Lesson 29 Uniform Motion Problems DDk 1 2
Lesson 29: Uniform Motion Problems: D+D=k 1 2
We remember that the key equation in a uniform motion problem is the equation that describes the distances that have been traveled. In all the problems thus far, the persons objects have traveled equal distances, so the distance diagrams for these problems have been equal.
In some problems the sum of one distance and another distance equals a certain number.
Example: Napoleon walked part of the 60 miles to the site of the battle and rode the rest of the way on a caisson. He walked at 3 mph and rode at 9 mph. if the total time of the trip was 8 hours, for how long did he walk?
Answer: DW+ DR= 60 RWTW + RR T R = 60 R=3 W RR = 9 TW + T =8 R TW = 2 hours
Example: Edward Longshanks and Queen Eleanor were 54 miles apart at dawn. Edward began the journey to the meeting place at 8 am at 3 mph; 2 hours later the queen set out to meet him. If they met at 4 pm, how fast did the queen travel?
Answer: DL + D Q= 54 RL TL + RQ T Q = 54 TL = 8 TQ = 6 RL = 3 RQ = 5 mph
Example: At noon Rocketman whizzed off toward Rocketland; 1 hour later Moonfa whizzed off in the opposite direction at a speed 200 kph less that of Rocketman. If they were 11, 800 kilometers apart at 5 pm, how fast did each travel?
Answer: DR+ DM= 11, 800 RRTR + RM MT = 11, 800 TR = 5 TM = 4 RM = R R – 200 RM = 1200 kph RR = 1400 kph
HW: Lesson 29 #1 -30
- Slides: 10