This activity is intended to supplement Algebra I, Chapter 7, Lesson 2.
Time Required: 15 minutes
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
Slope and graphically solving equation
Teacher Preparation and Notes
The student worksheet provides instructions and question to guide the inquiry and focus the observations.
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 ; Upstream rate
Substituting this into either equation gives a distance of . Alton and Barnhart are about apart along the Mississippi in the St. Louis area.
Problem 2 – Distance-Time Graph, Explore Slopes
Using for this situation gives the following equations:
where is the rate (speed) of the steam engine and is the rate (velocity) of Velma’s walking.
Students should set up lists and 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 and Velma’s walking so that the lines go through the points and . The slopes of the lines represent the rates of and . 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 or ).
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 and Velma walks .
So, the speed of the airplane in still air is .
2. This problem can only be solved algebraically.
So, the two cars will be apart in .