Presentation on theme: "Math 8H Problem Solving Day 2 Rate Time = Distance Algebra 1 Glencoe McGraw-Hill JoAnn Evans."— Presentation transcript:
Math 8H Problem Solving Day 2 Rate Time = Distance Algebra 1 Glencoe McGraw-Hill JoAnn Evans
Two Types of rate time = distance Problems: d + d = d Distance + distance = total vehicle 1 vehicle 2 distance The two vehicles will either be starting at opposite ends and traveling toward each other -or- they will start at the same place and travel in opposite directions until they reach a given distance apart. d = d distance of = distance of vehicle 1 vehicle 2 This is the catch up formula. The first vehicle will start, then the second vehicle will begin at a later time, traveling at a faster speed, until it catches up with the first vehicle. -or- It is the round trip formula.
Example 1: d + d = total distance E trainA train 480 km apart Two trains left the same station at the same time and traveled in opposite directions. The E train averaged 130 km/h and the A trains speed was 110 km/h. In how many hours were they 480 km apart? start The problem gives the distance and the rates. You need to find the time. Let t = time
Verbal Sentence: E trains distance + A trains distance = total distance d + d = d Remember: d = rt rt + rt = d Equation: 130t + 110t = 480 Solution: The trains were 480 km apart in 2 hrs.
Example 2: d + d = total distance Faster CarSlower car 700 km apart Two cars traveled in opposite directions from the same starting point. The rate of one car was 20 km/h less than the rate of the other car. After 5 hours, the cars were 700 km apart. Find the rate of each car. start The problem gives the distance and the time. You need to find the rate. Let r = rate of faster car Let r – 20 = rate of slower car
Verbal Sentence: faster cars dist. + slower cars dist. = total distance d + d = d Remember: d = rt rt + rt = d Equation: r5 + (r-20)5 = 700 Solution: The faster car is 80 km/h and the slow is 60 km/h.
Example 3: d = d Sam started out in his car traveling at 60 mph. Two hours later, Jenny left from the same point. She traveled along the same road at 75 mph. After how many hours will she catch up with Sam? The problem gives the rate. You need to find the time. Let t = Jennys time Let t + 2 = Sams time (he started 2 hrs earlier) start Sams car Jennys car Point where Jenny caught up
Verbal Sentence: Jennys distance = Sams distance d = d Remember: d = rt rt = rt Equation: 75t = 60(t + 2) Solution: It took Jenny 8 hours to catch up with Sam.
Example 4: d = d The Wagner family drove to the beach at 75 km/h. They returned later in heavy traffic, slowed to 50 km/h. It took 1 hour longer to return home than it did to get to the beach. How long did it take to get home? The problem gives the rate. You need to find the time. Let t = time to return home Let t - 1 = time it took to get to beach to the beach return home
Verbal Sentence: distance home = distance to beach Remember: d = rt rt = rt Equation: 50t = 75(t – 1) Solution: It took 3 hours to return home. d = d