Math 8

Lesson 4.5

Introduction

Course Objectives

This lesson will address the following course outcomes:

· 20. Translate problems from a variety of contexts into mathematical representation and vice versa (linear, exponential, simple quadratics).

· 21. Describe the behavior of common types of functions using words, algebraic symbols, graphs, and tables. Include descriptions of the dependent and independent variables.

Specific Objectives

Students will understand that

· a variable is a symbol that is used in algebra to represent a quantity that can change.

· many variables can be present in a scenario or experiment, but some can be held fixed in order to analyze the effect that the change in one variable has on another.

· verbal, numerical, algebraic, and graphical representations provide different yet consistent information on quantifiable situations.

Students will be able to

· evaluate an expression.

· informally describe the change in one variable as another variable changes. verbally describe useful conclusions after exploration of a quantifiable situation.

· create a table of values from a given formula.

· create a graph using the table of values.

· compare the features of a quadratic graph and a linear graph.

You have now explored linear and exponential models in some depth. In this lesson, we will explore another type of model.

Intuition

Problem Situation: Calculating the Braking Distance of a Car

Experts agree that driving defensively saves lives. Knowing how far it takes your vehicle to come to a complete stop is one aspect of safe driving. For example, when you are going only 45 miles per hour (mph), you are traveling about 66 feet every second. This means that to be a safe driver, you need to drive “in front of you” (i.e., you need to know what is going on ahead of you so that you can react accordingly). In this lesson, you will learn more about what it takes to drive defensively by examining the braking distance of a vehicle. Braking distance is the distance a car travels in the time between when the brake is applied and when it comes to a full stop.  For this problem, we will be looking at minimum braking distance when the brakes are fully applied and the car is skidding.

#1 Points possible: 5. Total attempts: 5

What are some variables that might affect the braking distance of a car?

Come up with your own list, and after submitting it, compare it to ours.

#2 Points possible: 5. Total attempts: 5

For this lesson, you will focus on examining how speed affects braking distance.

Make a prediction on what will happen to braking distance if you were to double the speed of the car before it applies the brakes.

· The braking distance would be shorter

· The braking distance would be the same

· The braking distance would be longer, but less than twice as long

· The braking distance would be twice as long

· The braking distance would be more than twice as long

Formula

The formula for the braking distance of a car is   d=V22g(f+G)d=V22g(f+G)

V = initial velocity of the car (feet per second). That is, the velocity of the car when the brakes were applied, when t = 0.

d = braking distance (feet)

G = roadway grade (percent written in decimal form). Note: There are no units for this variable.

f = coefficient of friction between the tires and the roadway (0 < f < 1). Note: Good tires on good pavement provide a coefficient of friction of about 0.8 to 0.85.

g = acceleration due to gravity (32.2 ft/sec2).  This is a constant.

Since g is a constant, this formula has four variables. To understand the relationships between the variables, you will hold two of them fixed. That leaves you with two variables—one that will affect the other. Since you want to see how speed affects braking distance, you will hold the other two variables, f and G, fixed.

#3 Points possible: 5. Total attempts: 5

Let f = 0.8 and G = 0.05.  Write a simplified form of the formula for the braking distance d using these values for the two variables.  Your formula should involve the remaining variable V.

d =

Do not round the value in the denominator.  If you have trouble, you will get a hint after two attempts.

While the original equation involved four variables, this simplified formula just involves two, allowing us to compare how changing the initial velocity changes braking distance.

#4 Points possible: 16. Total attempts: 5

For each of the velocities in the table, given in miles per hour, first convert them to feet per second.  Then, use your simplified formula from above to determine the braking distance.  Give answers to two decimal places.

Velocity (miles/hr)	Velocity (ft/sec)	Braking Distance (ft)
10
20
40
80

#5 Points possible: 12. Total attempts: 5

Suppose the speed doubles from 10mph to 20mph.  Complete the two sentences below, giving answers to one decimal place.

The braking distance at 20mph is  ft longer than the braking distance at 10mph

The braking distance at 20mph is  times as long as the braking distance at 10mph

 

Suppose the speed doubles from 40mph to 80mph.  Complete the two sentences below, giving answers to one decimal place.

The braking distance at 80mph is  ft longer than the braking distance at 40mph

The braking distance at 80mph is  times as long as the braking distance at 40mph

 

#6 Points possible: 5. Total attempts: 5

Now we can return to a question we asked you to make a prediction about earlier.

Based on the pattern from your calculations, what will happen to braking distance if you were to double the speed of the car before it applies the brakes?

· The braking distance would be shorter

· The braking distance would be the same

· The braking distance would be twice as long

· The braking distance would be three times as long

· The braking distance would be four times as long

· The braking distance would be five times as long

#7 Points possible: 5. Total attempts: 5

Now try to extend that idea to answer to complete the sentence below.

If the speed were to triple, the braking distance would be  times as long

#8 Points possible: 5. Total attempts: 5

Plot the data from your table of values you calculated earlier, with velocity in miles per hour on the horizontal axis, and braking distance on the vertical axis.

 

Clear All Draw: 

#9 Points possible: 5. Total attempts: 5

What best describes the shape of the graph of the data?

· Linear – line shaped

· Curving upwards

#10 Points possible: 5. Total attempts: 5

Look at your simplified formula from earlier.  What family of equations does this formula belong to?

· Linear

· Exponential

· Quadratic

· None of these

HW 4.5

#1 Points possible: 5. Total attempts: 5

Which of the following was one of the main mathematical ideas of the lesson?

· Braking distance is affected by many factors.

· When using variables, it is only important to know what numbers to substitute in for them.

· When using variables, it is important to know what they represent and what units should be used with them.

· A subscript is a label on a variable.

#2 Points possible: 12. Total attempts: 5

In the lesson, you investigated the relationship between velocity and braking distance. You will now investigate the relationship between the coefficient of friction and the braking distance. Recall that the formula for the braking distance of a car is d=V22g(f+G)d=V22g(f+G)

a. Which of the variables in the formula represents a constant?

· f

· d

· V

· G

· g

b. To investigate the relationship between the coefficient of friction and the braking distance, you need to hold the other variables fixed. Let G = 0.02. Which of the following is a correct interpretation of the value G = 0.02?

· The grade of a road is 2%, which is a vertical increase of 2 feet for every 1 foot of horizontal increase.

· The grade of a road is 0.02%, which is a vertical increase of 0.02 feet for every 1 foot of horizontal increase.

· The grade of a road is 2%, which is a vertical increase of 2 feet for every 100 feet of horizontal increase.

· The grade of a road is 0.02%, which is a vertical increase of 0.02 feet for every 100 feet of horizontal increase.

c. Let V = 72 mph and use the value of G in Part (b). Which of the following expressions represents the simplified form of the formula using these values?

· d=11,151.3664.4f+1.288 ftd=11,151.3664.4f+1.288 ft

· d=11,151.3664.4f+8,657.89 ftd=11,151.3664.4f+8,657.89 ft

· d=11,151.3664.4f+0.02 ftd=11,151.3664.4f+0.02 ft

#3 Points possible: 18. Total attempts: 5

Use the formula you found in part c of the previous question.

a. Create a table of values for f and d (in feet). Use the values of f given in the table. Perform one of the calculations on paper showing the units. You may then use technology to complete the table.  Round answers to the nearest tenth.

f	d (feet)
0.30
0.50
0.70
0.90

b.

c. The four values of f correspond to the coefficient of friction for four road conditions:  an icy road, a very good road with great tires, an asphalt road with fair tires, and a wet road with fair tires.  Match the coefficients of friction to the appropriate conditions by looking at the braking distance required.

i. Icy road, f =

ii. Very good road with great tires, f =

iii. Asphalt road with fair tires, f =

iv. Wet road with fair tires, f =

d. In the table, the coefficient of friction, f, is increasing at a constant rate, since each value is 0.2 more than the previous value.  How is d changing as f increases at a constant rate?

. The stopping distance is decreasing

. The stopping distance is constant

. The stopping distance is increasing

#4 Points possible: 5. Total attempts: 5

In statistics, the formula E=1.96⋅√ˆp(1−ˆp)nE=1.96⋅p^(1-p^)n is used to calculate the margin of error E (at a 95% confidence level) for a survey of n people where ˆpp^  is the proportion of the people who responded a particular way.

In a particular survey of 1300 people, the proportion who favored stiffer penalties for drunk driving was 71%, so ˆp=0.71p^=0.71 .  Determine the margin of error, E.  Report the answer rounded to 3 decimal places.

Margin of error:

#5 Points possible: 20. Total attempts: 5

Newton’s law of gravitational force (measured in Newtons) between two objects r meters apart with masses m1 kilograms and m2 kilograms is given by the formula F=Gm1m2r2F=Gm1m2r2 , where G is a constant approximately equal to 6.674×10−116.674×10-11 .  The earth and the moon are about 384,000 kilometers apart.  The mass of the moon is about 73,480,000,000,000,000,000,000 kg, and the mass of the earth is about 5,972,200,000,000,000,000,000,000 kg.

a. The size of the numbers in this question make them hard to work with.  Rewrite them using scientific notation.  Don’t forget to check the units, and make any necessary conversions. The distance between the earth and the moon:       meters Mass of the moon:       kilograms Mass of the earth:       kilograms

b. Calculate the gravitational force.  Give your answer in scientific notation.   Note: Since the number 384,000 only has three significant digits (numbers before trailing zeros), it is appropriate to round your final answer so that it also has three significant digits (2 decimal places, in scientific notation).     Newtons

#6 Points possible: 20. Total attempts: 5

The tuition at a daycare center is based on family income.  A reduced tuition has a subsidy.  There are three levels of tuition:

· Full subsidy – the family does not pay any tuition

· Partial subsidy – the family pays part of the tuition

· No subsidy – the family pays the full tuition

The data for the daycare center, showing how many students there are for each age level and tuition level, is given below.  Answer the questions below. Round to the nearest whole percent.

		Tuition Level
		Full Subsidy	Partial Subsidy	No Subsidy	Total
Age  Level	3 year-olds	17	13	8
	4 year-olds	22	14	15
	5 year-olds	15	16	11
	Total

a. Complete the last column and last row.

b. What percentage of 3 year-olds received a full or partial subsidy?   %

c. What percentage of those who receive no subsidy are 5 years old?   %

d. What percentage of the students are 3 years old?   %

e. The daycare center’s funding for one term comes from federal funding for the subsidy and the tuition paid by families based on the formula below.  Find the funding for the center.

· Funding = 1,530F + 1,750P + 1,875N where

· F = number of children receiving a full subsidy

· P = number of children receiving a partial subsidy

· N = number of children receiving no subsidy Funding for the center = $

#7 Points possible: 10. Total attempts: 5

In Lesson 2.1, you used a formula that was written as steps in a form to calculate taxes for different people.  Formulas are often written in this way.  One example is the Expected Family Contribution (EFC) Formula, which is used to determine if a college student is eligible for financial aid.  The EFC has many different sections that each use different calculations.  One section of the 2010-11 form is shown below.

Student’s Contribution from Assets
Cash, savings, and checking
Net worth of investments	+
Net worth of business and/or investment farm	+
Net worth (sum of lines 45 through 47)
Assessment rate	×	0.20
Students Contribution from Assets	=

 

a. Calculate the Student’s Contribution from Assets given the following information, to the nearest dollar. Cash:  $500 Savings:  $1,240 Investments: $0 Business:  $0 Checking:  $732 Student’s Contribution from Assets:  $

b. Write a formula (equation) that summarizes the calculation in this form using the following variables: C = Cash including savings and checking Ni = Net worth of investment Nb = Net worth of business or farm S = Student’s contribution from assets