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8.1: Boats in Motion

Difficulty Level: At Grade Created by: CK-12

This activity is intended to supplement Algebra I, Chapter 7, Lesson 2.

Problem 1 - Solving for Two Unknowns

From the town of Alton to Barnhart, the Mississippi River has an average surface speed of about \begin{align*}2 \ mph\end{align*}. Suppose it takes a boat 3 hours to travel downstream, but 5 hours to travel upstream the same distance between the two towns.

1. Let \begin{align*}r\end{align*} be the rate of the boat in still water. How could the rate upstream and the rate downstream be expressed?

2. Use the above information to fill in the blank spaces.

\begin{align*}\text{distance} = \text{rate} \times \text{time}\end{align*}

down \begin{align*}d = \underline{\;\;\;\;\;\;\;\;} \times \underline{\;\;\;\;\;\;\;\;}\end{align*}

up \begin{align*}d = \underline{\;\;\;\;\;\;\;\;} \times \underline{\;\;\;\;\;\;\;\;}\end{align*}

Note: The solution occurs when the distance \begin{align*}d\end{align*} in both equations is the same.

3. Set the equations equal to each other and solve for \begin{align*}r\end{align*} algebraically. Show your work here.

Solve the system of equations graphically. Press \begin{align*}Y=\end{align*} and enter both equations.

Note: \begin{align*}x\end{align*} should replace \begin{align*}r\end{align*}.

The solution is the intersection point. Adjust the window so that you can see the intersection of the two lines.

Find this intersection point by pressing \begin{align*}2^{nd}\end{align*} [TRACE] and selecting intersect. Select the first line, the second line, and make a guess. The coordinates of the intersection point will appear.

4. How does this point compare with your solution from Question 3?

Record the rate of the boat in still water and the distance between the towns. Include units.

Problem 2 - Distance-time graph, explore slopes

Velma is riding on a steam engine locomotive. As she walks forward, she travels \begin{align*}1.1 \ miles\end{align*} in \begin{align*}2 \ minutes\end{align*}. As she walks back to her seat she travels \begin{align*}0.9 \ miles\end{align*} in \begin{align*}2 \ minutes\end{align*}. How fast is the train moving and how fast was Velma walking inside the train?

To explore this situation graphically, we need to set up the graph.

Create a scatter plot of the data points.

Press STAT[ENTER] and enter the two pieces of the data (2, 1.1) and (2, 0.9) into lists \begin{align*}L1\end{align*} and \begin{align*}L2\end{align*}.

Press \begin{align*}2^{nd}\end{align*} [Y=]ENTER, choose scatter plot, \begin{align*}L1\end{align*} for X list and \begin{align*}L2\end{align*} for Y list.

Press WINDOW. Set \begin{align*}x\end{align*} to \begin{align*}[-0.5, 2.5]\end{align*} and \begin{align*}y\end{align*} to \begin{align*}[-0.2, 1.5]\end{align*}.

Graph the two \begin{align*}d = r \cdot t\end{align*} equations

Press APPS and select Transform. Enter the equations on the \begin{align*}Y=\end{align*} screen as shown at the right.

Then press WINDOW and arrow to SETTINGS.

Set \begin{align*}A=0.1\end{align*}, \begin{align*}B=0.1\end{align*} and \begin{align*}\text{Step} = 0.01\end{align*}

In these equations, \begin{align*}A\end{align*} represents the rate of the train and \begin{align*}B\end{align*} represent the rate of Velma walking.

Press GRAPH. Use the arrow keys to adjust the values of \begin{align*}A\end{align*} and \begin{align*}B\end{align*} so the line goes through the point. The \begin{align*}Y1\end{align*} line should go through the top point and the \begin{align*}Y2\end{align*} line should go through the bottom point.

5. What does the slope of the line in the distance-time graph represent?

6. Apply \begin{align*}d = r \cdot t\end{align*} to this situation. \begin{align*}r = ?\end{align*}

(Hint: \begin{align*}r\end{align*} depends on \begin{align*}A\end{align*}, the speed of the steam engine, and \begin{align*}B\end{align*}, the velocity of Velma.)

\begin{align*}\text{distance} = \text{rate} \times \text{time}\end{align*}

forward \begin{align*}\underline{\;\;\;\;} = \underline{\;\;\;\;\;\;\;\;} \times \underline{\;\;\;\;\;}\end{align*}

back \begin{align*}\underline{\;\;\;\;} = \underline{\;\;\;\;\;\;\;\;} \times \underline{\;\;\;\;\;}\end{align*}

7. Algebraically solve the equation.

Problem 3 - Extension/Homework

1. An airplane flew \begin{align*}3 \ hrs\end{align*} with a tail wind of \begin{align*}20 \ km/h\end{align*}. The return flight with the same wind took \begin{align*}3.5 \ hrs\end{align*}. Find the speed of the airplane in still air. Fill in the chart are answer and solve.

\begin{align*}\text{distance} = \text{rate} \times \text{time}\end{align*}

west \begin{align*} \underline{\;\;\;\;\;} = (r + \underline{\;\;\;\;\;}) \times \underline{\;\;\;\;\;}\end{align*}

east \begin{align*} \underline{\;\;\;\;\;} = (r - \underline{\;\;\;\;\;}) \times \underline{\;\;\;\;\;}\end{align*}

2. Two cars leave town going in opposite directions. One travels \begin{align*}50 \ mph\end{align*} and the other travels \begin{align*}30 \ mph\end{align*}. In how many hours will they be \begin{align*}160 \ miles\end{align*} apart?

\begin{align*}\text{distance} = \text{rate} \times \text{time}\end{align*}

slow car \begin{align*}\underline{\;\;\;\;\;} = \underline{\;\;\;\;\;} \times \underline{\;\;\;\;\;}\end{align*}

fast car \begin{align*}\underline{\;\;\;\;\;} = \underline{\;\;\;\;\;} \times \underline{\;\;\;\;\;}\end{align*}

Hint: Distances will add up to 160

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