Objective: Given the story and condition of the centipede and the bug, the student will be able to explain the relationship between distance, rate and time with 100% accuracy. The Bug & The Centipede

One sunny day, Senn T. Pede and Meladie Bug start out from the elm tree and move toward the rose bush which is 45 feet away. Senn crawls at 5 feet per minute and Ladie crawls at 3 feet per minute

Use the Diagram to show where each critter is exactly 3 minutes after they start their journey. Place the letter “S” at Senn’s location and the letter “L” at Ladie’s location S

Use the Diagram to show where each critter is exactly 3 minutes after they start their journey. Place the letter “S” at Senn’s location and the letter “L” at Ladie’s location LS 9

Complete the following table to show how far Senn and Ladie are from the elm tree for each time value Time, in minutes, since the friends leave the elm tree Senn’s distance, in feet, from the elm tree Ladie’s distance, in feet, from the elm tree

Complete the following table to show how far Senn and Ladie are from the elm tree for each time value Time, in minutes, since the friends leave the elm tree Senn’s distance, in feet, from the elm tree Ladie’s distance, in feet, from the elm tree

Describe a pattern you see in Senn’s row of the table. You may use words or mathematical symbols. Time, in minutes, since the friends leave the elm tree Senn’s distance, in feet, from the elm tree Ladie’s distance, in feet, from the elm tree …add five feet for every minute… …multiply five feet times every minute…

4. How far will Senn be from the elm tree after 7 minutes? …add five feet for every minute… = …or multiply five ft/min times 7 minutes… (5) (7) = 35 feet away from the elm tree

5. How long will it take Senn to crawl 40 feet? …if Senn can crawl five feet in one minute, then how many 5 foot increments will go into 40 feet? …or divide five ft/min into 40 feet… (40  5) = 8 minutes

6. Describe a pattern you see in Ladie’s row on the table (from problem #3). You may use words or mathematical symbols. …add three feet for every minute… …multiply three feet times every minute…

7. How far will Ladie be from the Rose Bush after 8 minutes? …add three feet for every minute… = …or multiply three ft/min times 8 minutes… (3) (8) = 24 feet away from the elm tree

8. How long will it take Ladie to crawl 36 feet? …if Lady can crawl three feet in one minute, then how many 3 foot increments will go into 36 feet? …or divide three ft/min into 36 feet… (36  3) = 12 minutes

9. After Senn reaches the rose bush, how long will he have to wait for Ladie to arrive? Explain how you determined your answer. We will need to find out how long it takes each critter, independently, to reach the rose bush… then find the difference in time. Senn’s time : 45 feet  5 feet/min = 45  5 = 9 min Ladie’s time : 45 feet  3 feet/min = 45  3 = 15 min 15 min – 9 min = 6 minutes.

On another morning, Senn and Ladie decide to head to the lake for a picnic. The lake is much further away than the rose bush. They will need to come up with a much better way of finding how long it will take them to reach their destination without having to work out the numbers on a lengthy table.

10. If Senn crawls for “m” minutes, and if “d” represents the distance, in feet, that he has traveled in those “m” minutes, write a rule that shows Senn the relationship between “m” and “d” for his pace.

Senn’s rate is 5 feet per minute = 5 m

10. If Senn crawls for “m” minutes, and if “d” represents the distance, in feet, that he has traveled in those “m” minutes, write a rule that shows Senn the relationship between “m” and “d” for his pace. Senn’s rate is 5 feet per minute d = 5 m

11. Use your rule from part (a) to determine how far Senn can crawl in one hour. Senn’s rule: d = 5 m How many minutes in one hour? d = (5) (60) Senn’s distance = 300 feet

12. How far does Senn Crawl between time m = 20 and m = 21 minutes. The time between minute 20 and minute 21 is? One minute!!! d = (5) (1) Senn’s distance = 5 feet in that moment

13. How far does Senn Crawl between time m = 25 and m = 26 minutes. The time between minute 25 and minute 26 is? One minute!!! d = (5) (1) Senn’s distance = 5 feet in that moment

14. If Ladie crawls for “m” minutes, and if “d” represents the distance, in feet, that he has traveled in those “m” minutes, write a rule that shows Ladie the relationship between “m” and “d” for her pace.

Ladie’s rate is 3 feet per minute = 3 m

14. If Ladie crawls for “m” minutes, and if “d” represents the distance, in feet, that he has traveled in those “m” minutes, write a rule that shows Ladie the relationship between “m” and “d” for her pace. Ladie’s rate is 3 feet per minute d = 3 m

15. Use your rule from part (a) to determine how far Ladie can crawl in one hour. Ladies’s rule: d = 3 m How many minutes in one hour? d = (3) (60) Ladie’s distance = 180 feet

16. How far does Lady Crawl between time m = 32 and m = 33 minutes. The time between minute 32 and minute 33 is? One minute!!! d = (3) (1) Ladie’s distance = 3 feet in that moment

17. How far does Ladie Crawl between time m = 32 and m = 34 minutes. The time between minute 32 and minute 34 is? Two minutes!!! d = (3) (2) Ladie’s distance = 6 feet in that moment

18. If the lake is 540 feet from the elm tree, how long will it take each critter to reach the lake? Show your work or explain how you got your result. Ladie’s rule: d = 3 m 540 = 3 m m = 540  3 m = 180 min Senn’s rule: d = 5 m 540 = 5 m m = 540  5 m = 108 min

Senn and Ladie have a turtle friend, Archimedes, who sometimes goes along on their adventures.

19. If Archimedes crawls for “m” minutes, and if “d” represents the distance, in feet, that he has traveled in those “m” minutes, the rule that shows the relationship between “m” and “d” for his pace is d = 7.5 (m). Use appropriate units to describe what the coefficient 7.5 tells you about Archimedes’ travels. Archi’s rate is 7.5 feet per minute d = 7.5 m

20. Explain what question the solution to the question 540 = 7.5 (m) would answer. Archi’s rule: d = 7.5 m 540 = 7.5 m m = 540  7.5 m = 72 min

21. What question is answered by the solution to the question d = (7.5)(540)? d = (7.5) (540) Is asking for the distance traveled if the critter has a speed of 7.5 feet per minute and travels for 540 minutes. d = 4050 feet

Distance in Feet Time In Minutes

Distance in Feet Time In Minutes

Distance in Feet Time In Minutes Distance in Feet Time In Minutes

Distance in Feet Time In Minutes

Distance in Feet Time In Minutes

Distance in Feet Time In Minutes

Compare the critters’ progress and answer the following: How can you tell which critter is fastest by the graph? Which critter is fastest? How long would it take each creature to reach a distance 15 feet away? How long would it take each creature to reach a distance 30 feet away? Approximately how long would it take each creature to reach a distance 18 feet away?