Algebra 1 Section 3 7 Uniform Motion Uniform

Algebra 1 Section 3. 7

Uniform Motion Uniform motion problems involve one or more objects traveling at constant speed. Formula: d = rt

d = rt rate × time = distance r t = d 50 50 3 t 45 5 = 150 = 50 t = 225 Be sure to use the right units!

Example 1 Both groups travel the same distance. The high school travels at 5 mi/hr for an “unknown” time [x]. The junior high travels at 3 mi/hr for one hour longer than the high school [x + 1].

Example 1 r junior high school 3 5 t = d t + 1 = 3(t + 1) t = 5 t Since the distances are equal: 3(t + 1) = 5 t

Example 1 3(t + 1) = 5 t 3 t + 3 = 5 t 3 = 2 t t = 1½ The high-school class needs 1½ hours to catch up with the junior-high class.

Solving Uniform Motion Problems 1. 2. 3. 4. Read the problem carefully. Plan a strategy. Solve an equation. Check.

Example 2 Northbound train Total distance: 333 mi. rate = r + 5 Mattoon Southbound train rate = r For both trains, t = 3 hours

Example 2 r northbound r + 5 southbound r t = d 3 3 = 3(r + 5) 3 r = Since total distance is 333 mi: 3(r + 5) + 3 r = 333

Example 2 3(r + 5) + 3 r = 333 3 r + 15 + 3 r = 333 6 r + 15 = 333 6 r = 318 r = 53

Example 2 r = 53 The southbound train travels at a rate [r ] of 53 mi/hr. The northbound train travels at a rate [r + 5] of 58 mi/hr.

Example 3 Both runners run for the same amount of time [t ]. Jon travels 400 m more than Paul. Jon’s distance Paul’s distance 400

Example 3 Jon Paul r t = d 375 t t = = 375 t 325 t Since Jon’s distance is 400 m greater than Paul’s distance: 375 t = 325 t + 400

Example 3 375 t = 325 t + 400 50 t = 400 t = 8 minutes

Example 3 t = 8 minutes Both runners ran for 8 minutes. Jon ran 375(8) = 3000 m, which equals 7. 5 laps. Paul ran 325(8) = 2600 m, which equals 6. 5 laps.

Homework: pp. 132 -134