CHAPTER 7 Systems of Equations Slide 2Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 7.1Systems of Equations in Two Variables 7.2The Substitution Method 7.3The Elimination Method 7.4Applications and Problem Solving 7.5Applications with Motion

OBJECTIVES 7.5 Applications with Motion Slide 3Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. aSolve motion problems using the formula d = rt.

7.5 Applications with Motion The Motion Formula Slide 4Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. Distance = Rate(or speed)  Time d = rt

EXAMPLE 7.5 Applications with Motion a Solve motion problems using the formula d = rt. AApplications with Motion (continued) Slide 5Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. Two cars leave Toledo at the same time traveling in opposite directions. One car travels at 60 mph and the other at 45 mph. In how many hours will they be 225 miles apart?

EXAMPLE 7.5 Applications with Motion a Solve motion problems using the formula d = rt. AApplications with Motion (continued) Slide 6Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 45 mph 60 mph 225 miles 1. Familiarize. Make a drawing.

EXAMPLE 7.5 Applications with Motion a Solve motion problems using the formula d = rt. AApplications with Motion (continued) Slide 7Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. Distance (in miles) Speed (in miles per hour) Time (in hours) Fast Cardistance of fast car 60t Slow Cardistance of slow car 45t Use a table.

EXAMPLE 7.5 Applications with Motion a Solve motion problems using the formula d = rt. AApplications with Motion (continued) Slide 8Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 2. Translate. Distance of fast car + Distance of slow car = 225 d = rt from the table we get: 60t + 45t = 225

EXAMPLE 3. Solve. 7.5 Applications with Motion a Solve motion problems using the formula d = rt. AApplications with Motion (continued) Slide 9Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE 4. Check. Distance of fast car + Distance of slow car = State. In 1 7/8 of an hour the cars will be 225 miles apart. 7.5 Applications with Motion a Solve motion problems using the formula d = rt. AApplications with Motion Slide 10Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE Because of a tail wind, a jet is able to fly 20 mph faster than another jet that is flying into the wind. In the same time that it takes the first jet to travel 90 miles the second jet travels 80 miles. How fast is each jet traveling? 7.5 Applications with Motion a Solve motion problems using the formula d = rt. BApplications with Motion (continued) Slide 11Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. r r  20

EXAMPLE 1. Familiarize. We try a guess. If the first jet is traveling 300 mph because of a tail wind the second jet would be traveling 300  20 or 280 mph. At 300 mph the first jet would travel 90/300, or 3/10 hr. At 280 mph, the other jet would travel 80/280 = 2/7 hr. Since both planes spend the same amount of time traveling, we see that our guess is incorrect. Let’s set up a table. 7.5 Applications with Motion a Solve motion problems using the formula d = rt. BApplications with Motion (continued) Slide 12Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE 7.5 Applications with Motion a Solve motion problems using the formula d = rt. BApplications with Motion (continued) Slide 13Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. Distance (in miles) Speed (in miles per hour) Time (in hours) Jet 180r Jet 290r + 20 Distance = Rate  Time

EXAMPLE 7.5 Applications with Motion a Solve motion problems using the formula d = rt. BApplications with Motion (continued) Slide 14Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. Distance (in miles) Speed (in miles per hour) Time (in hours) Jet 180r80/r Jet 290r /(r + 20) The times must be the same 2. Translate. Fill in the blank column in the table. time = distance/rate.

EXAMPLE Since the times must be the same for both planes, we have the equation 7.5 Applications with Motion a Solve motion problems using the formula d = rt. BApplications with Motion (continued) Slide 15Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE 3. Carry out. To solve the equation, we first multiply both sides by the LCM of the denominators r(r + 20). Simplifying Using the distributive law Subtracting 80r from both sides Dividing both sides by Applications with Motion a Solve motion problems using the formula d = rt. BApplications with Motion (continued) Slide 16Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE Now we have a possible solution. The speed of one jet is 160 mph and the speed of the other jet is 180 mph. 4. Check. Reread the problem to confirm that we were to find the speeds. At 160 mph the jet would cover 80 miles in ½ hour and at 180 mph the other jet would cover 90 miles in ½ hour. Since the times are the same, the speeds check. 5. State. One jet is traveling at 160 mph and the second jet is traveling at 180 mph. 7.5 Applications with Motion a Solve motion problems using the formula d = rt. BApplications with Motion Slide 17Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

7.5 Applications with Motion More Tips for Solving Motion Problems Slide 18Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 1. Translating to a system of equations eases the solution of many motion problems. 2. At the end of the problem, always ask yourself, “Have I found what the problem asked for?” You might have solved for a certain variable but still not have answered the question of the original problem.