lesson 8 homework module 1 grade 5

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5th grade (Eureka Math/EngageNY)

Unit 1: module 1: place value and decimal fractions, unit 2: module 2: multi-digit whole number and decimal fraction operations, unit 3: module 3: addition and subtractions of fractions, unit 4: module 4: multiplication and division of fractions and decimal fractions, unit 5: module 5: addition and multiplication with volume and area, unit 6: module 6: problem solving with the coordinate plane.

Common Core Grade 5 Math (Worksheets, Homework, Lesson Plans)

Looking for video lessons that will help you in your Common Core Grade 5 Math classwork or homework? Looking for Common Core Math Worksheets and Lesson Plans that will help you prepare lessons for Grade 5 students?

The following lesson plans and worksheets are from the New York State Education Department Common Core-aligned educational resources. The Lesson Plans and Worksheets are divided into six modules.

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Grade 5 Homework, Lesson Plans And Worksheets

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Eureka Math Grade 8 Module 5 Lesson 1 Answer Key

Engage ny eureka math 8th grade module 5 lesson 1 answer key, eureka math grade 8 module 5 lesson 1 example answer key.

→ Let x represent the time it takes to travel y feet. \(\frac{256}{4}\) = \(\frac{y}{x}\) y = \(\frac{256}{4}\) x y = 64x

→ What are some of the predictions that this equation allows us to make? After one second, or when x = 1, the distance traveled is 64 feet. Accept any reasonable predictions that students make.

→ Use your equation to complete the table. → What is the average speed of the moving object from 0 to 3 seconds? The average speed is 64 feet per second. We know that the object has a constant rate of change; therefore, we expect the average speed to be the same over any time interval.

→ According to the data, how many feet did the stone drop in 3 seconds? The stone dropped 144 feet.

→ How can that be? It must be that our initial assumption of constant rate was incorrect. What predictions can we make now? After one second, x = 1; the stone dropped 16 feet, etc. → Let’s make a prediction based on a value of x that is not listed in the table. How far did the stone drop in the first 3.5 seconds? What have we done in the past to figure something like this out?

Eureka Math Grade 8 Module 5 Lesson 1 Exercise Answer Key

Exercise 2. Name a prediction that would require more information. Answer: Sample student response: We would need more information to predict the distance traveled after 3.75 seconds.

Exercise 3. What is the average speed of the object between 0 and 3 seconds? How does this compare to the average speed calculated over the same interval in Example 1? \(\text { Average Speed } = \frac{\text { distance traveled over a given time interval }}{\text { time interval }}\) Answer: The average speed is 48 feet per second: \(\frac{144}{3}\) = 48. This is different from the average speed calculated in Example 1. In Example 1, the average speed over an interval of 3 seconds was 64 feet per second.

Exercise 4. Take a closer look at the data for the falling stone by answering the questions below. a. How many feet did the stone drop between 0 and 1 second? Answer: The stone dropped 16 feet between 0 and 1 second.

b. How many feet did the stone drop between 1 and 2 seconds? Answer: The stone dropped 48 feet between 1 and 2 seconds.

c. How many feet did the stone drop between 2 and 3 seconds? Answer: The stone dropped 80 feet between 2 and 3 seconds.

d. How many feet did the stone drop between 3 and 4 seconds? Answer: The stone dropped 112 feet between 3 and 4 seconds.

e. Compare the distances the stone dropped from one time interval to the next. What do you notice? Answer: Over each interval, the difference in the distance was 32 feet. For example, 16+32 = 48, 48+32 = 80, and 80+32 = 112.

Exercise 5. What is the average speed of the stone in each interval 0.5 second? For example, the average speed over the interval from 3.5 seconds to 4 seconds is \(\frac{\text { distance traveled over a given time interval }}{\text { time interval }}\) = \(\frac{256-196}{4-3.5}\) = \(\frac{60}{0.5}\) = 120;120 feet per second Repeat this process for every half-second interval. Then, answer the question that follows. a. Interval between 0 and 0.5 second: Answer: \(\frac{4}{0.5}\) = 8;8 feet per second

b. Interval between 0.5 and 1 second: Answer: \(\frac{12}{0.5}\) = 24;24 feet per second

c. Interval between 1 and 1.5 seconds: Answer: \(\frac{20}{0.5}\) = 40;40 feet per second

d. Interval between 1.5 and 2 seconds: Answer: \(\frac{28}{0.5}\) = 56;56 feet per second

e. Interval between 2 and 2.5 seconds: Answer: \(\frac{36}{0.5}\) = 72;72 feet per second

f. Interval between 2.5 and 3 seconds: Answer: \(\frac{44}{0.5}\) = 88;88 feet per second

g. Interval between 3 and 3.5 seconds: Answer: \(\frac{52}{0.5}\) = 104;104 feet per second

h. Compare the average speed between each time interval. What do you notice? Answer: Over each interval, there is an increase in the average speed of 16 feet per second. For example, 8 + 16 = 24, 24 + 16 = 40, 40 + 16 = 56, and so on.

Eureka Math Grade 8 Module 5 Lesson 1 Problem Set Answer Key

b. Suppose that after 1 second, the ball is at its highest point of 22 feet (above point C) and has traveled a horizontal distance of 44 feet. What are the approximate coordinates of the ball at the following values of t: 0.25, 0.5, 0.75, 1, 1.25, 1.5, 1.75, and 2. Answer: Most answers will vary because students are approximating the coordinates. The coordinates that must be correct because enough information was provided are denoted by a *. At t = 0.25, the coordinates are approximately (11, 10). At t = 0.5, the coordinates are approximately (22, 18). At t = 0.75, the coordinates are approximately (33, 20). *At t = 1, the coordinates are approximately (44, 22). At t = 1.25, the coordinates are approximately (55, 19). At t = 1.5, the coordinates are approximately (66, 14). At t = 1.75, the coordinates are approximately (77, 8). *At t = 2, the coordinates are approximately (88, 0).

c. Use your answer from part (b) to write two predictions. Answer: Sample predictions: At a distance of 44 feet from where the ball was thrown, it is 22 feet in the air. At a distance of 66 feet from where the ball was thrown, it is 14 feet in the air.

d. What is happening to the ball when it has coordinates (88,0)? Answer: At point (88,0), the ball has traveled for 2 seconds and has hit the ground at a distance of 88 feet from where the ball began.

e. Why do you think the ball is at point (0, 6) when t = 0? In other words, why isn’t the height of the ball 0? Answer: The ball is thrown from point A to point B. The fact that the ball is at a height of 6 feet means that the person throwing it must have released the ball from a height of 6 feet.

f. Does the graph allow us to make predictions about the height of the ball at all points? Answer: While we cannot predict exactly, the graph allows us to make approximate predictions of the height for any value of horizontal distance we choose.

Eureka Math Grade 8 Module 5 Lesson 1 Exit Ticket Answer Key

b. What is the average speed of the ball over the first 0.25 seconds? What is the average speed of the ball over the next 0.25 seconds (from 0.25 to 0.5 seconds)? Answer: \(\frac{\text { distance traveled over a given time interval }}{\text { time interval }}\) = \(\frac{3-0}{0.25-0}\) = \(\frac{3}{0.25}\) = 12;12 feet per second \(\frac{\text { distance traveled over a given time interval }}{\text { time interval }}\) = \(\frac{4-3}{0.5-0.25}\) = \(\frac{1}{0.25}\) = 4;4 feet per second

c. Is the height of the ball changing at a constant rate? Answer: No, it is not. If the ball were traveling at a constant rate, the average speed would be the same over any time interval.

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