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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.

ID:11237

Time Required: 15 minutes

Activity Overview

In this activity, students will explore the motion of a boat going up and down the river. They will be instructed to solve the resulting system of equations algebraically and graphically. The transformation graphing application will allow students to explore the slope of a distance-time graph. Additional problems are provided for further practice.

Topic: Linear Equations

• Motion, distance=rate×time\begin{align*}distance = rate \times time\end{align*}
• Slope and graphically solving equation

Teacher Preparation and Notes

• The student worksheet provides instructions and question to guide the inquiry and focus the observations.

Associated Materials

Problem 1 – Boat Motion & Graphically Solve

Ask students why the boat goes faster downstream than upstream. They should know that the boat goes with the current downstream making it travel more distance in less time. When the boat goes upstream, it has to fight the current.

1. Downstream rate =r+2\begin{align*}= r + 2\end{align*}; Upstream rate =r2\begin{align*}= r - 2\end{align*}

2. dd=3(r+2)=5(r2)\begin{align*}d & = 3(r + 2) \\ d & = 5(r - 2)\end{align*}

3. 3r+62rr=5r10=16=8 mph\begin{align*}3r + 6 & = 5r - 10 \\ 2r & = 16 \\ r & = 8\ mph\end{align*}

Substituting this into either equation gives a distance of 30 miles\begin{align*}30\ miles\end{align*}. Alton and Barnhart are about 30 miles\begin{align*}30\ miles\end{align*} apart along the Mississippi in the St. Louis area.

Problem 2 – Distance-Time Graph, Explore Slopes

Using d=rt\begin{align*}d = r \cdot t\end{align*} for this situation gives the following equations:

1.10.9=(A+B)2=(AB)2\begin{align*}1.1 & = (A + B) 2 \\ 0.9 & = (A - B) 2\end{align*}

where A\begin{align*}A\end{align*} is the rate (speed) of the steam engine and B\begin{align*}B\end{align*} is the rate (velocity) of Velma’s walking.

Students should set up lists L1\begin{align*}L1\end{align*} and L2\begin{align*}L2\end{align*} and the scatter plot settings as shown at right.

On the graph, students are to use the arrows keys to change the rates of the train (A)\begin{align*}(A)\end{align*} and Velma’s walking (B)\begin{align*}(B)\end{align*} so that the lines go through the points (2,1.1)\begin{align*}(2, 1.1)\end{align*} and (2,0.9)\begin{align*}(2, 0.9)\end{align*}. The slopes of the lines represent the rates of A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*}. Using the Transformation Graphing App, students will only be able to graph one equation at a time, but they should see that once one equation has been graphed correctly, the other will be correct as well (no need to change the values of A\begin{align*}A\end{align*} or B\begin{align*}B\end{align*}).

To solve this system of equations, students are to distribute and then add the first equation to the second.

The solution is the train moves at 0.5 miles/min\begin{align*}0.5\ miles/min\end{align*} and Velma walks 0.05 miles/min\begin{align*}0.05\ miles/min\end{align*}.

Extension/Homework

1. DDSo 3r+600.5rr=(r+20)3=(r20) 3.5=3.5r70=130=260 km/h\begin{align*}D &= (r + 20) \cdot 3\\ D &= (r - 20) \ 3.5\\ \text{So} \ 3r + 60 & = 3.5r - 70\\ 0.5r & = 130\\ r &= 260\ km/h\end{align*}

So, the speed of the airplane in still air is 260 km/hr\begin{align*}260\ km/hr\end{align*}.

2. This problem can only be solved algebraically.

dslow30t+50tt=30tdfast=50t=160=2 hours\begin{align*}d_{slow} & = 30t \quad d_{fast} = 50t\\ 30t + 50t & = 160\\ t &= 2 \ hours\end{align*}

So, the two cars will be 160 miles\begin{align*}160\ miles\end{align*} apart in 2 hours\begin{align*}2\ hours\end{align*}.

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