[\begin{equation} b\end{equation}\] is the y-intercept, or the point where the line crosses the y-axis.

For the cattle drive problem, we formulated an equation \[\begin{equation} d = 11t\end{equation}\]. This equation is similar to the slope-intercept form, with \[\begin{equation} d\end{equation}\] corresponding to \[\begin{equation} y\end{equation}\], \[\begin{equation} 11\end{equation}\] as our \[\begin{equation} m\end{equation}\] (rate of change or slope), and \[\begin{equation} t\end{equation}\] to \[\begin{equation} x\end{equation}\]. Since there is no y-intercept in this particular equation (or it is 0), the line will pass through the origin, representing that no distance is covered at time zero.

Distance-Time Graphs

Distance-time graphs are graphical representations that show how distance varies with time. They are a key tool in understanding motion and speed over time in algebra and physics.

When graphing our cattle drive problem, we plot time (\[\begin{equation} t\end{equation}\]) on the x-axis and distance (\[\begin{equation} d\end{equation}\]) on the y-axis. Each point on the graph represents the distance covered by the cattle at a certain time. The graph of a constant rate of change, like 11 miles per day, will be a straight line, as the relationship between \[\begin{equation} t\end{equation}\] and \[\begin{equation} d\end{equation}\] is linear. The steeper the line, the faster the rate of change, indicating quicker travel over time.

Problem-Solving in Algebra

Algebra is full of real-world applications, and problem-solving often involves creating and manipulating algebraic equations to find unknown values.

In solving the Chisholm Trail problem, we had to understand the relationship between distance and time in order to formulate an equation. We then used this equation to solve for unknown variables like the time required (\[\begin{equation} t\end{equation}\]) to cover a given distance (\[\begin{equation} d\end{equation}\]). Algebraic problem-solving skills enable us to handle tasks ranging from simple calculations to complex real-world scenarios. By breaking down the problem into smaller steps, identifying the known and unknown variables, formulating equations, and solving them, students can adeptly navigate through various algebraic challenges.