Algebra 1 Section 13.7

Motion Problems The basic equation used here is r • t = d.

Example 1 r • t = d Upstream Downstream d d The variable here is d, the distance traveled up and down the river.

Example 1 r • t = d Upstream Downstream 2 d 8 d The upstream rate is 5 – 3 = 2; the downstream rate is 5 + 3 = 8 (mi/hr).

Example 1 r • t = d Upstream Downstream 2 d/2 d 8 d/8 d Time is calculated by dividing the distance by the rate.

Example 1 r • t = d 2 d/2 d 8 d/8 d d 2 8 + = 5 Upstream Downstream 2 d/2 d 8 d/8 d d 2 8 + = 5 Using the times given in the problem, we observe they spent 5 hours paddling.

Example 1 d 2 8 + = 5 8 ( )8 4d + d = 40 5d = 40 d = 8

Example 1 d = 8 The distance up the river is 8 miles. Upstream time is d/2 = 4 hr. They stopped for lunch at 1:00 PM. Downstream time is d/8 = 1 hr.

Example 2 r • t = d r 6r Let r = the speed of the car By car By plane r 6r Let r = the speed of the car 6r = the speed of the plane

Example 2 r • t = d By car By plane r 780 6r 780 In both cases, the distance is 780 miles.

Example 2 r • t = d By car By plane r 780/r 780 6r 780/6r 780 Time is calculated by dividing the distance by the rate.

Example 2 r • t = d By car By plane r 780/r 780 6r 780/6r 780 It is helpful to think of 12½ as 25/2 when writing the equation. 780 r 6r = + 25 2

Example 2 780 r 6r = + 25 2 The LCD is 6r. 780 r 6r = + 25 2 The LCD is 6r. 780 r 6r 6r ( ) = 6r ( ) + 6r ( ) 25 2 4680 = 780 + 75r

Example 2 4680 = 780 + 75r 3900 = 75r r = 52 mi/hr 6r = 312 mi/hr The time for the car is 15 hr. The time for the plane is 2½ hr.

Work Problems The amount of work done in one hour is the reciprocal of the number of hours worked. If a job takes 4 hours, then ¼ can be completed in one hour.

Example 3 Let m = the number of minutes it should take them to do the job together. Allison can do 1 chapter per 60 minutes. Sean can do 1 chapter per 75 minutes.

Example 3 1 60 75 + = 1 m The LCD is 300m. 5m + 4m = 300 9m = 300 + = 1 m The LCD is 300m. 5m + 4m = 300 9m = 300 100 3 m = = 33⅓ min

Example 3 Always check to see that your solution is reasonable. Notice that the answer is not the average of half of each person’s times.

Example 4 Let h = the number of hours it should take for Ned to do the job. Mr. Mathews can do 1 job per 10 hours. Together, they can do 1 job per 4 hours.

Example 4 1 10 h + = 1 4 The LCD is 20h. 2h + 20 = 5h 20 = 3h + = 1 4 The LCD is 20h. 2h + 20 = 5h 20 = 3h 20 3 h = = 6⅔ hr

Homework: pp. 563-566