physic orientation

course phy 121

11:28 pm on august 26,2009

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

001. typewriter notation

Note that there are six questions in this exercise. Be sure to continue scrolling down until you get to the end of the exercise.

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Question: `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4). Then evaluate each expression for x = 2.

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Your solution:

In x-2/x+4 it wants you to divide 2/x and subtract it from x and add 4 the other problem it wants you to subtract 2 from x and add 4 to x then divide the answers from each

x=2 2-2/2+4= 2-1+4= 1+4=5

(2-2)/(2+4) = 0/6

confidence rating:

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Given Solution:

`aThe order of operations dictates that grouped expressions must be evaluated first, that exponentiation must be done before multiplication or division, which must be done before addition or subtraction.

It makes a big difference whether you subtract the 2 from the x or divide the -2 by 4 first. If there are no parentheses you have to divide before you subtract. Substituting 2 for x we get

2 - 2 / 2 + 4

= 2 - 1 + 4 (do multiplications and divisions before additions and subtractions)

= 5 (add and subtract in indicated order)

If there are parentheses you evaluate the grouped expressions first:

(x - 2) / (x + 4) = (2 - 2) / ( 2 + 4 ) = 0 / 6 = 0.

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `q002. Explain the difference between 2 ^ x + 4 and 2 ^ (x + 4). Then evaluate each expression for x = 2.

Note that a ^ b means to raise a to the b power. This process is called exponentiation, and the ^ symbol is used on most calculators, and in most computer algebra systems, to represent exponentiation.

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Your solution:

2^2+4= 4+4=8 2^(2+4)= 2^6= 64

confidence rating:

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Given Solution:

`a2 ^ x + 4 indicates that you are to raise 2 to the x power before adding the 4.

2 ^ (x + 4) indicates that you are to first evaluate x + 4, then raise 2 to this power.

If x = 2, then

2 ^ x + 4 = 2 ^ 2 + 4 = 2 * 2 + 4 = 4 + 4 = 8.

and

2 ^ (x + 4) = 2 ^ (2 + 4) = 2 ^ 6 = 2*2*2*2*2*2 = 64.

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Self-critique (if necessary):

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Question: `q003. What is the numerator of the fraction in the expression x - 3 / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x? What is the denominator? What do you get when you evaluate the expression for x = 2?

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Your solution:

the numerator is the top so 3 is the numerator the denominator is { (2x-5)^2*3x+1}

x=2 -3/[(2*2-5)^2*3(2)+1] -2+7(2)

2- 3/ (-1^2) *6+1] -2+14

2- 3/ 7-2+14 I think we need a fraction so 2 -2 +14 -3/7 = 14- 3/7 14= 98/7 98/7-3/7= 95/7

confidence rating:

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Given Solution:

`aThe numerator is 3. x isn't part of the fraction. / indicates division, which must always precede subtraction. Only the 3 is divided by [ (2x-5)^2 * 3x + 1 ] and only [ (2x-5)^2 * 3x + 1 ] divides 3.

If we mean (x - 3) / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x we have to write it that way.

The preceding comments show that the denominator is [ (2x-5)^2 * 3x + 1 ]

Evaluating the expression for x = 2:

- 3 / [ (2 * 2 - 5)^2 * 3(2) + 1 ] - 2 + 7*2 =

2 - 3 / [ (4 - 5)^2 * 6 + 1 ] - 2 + 14 = evaluate in parenthese; do multiplications outside parentheses

2 - 3 / [ (-1)^2 * 6 + 1 ] -2 + 14 = add inside parentheses

2 - 3 / [ 1 * 6 + 1 ] - 2 + 14 = exponentiate in bracketed term;

2 - 3 / 7 - 2 + 14 = evaluate in brackets

13 4/7 or 95/7 or about 13.57 add and subtract in order.

The details of the calculation 2 - 3 / 7 - 2 + 14:

Since multiplication precedes addition or subtraction the 3/7 must be done first, making 3/7 a fraction. Changing the order of the terms we have

2 - 2 + 14 - 3 / 7 = 14 - 3/7 = 98/7 - 3/7 = 95/7.

COMMON STUDENT QUESTION: ok, I dont understand why x isnt part of the fraction? And I dont understand why only the brackets are divided by 3..why not the rest of the equation?

INSTRUCTOR RESPONSE: Different situations give us different algebraic expressions; the situation dictates the form of the expression.

If the above expression was was written otherwise it would be a completely different expression and most likely give you a different result when you substitute.

If we intended the numerator to be x - 3 then the expression would be written (x - 3) / [(2x-5)^2 * 3x + 1 ] - 2 + 7x, with the x - 3 grouped.

If we intended the numerator to be the entire expression after the / the expression would be written x - 3 / [(2x-5)^2 * 3x + 1 - 2 + 7x ].

STUDENT COMMENT: I wasn't sure if the numerator would be 3 or -3. or is the subtraction sign just that a sign in this case?

INSTRUCTOR RESPONSE: In this case you would regard the - sign as an operation to be performed between the value of x and the value of the fraction, rather than as part of the numerator. That is, you would regard x - 3 / [ (2x-5)^2 * 3x + 1 ] as a subtraction of the fraction 3 / [ (2x-5)^2 * 3x + 1 ] from the term x.

STUDENT QUESTION: There was another question I had about this problem that wasn’t addressed. At the end when you changed the order of operation from

2 – 2 + 14 – 3/7 = 14 – 3/7

where did the 98/7 – 3/7 come into play before the end solution of 95/7? I must have forgotten how to do this part.

INSTRUCTOR RESPONSE: It's not clear how you can get 95/7 without this step.

To do the subtraction 14 - 3/7 both terms must be expressed in terms of a common denominator. The most convenient common denominator is 7.

So 14 must be expressed with denominator 7. This is accomplished by multiplying 14 by 7 / 7, obtaining 14 * 7 / 7 = 98 / 7. Since 7/7 = 1, we have just multiplied 14 by 1. We chose to use 7 / 7 in order to give us the desired denominator 7.

Thus our subtraction is

14 - 3/7 =

98/7 - 3/7 =

(98 - 3) / 7 =

95 /7.

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `q004. Explain, step by step, how you evaluate the expression (x - 5) ^ 2x-1 + 3 / x-2 for x = 4.

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Your solution:

plug in 4 for x = (4-5)^2*4-1 +3 /4-2

1^2=1 1*4-1+3/4-2 = 4-1+3/4-2= 4-1-2+3/4= 1 3/4

you subtract 4-5 and get -1 and raise -1 to the 2nd and get 1 multiply 1*4 and get 4 since 3/4 is a fraction you take 4-1-2 and get 1 and add 3/4 so it is 1 3/4

confidence rating:

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Given Solution:

`aWe get

(4-5)^2 * 4 - 1 + 3 / 4 - 2

= (-1)^2 * 4 - 1 + 3 / 4 - 2 evaluating the term in parentheses

= 1 * 4 - 1 + 3 / 4 - 2 exponentiating (2 is the exponent, which is applied to -1 rather than multiplying the 2 by 4

= 4 - 1 + 3/4 - 2 noting that 3/4 is a fraction and adding and subtracting in order we get

= 1 3/4 = 7 /4 (Note that we could group the expression as 4 - 1 - 2 + 3/4 = 1 + 3/4 = 1 3/4 = 7/4).

COMMON ERROR:

(4 - 5) ^ 2*4 - 1 + 3 / 4 - 2 =

-1 ^ 2*4 - 1 + 3 / 4-2 =

-1 ^ 8 -1 + 3 / 4 - 2.

INSTRUCTOR COMMENTS:

There are two errors here. In the second step you can't multiply 2 * 4 because you have (-1)^2, which must be done first. Exponentiation precedes multiplication.

Also it isn't quite correct to write -1^2*4 at the beginning of the second step. If you were supposed to multiply 2 * 4 the expression would be (-1)^(2 * 4).

Note also that the -1 needs to be grouped because the entire expression (-1) is taken to the power. -1^8 would be -1 because you would raise 1 to the power 8 before applying the - sign, which is effectively a multiplication by -1.

STUDENT QUESTION: if it's read (-1)^8 it would be 1 or would you apply the sign afterward even if it is grouped and it be a -1?

INSTRUCTOR RESPONSE: The 8th power won't occur in this problem, of course, but you ask a good question.

-1^8 would require raising 1 to the 8th power, then applying the negative sign, and the result would be -1.

(-1)^8 would be the 8th power of -1, which as you see would be 1.

STUDENT COMMENT: I think it would be easier to visualize what your trying to raise to the exponent if you actually put parenthesis around the 2, that part seems to get tricky on the computer.

INSTRUCTOR RESPONSE: The expression was intentionally written to be misleading and make the point that, to avoid ambiguity, order of operations apply strictly, no matter what the expression looks like.

Normally, for clarity, the parentheses would be included. They aren't necessary, but when helpful it's a good idea to include them. You can, of course, have too many parentheses in an expression, making it harder than necessary to sort out. In practice we try to strike a balance.

The original expression was

(x - 5) ^ 2x-1 + 3 / x-2

White spaces make no difference in how an expression is evaluated, but they can help show the structure; e.g.,

(x - 5)^2 * x - 1 + 3 / x -2

is a visual improvement over the original. The * between the 2 and the x is not strictly necessary, but is also helpful.

((((x - 5) ^ 2)) * x) - 1 + (3 / x) - 2

verges on having too many parentheses at the beginning; it does help clarify the 3 / x.

STUDENT COMMENT

Although I read through your explanation and do see the point you are making, that 2x is actually 2 * x, I still think that

(-1) should be raised to 2x rather than 2. Kaking the answer -11/4, not 7/4.

INSTRUCTOR RESPONSE

When the expression (x - 5) ^ 2x-1 + 3 / x-2 is copied and pasted into a computer algebra system it is translated as

This notation is universal and unambiguous. Any deviation from strict interpretation (which does occur among some authors and among manufacturers of some calculators) tends to result in ambiguity and confusion.

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `q005. At the link

http://www.vhcc.edu/dsmith/genInfo/introductory problems/typewriter_notation_examples_with_links.htm

(copy this path into the Address box of your Internet browser; alternatively use the path

http://vhmthphy.vhcc.edu/ > General Information > Startup and Orientation (either scroll to bottom of page or click on Links to Supplemental Sites) > typewriter notation examples

and you will find a page containing a number of additional exercises and/or examples of typewriter notation.Locate this site, click on a few of the links, and describe what you see there.

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Your solution:

I saw examples of notation and problems that show how to do the problem is order. It tells about parenthesis and exponents and how you need to do these in a certain order. Physic students need to know how to do these problems but some math classes only need to know the basics.

confidence rating:

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Given Solution:

`aYou should see a brief set of instructions and over 30 numbered examples. If you click on the word Picture you will see the standard-notation format of the expression. The link entitled Examples and Pictures, located in the initial instructions, shows all the examples and pictures without requiring you to click on the links. There is also a file which includes explanations.

The instructions include a note indicating that Liberal Arts Mathematics students don't need a deep understanding of the notation, Mth 173-4 and University Physics students need a very good understanding,

while students in other courses should understand the notation and should understand the more basic simplifications.

There is also a link to a page with pictures only, to provide the opportunity to translated standard notation into typewriter notation.

end program

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Self-critique (if necessary):

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Question: `q006 Standard mathematics notation is easier to look at; it's easier to see the meaning of the expressions.

However it's very important to understand order of operations, and students do get used to this way of doing it.

You should of course write everything out in standard notation when you work it on paper.

It is likely that you will at some point use a computer algebra system, and when you do you will probably have to enter expressions using a keyboard, so it is well worth the trouble to get used to this notation.

Indicate your understanding of why it is important to understand this notation.

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Your solution:

Standard notation is easier to understand when you write it out but you need to understand the order of operations. When you write it out you can use standard notation but you need to get used to the computer algebra system. When you use a keyboard it will make it easier if you understand this notation.

confidence rating:

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If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

002. Describing Graphs

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Question: `q001. You will frequently need to describe the graphs you have constructed in this course. This exercise is designed to get you used to some of the terminology we use to describe graphs. Please complete this exercise and email your work to the instructor. Note that you should do these graphs on paper without using a calculator. None of the arithmetic involved here should require a calculator, and you should not require the graphing capabilities of your calculator to answer these questions.

Problem 1. We make a table for y = 2x + 7 as follows: We construct two columns, and label the first column 'x' and the second 'y'. Put the numbers -3, -2, -1, -, 1, 2, 3 in the 'x' column. We substitute -3 into the expression and get y = 2(-3) + 7 = 1. We substitute -2 and get y = 2(-2) + 7 = 3. Substituting the remaining numbers we get y values 5, 7, 9, 11 and 13. These numbers go into the second column, each next to the x value from which it was obtained. We then graph these points on a set of x-y coordinate axes. Noting that these points lie on a straight line, we then construct the line through the points.

Now make a table for and graph the function y = 3x - 4.

Identify the intercepts of the graph, i.e., the points where the graph goes through the x and the y axes.

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Your solution:

it has to go through the x and y axis at the same number so y= 0 and x= 0

then y=0 0=3x-4 x=4/3 the coordinates for x= ( 4/3,0)

y= 3*0-4 = -4 the coordinates for y=( 0,-4)

confidence rating:

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Given Solution:

`aThe graph goes through the x axis when y = 0 and through the y axis when x = 0.

The x-intercept is therefore when 0 = 3x - 4, so 4 = 3x and x = 4/3.

The y-intercept is when y = 3 * 0 - 4 = -4. Thus the x intercept is at (4/3, 0) and the y intercept is at (0, -4).

Your graph should confirm this.

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Self-critique (if necessary):

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Question: `q002. Does the steepness of the graph in the preceding exercise (of the function y = 3x - 4) change? If so describe how it changes.

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Your solution:

It appears to be straight down

confidence rating:

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Given Solution:

`aThe graph forms a straight line with no change in steepness.

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Question: `q003. What is the slope of the graph of the preceding two exercises (the function ia y = 3x - 4;slope is rise / run between two points of the graph)?

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Your solution:

since it is 3x then it is 3

confidence rating:

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Given Solution:

`aBetween any two points of the graph rise / run = 3.

For example, when x = 2 we have y = 3 * 2 - 4 = 2 and when x = 8 we have y = 3 * 8 - 4 = 20. Between these points the rise is 20 - 2 = 18 and the run is 8 - 2 = 6 so the slope is rise / run = 18 / 6 = 3.

Note that 3 is the coefficient of x in y = 3x - 4.

Note the following for reference in subsequent problems: The graph of this function is a straight line. The graph increases as we move from left to right. We therefore say that the graph is increasing, and that it is increasing at constant rate because the steepness of a straight line doesn't change.

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Self-critique (if necessary):

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Question: `q004. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = 0 and x = 3.

Would you say that the graph is increasing or decreasing?

Does the steepness of the graph change and if so, how?

Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?

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Your solution:

if y=x^2 then x=0 so y=0^2= 0 and x=3 so y=3^2=9 the numbers are ( 0,0) (1,1) (2,4) (3,9)

x=1 y=1^2= y=1 x=2 y= 2^2=4

the graph is increasing and an increasing rate because the numbers are going up at an increasing rate not in a steady pattern

confidence rating:

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Given Solution:

`aGraph points include (0,0), (1,1), (2,4) and (3,9). The y values are 0, 1, 4 and 9, which increase as we move from left to right.

The increases between these points are 1, 3 and 5, so the graph not only increases, it increases at an increasing rate

STUDENT QUESTION: I understand increasing...im just not sure at what rate...how do you determine increasing at an increasing rate or a constant rate?

INSTRUCTOR RESPONSE: Does the y value increase by the same amount, by a greater amount or by a lesser amount every time x increases by 1?

In this case the increases get greater and greater. So the graph increases, and at an increasing rate. *&*&.

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Self-critique (if necessary):

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Question: `q005. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = -3 and x = 0.

Would you say that the graph is increasing or decreasing?

Does the steepness of the graph change and if so, how?

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Your solution:

y=x^2 x= -3,-2,-1,0 so y=9,4,1,0 the sets are (-3,9) (-2,4) (-1,1) (0,0)

it is decreasing

it is decreasing in steepness

decreasing at a decreasing rate

confidence rating:

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Given Solution:

`aFrom left to right the graph is decreasing (points (-3,9), (-2,4), (-1,1), (0,0) show y values 9, 4, 1, 0 as we move from left to right ). The magnitudes of the changes in x from 9 to 4 to 1 to 0 decrease, so the steepness is decreasing.

Thus the graph is decreasing, but more and more slowly. We therefore say that the graph is decreasing at a decreasing rate.

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Self-critique (if necessary):

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Question: `q006. Make a table of y vs. x for y = `sqrt(x). [note: `sqrt(x) means 'the square root of x']. Graph y = `sqrt(x) between x = 0 and x = 3.

Would you say that the graph is increasing or decreasing?

Does the steepness of the graph change and if so, how?

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Your solution:

y=sqrt(0)=0 y=sqrt(1)=1 y=sqrt(2)=1.41 y=sqrt(3)=1.73 so the stes are (0,0) (1,1) (2,1.41) (3,1.73)

increasing graph

steepness is decreasing

it is increasing at a decreasing rate

confidence rating:

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Given Solution:

`aIf you use x values 0, 1, 2, 3, 4 you will obtain graph points (0,0), (1,1), (2,1.414), (3. 1.732), (4,2). The y value changes by less and less for every succeeding x value. Thus the steepness of the graph is decreasing.

The graph would be increasing at a decreasing rate.

If the graph represents the profile of a hill, the hill starts out very steep but gets easier and easier to climb. You are still climbing but you go up by less with each step, so the rate of increase is decreasing.

If your graph doesn't look like this then you probably are not using a consistent scale for at least one of the axes. If your graph isn't as described take another look at your plot and make a note in your response indicating any difficulties.

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Question: `q007. Make a table of y vs. x for y = 5 * 2^(-x). Graph y = 5 * 2^(-x) between x = 0 and x = 3.

Would you say that the graph is increasing or decreasing?

Does the steepness of the graph change and if so, how?

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Your solution:

y=5*2^(-x) x=( 0) (1) (2) (3) y=5*2^(0)= 5*1=5 (0,5)

y=5*2^(-1)= 5*1/2= 5/2= 2.5 (1,2.5)

y=5*2^(-2)= 5*1/4=5/4=1.25 (2,1.25)

y=5*2^(-3)=5*1/8=5/8=.625 (3,.625)

graph is decreasing

steepness is decreasing and it is decreasing at a decreasing rate

confidence rating:

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Given Solution:

`a** From basic algebra recall that a^(-b) = 1 / (a^b).

So, for example:

2^-2 = 1 / (2^2) = 1/4, so 5 * 2^-2 = 5 * 1/4 = 5/4.

5* 2^-3 = 5 * (1 / 2^3) = 5 * 1/8 = 5/8. Etc.

The decimal equivalents of the values for x = 0 to x = 3 will be 5, 2.5, 1.25, .625. These values decrease, but by less and less each time.

The graph is therefore decreasing at a decreasing rate. **

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Question: `q008. Suppose you stand still in front of a driveway. A car starts out next to you and moves away from you, traveling faster and faster.

If y represents the distance from you to the car and t represents the time in seconds since the car started out, would a graph of y vs. t be increasing or decreasing?

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Your solution:

the graph would be increasing because you would start out at 0 since the car is next to you then 2 sec it would be like 10 feet away then 4 seconds it would be like 60 feet away or something like that so

it would increase at an increasing rate

confidence rating:

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Given Solution:

`a** The speed of the car increases so it goes further each second. On a graph of distance vs. clock time there would be a greater change in distance with each second, which would cause a greater slope with each subsequent second. The graph would therefore be increasing at an increasing rate. **

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Self-critique (if necessary):

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Self-critique Rating:

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

006. Physics

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Question: `q001. There are two parts to this problem. Reason them out using common sense.

If the speed of an automobile changes by 2 mph every second, then how long will it take the speedometer to move from the 20 mph mark to the 30 mph mark?

Given the same rate of change of speed, if the speedometer initially reads 10 mph, what will it read 7 seconds later?

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Your solution:

the difference between 20 and 30mph is 10 mph and it changes 2 mph every second then you need 5 changes so 5 seconds

If you start out at 10 and you look at it in 7 seconds then it should read 2*4=14 14+10= 24mph

confidence rating:

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Given Solution:

`aIt will take 5 seconds to complete the change. 30 mph - 20 mph = 10 mph change at 2 mph per second (i.e., 2 mph every second) implies 5 seconds to go from 20 mph to 30 mph

Change in speed is 2 mph/second * 7 seconds = 14 mph Add this to the initial 10 mph and the speedometer now reads 24 mph.

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Question: `q002. An automobile traveling down a hill passes a certain milepost traveling at a speed of 10 mph, and proceeds to coast to a certain lamppost further down the hill, with its speed increasing by 2 mph every second. The time required to reach the lamppost is 10 seconds. It then repeats the process, this time passing the milepost at a speed of 20 mph.

Will the vehicle require more or less than 10 seconds to reach the lamppost?

Since its initial speed was 10 mph greater than before, does it follow that its speed at the lamppost will be 10 mph greater than before?

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Your solution:

Less time because it is going faster.

No because it will not have the same amount of time to pick up speed

confidence rating:

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Given Solution:

`aIf it starts coasting down the same section of road at 20 mph, and if velocity changes by the same amount every second, the automobile should always be traveling faster than if it started at 10 mph, and would therefore take less than 10 seconds.

The conditions here specify equal distances, which implies less time on the second run. The key is that, as observed above, the automobile has less than 10 seconds to increase its speed. Since its speed is changing at the same rate as before and it has less time to change it will therefore change by less.

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Self-critique (if necessary):

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Question: `q003. The following example shows how we can measure the rate at which an automobile speeds up: If an automobile speeds up from 30 mph to 50 mph as the second hand of a watch moves from the 12-second position to the 16-second position, and its speed changes by 20 mph in 4 seconds. This gives us an average rate of velocity change equal to 20 mph / 4 seconds = 5 mph / second.

We wish to compare the rates at which two different automobiles increase their speed:

Which automobile speeds up at the greater rate, one which speeds up from 20 mph to 30 mph in five seconds or one which speeds up from 40 mph to 90 mph in 20 seconds?

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Your solution:

if you go from 20 to 30 in 5 seconds then you go 10mph in 5 seconds or 2 mph every second

if you go from 40 to 90 then you go 50mph in 20 seconds or 2.5mph every second

car b goes faster

confidence rating:

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Given Solution:

The first automobile's speed changes from 20 mph to 30mph, a 10 mph difference, which occurs in 5 seconds. So the rate of chage in 10 mph / (5 sec) = 2 mph / sec. = rate of change of 2 mph per second.

The second automobile's speed changes from 40 mph to 90 mph, a 50 mph difference in 20 seconds so the rate of change is 50 mph / (20 sec) = 2.5 mph per second.

Therefore, the second auto is increasing its velocity ar a rate which is .5 mph / second greater than that of the first.

Self-critique:

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Question: `q004. If an automobile of mass 1200 kg is pulled by a net force of 1800 Newtons, then the number of Newtons per kg is 1800 / 1200 = 1.5. The rate at which an automobile speeds up is determined by the net number of Newtons per kg. Two teams pulling on ropes are competing to see which can most quickly accelerate their initially stationary automobile to 5 mph. One team exerts a net force of 3000 Newtons on a 1500 kg automobile while another exerts a net force of 5000 Newtons on a 2000 kg automobile.

Which team will win and why?

If someone pulled with a force of 500 Newtons in the opposite direction on the automobile predicted to win, would the other team then win?

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Your solution:

3000/1500kg= 2 newtons per kg

5000/2000=2.5 newtons per kg

team b beause they have more newton per kg

if you take 5000-500 then team b has 4500/2000=2.25 so they still win

confidence rating:

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Given Solution:

`aThe first team's rate is 3000 Newtons divided by 1500 kg or 2 Newtons per kg, while the second team's rate is 5000 Newtons divided by 2000 kg or 2.5 Newtons per kg. The second team therefore increases velocity more quickly. Since both start at the same velocity, zero, the second team will immediately go ahead and will stay ahead.

The second team would still win even if the first team was hampered by the 500 Newton resistance, because 5000 Newtons - 500 Newtons = 4500 Newtons of force divided by 2000 kg of car gives 2.25 Newtons per kg, still more than the 2 Newtons / kg of the first team

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Self-critique (if necessary):

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Question: `q005. Both the mass and velocity of an object contribute to its effectiveness in a collision. If a 250-lb football player moving at 10 feet per second collides head-on with a 200-lb player moving at 20 feet per second in the opposite direction, which player do you precidt will be moving backward immediately after the collision, and why?

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Your solution:

mass * speed

250*10=2500 for the first 200*20=4000 for the second so the first will fall backwards

confidence rating:

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Given Solution:

`aGreater speed and greater mass both provide advantages. In this case the player with the greater mass has less speed, so we have to use some combination of speed and mass to arrive at a conclusion.

It turns out that if we multiply speed by mass we get the determining quantity, which is called momentum. 250 lb * 10 ft/sec = 2500 lb ft / sec and 200 lb * 20 ft/sec = 4000 lb ft / sec, so the second player will dominate the collision.

In this course we won't use pounds as units, and in a sense that will become apparent later on pounds aren't even valid units to use here. However that's a distinction we'll worry about when we come to it.

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Self-critique (if necessary):

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Question: `q006. Two climbers eat Cheerios for breakfast and then climb up a steep mountain as far as they can until they use up all their energy from the meal. All other things being equal, who should be able to climb further up the mountain, the 200-lb climber who has eaten 12 ounces of Cheerios or the 150-lb climber who has eaten 10 ounces of Cheerios?

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Your solution:

the 200 lb has 12 ounces so he has 12/200= .06

the 150lb has 10 ounces so he has 10/150 =.067

the second guy ate more per pound of body weight so he has more breakfast and more energy

confidence rating:

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Given Solution:

`aThe comparison we make here is the number of ounces of Cheerios per pound of body weight. We see that the first climber has 12 oz / (200 lb) = .06 oz / lb of weight, while the second has 10 0z / (150 lb) = .067 oz / lb. The second climber therefore has more energy per pound of body weight.

It's the ounces of Cheerios that supply energy to lift the pounds of climber. The climber with the fewer pounds to lift for each ounce of energy-producing Cheerios will climb further.

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Self-critique (if necessary):

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Question: `q007. Two automobiles are traveling up a long hill with an steepness that doesn't change until the top, which is very far away, is reached. One automobile is moving twice as fast as the other. At the instant the faster automobile overtakes the slower their drivers both take them out of gear and they coast until they stop.

Which automobile will take longer to come to a stop? Will that automobile require about twice as long to stop, more than twice as long or less than twice as long?

Which automobile will have the greater average coasting velocity? Will its average coasting velocity by twice as great as the other, more than twice as great or less than twice as great?

Will the distance traveled by the faster automobile be equal to that of the slower, twice that of the slower or more than twice that of the slower?

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Your solution:

If is going twice as fast then it should take twice as long to stop

the faster would have more velocity and it should be twice

it is going faster and with more velocity so the distance of the faster one should be more but I dont know how much more

confidence rating:

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Given Solution:

`aIt turns out that, neglecting air resistance, since the slope is the same for both, both automobiles will change velocity at the same rate. So in this case the second would require exactly twice as long.

If you include air resistance the faster car experiences more so it actually takes a bit less than twice as long as the slower.

For the same reasons as before, and because velocity would change at a constant rate (neglecting air resistance) it would be exactly twice as great if air resistance is neglected.

Interestingly if it takes twice as much time and the average velocity is twice as great the faster car travels four times as far.

If there is air resistance then it slows the faster car down more at the beginning than at the end and the average velocity will be a bit less than twice as great and the coasting distance less than four times as far.

STUDENT COMMENT: I do not understand why the car would go four times as far as the slower car.

INSTRUCTOR RESPONSE: The faster car takes twice as long to come to rest, and have twice the average velocity.

If the car traveled at the same average velocity for twice as long it would go twice as far.

If it traveled at twice the average velocity for the same length of time it would go twice as far.

However it travels at twice the average velocity for twice as long, so it goes four times as far.

STUDENT COMMENT:

it’s hard to know this stuff without having first discussed it in notes or read it in the book, or

have an equation handy. I guess this will all come with the class.

INSTRUCTOR RESPONSE

One purpose of this and similar exercises is to get students into the habit of thinking for themselves, as opposed to imitating what they see done in a textbook. You're doing some good thinking. When you get to the text and other materials, ideally you'll be better prepared to understand them as a result of this process.

This works better for some students than others, but it's beneficial to just about everyone.

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Self-critique (if necessary):

I didnt know it would be four times as fast so if it is traveling twice as fast and it takes twice as long to stop then 2*2 =4 so 4 times the distance?

Right. Very good.

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Question: `q008. When a 100 lb person hangs from a certain bungee cord, the cord stretches by 5 feet beyond its initial unstretched length. When a person weighing 150 lbs hangs from the same cord, the cord is stretched by 9 feet beyond its initial unstretched length. When a person weighing 200 lbs hangs from the same cord, the cord is stretched by 12 feet beyond its initial unstretched length.

Based on these figures, would you expect that a person of weight 125 lbs would stretch the cord more or less than 7 feet beyond its initial unstretched length?

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Your solution:

if 100= 5 feet 150= 9 ft 200=12 ft

difference of 4 ft then 3 feet but a 50 lb difference in weight

difference between 100 and 150 is 4 but the difference between 150 and 200 is 3ft

stretching 7 feet at 125 would be wrong it should be more than 7

confidence rating:

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Given Solution:

`aFrom 100 lbs to 150 lbs the stretch increased by 4 feet, from 150 lbs to 200 lbs the increase was only 3 feet. Thus it appears that at least in the 100 lb - 200 lb rands each additional pound results in less increase in length than the last and that there would be more increase between 100 lb and 125 lb than between 125 lb and 150 lb. This leads to the conclusion that the stretch for 125 lb would be more than halfway from 5 ft to 9 ft, or more than 7 ft.

A graph of stretch vs. weight would visually reveal the nature of the nonlinearity of this graph and would also show that the stretch at 125 lb must be more than 7 feet (the graph would be concave downward, or increasing at a decreasing rate, so the midway stretch would be higher than expected by a linear approximation).

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Question: `q009. When given a push of 10 pounds, with the push maintained through a distance of 4 feet, a certain ice skater can coast without further effort across level ice for a distance of 30 feet. When given a push of 20 pounds (double the previous push) through the same distance, the skater will be able to coast twice as far, a distance of 60 feet. When given a push of 10 pounds for a distance of 8 feet (twice the previous distance) the skater will again coast a distance of 60 feet.

The same skater is now accelerated by a sort of a slingshot consisting of a bungee-type cord slung between two posts in the ice. The cord, as one might expect, exerts greater and greater force as it is pulled back further and further. Assume that the force increases in direct proportion to pullback (ie.g., twice the pullback implies twice the force).

When the skater is pulled back 4 feet and released, she travels 20 feet. When she is pulled back 8 feet and released, will she be expected to travel twice as far, more than twice as far or less than twice as far as when she was pulled back 4 feet?

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Your solution:

If you have a bungee and it is pulled back twice as far and the power increases as you pull back that would double the distance and if you double the distaance you go pulling back that doubles the distance of the skater so you need to four times it. She will go four times as far

confidence rating:

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Given Solution:

`aThe distance through which the force acts will be twice as great, which alone would double the distance; because of the doubled pullback and the linear proportionality relationship for the force the average force is also twice as great, which alone would double the distance. So we have to double the doubling; she will go 4 times as far

STUDENT COMMENT: I do not understand the linear proportionality relationship for the force.

If the skater is pulled back an extra four feet, does that mean that the amount of pounds propelling her is also doubled?

INSTRUCTOR COMMENT: That is so. However the force propelling her isn't the only thing that influences how far she slides. The distance through which the force is applied is also a factor.

Doubling the force alone would double the sliding distance.

Doubling the distance through which the force is applied would double the sliding distane.

Doubling both the applied force and the distance through which it is applied quadruples the sliding distance.

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Self-critique (if necessary):

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Question: `q010. Two identical light bulbs are placed at the centers of large and identically frosted glass spheres, one of diameter 1 foot and the other of diameter 2 feet.

To a moth seeking light from half a mile away, unable to distinguish the difference in size between the spheres, will the larger sphere appear brighter, dimmer or of the same brightness as the first?

To a small moth walking on the surface of the spheres, able to detect from there only the light coming from 1 square inch of the sphere, will the second sphere appear to have the same brightness as the first, twice the brightness of the first, half the brightness of the first, more than twice the brightness of the first, or less than half the brightness of the first?

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Your solution:

If you are 1/2 mile away it would probably appear the same because the spheres would not have an impact on how you see the light

the light with 2 ft diameter will appear dimmer than the 1 ft diameter light because the bulb in the first light with 1 foot of diameter has less area to light and the bulb in the second has 2 ft diameter so it seem like it has twice the area to light and would therefor appear dimmer

confidence rating:

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Given Solution:

`aBoth bulbs send out the same energy per second. The surface of the second bulb will indeed be dimmer than the first, as we will see below. However the same total energy per second reaches the eye (identically frosted bulbs will dissipate the same percent of the bulb energy) and from a great distance you can't tell the difference in size, so both will appear the same. The second sphere, while not as bright at its surface because it has proportionally more area, does have the extra area, and that exactly compensates for the difference in brightness. Specifically the brightness at the surface will be 1/4 as great (twice the radius implies 4 times the area which results in 1/4 the illumination at the surface) but there will be 4 times the surface area.

Just as a 2' x 2' square has four times the area of a 1' x 1' square, a sphere with twice the diameter will have four times the surface area and will appear 1 / 4 as bright at its surface. Putting it another way, the second sphere distributes the intensity over four times the area, so the light on 1 square inch has only 1 / 4 the illumination.

STUDENT COMMENT: I understand the first part of the problem about the distances. But the second part really confuses me. Looking straight down from the top of the spheres, the bulb is the same intensity and the frosted glass is exactly the same, so why would it seem dimmer? I would think that if a person was standing in front of the spheres, that person would be able to tell a difference, but not extremely close.

INSTRUCTOR RESPONSE: Imagine a light bulb inside a frosted glass lamp of typical size. Imagine it outside on a dark night. If you put your eye next to the glass, the light will be bright. Not as bright as if you put your eye right next to the bulb, but certainly bright. The power of the bulb is spread out over the lamp, but the lamp doesn't have that large an area so you detect quite a bit of light.

If you put the same bulb inside a stadium with a frosted glass dome over it, and put your eye next to the glass on a dark night, with just the bulb lit, you won't detect much illumination. The power of the bulb is distributed over a much greater area than that of the lamp, and you detect much less light.

STUDENT COMMENT:

I also didn’t get the second part of the question. I still don’t really see where the ¼ comes from.

INSTRUCTOR RESPONSE:

First you should address the explanation given in the problem:

'Just as a 2' x 2' square has four times the area of a 1' x 1' square, a sphere with twice the diameter will have four times the surface area and will appear 1 / 4 as bright at its surface. Putting it another way, the second sphere distributes the intensity over four times the area, so the light on 1 square inch has only 1 / 4 the illumination. '

Do you understand this explanation?

If not, what do you understand about it and what don't you understand?

This simple image of a 2x2 square being covered by four 1x1 squares is the most basic reason the larger sphere has four time the area of the smaller.

There is, however, an alternative explanation in terms of formulas:

The surface area of a sphere is 4 pi r^2.

If r is doubled, r^2 increases by factor 2^2 = 4.

So a sphere with double the radius has four time the area.

If the same quantity is spread out over the larger sphere, it will be 1/4 as dense on the surface.

STUDENT COMMENT:

I also have no clue why the extra area doesn’t take away some brightness.

INSTRUCTOR RESPONSE:

All the light produced by the bulb is passing through either of the spheres. From a distance you see all the light, whichever sphere you're looking at; you see just as much light when looking at one as when looking at the other.

From a distance you can't tell whether you're looking at the sphere with larger area but less intensity at its surface, or the sphere with lesser area and greater intensity at its surface.

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Self-critique (if necessary):

Im not sure if I understand correctly what I am doing. Did my explanation make any sense? After reading yours Im not sure if it does.

Your answer made perfect sense. It didn't include all the detail of the given solution but everything you said was right, and on the right track. Much better than the answers I usually get to this question.

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Question: `q011. The water in a small container is frozen in a freezer until its temperature reaches -20 Celsius. The container is then placed in a microwave oven, which proceeds to deliver energy at a constant rate of 600 Joules per second. After 10 seconds the ice is still solid and its temperature is -1 Celsius. After another 10 seconds a little bit of the cube is melted and the temperature is 0 Celsius. After another minute most of the ice is melted but there is still a good bit of ice left, and the ice and water combination is still at 0 Celsius. After another minute all the ice is melted and the temperature of the water has risen to 40 degrees Celsius.

Place the following in order, from the one requiring the least energy to the one requiring the most:

Increasing the temperature of the ice by 20 degrees to reach its melting point.

Melting the ice at its melting point.

Increasing the temperature of the water by 20 degrees after all the ice melted.

At what temperature does it appear ice melts, and what is the evidence for your conclusion?

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Your solution:

least energy is increasing water by 20 degrees after all ice is melted

then increasing the temp of the ice by 20 degrees

Melting ice at melting point is the step that requires the most energy

Ice melts at 0 degrees C

confidence rating:

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Given Solution:

`aSince the temperature is the same when a little of the ice is melted as when most of it is melted, melting takes place at this temperature, which is 0 Celsius.

The time required to melt the ice is greater than any of the other times so melting at 0 C takes the most energy. Since we don't know how much ice remains unmelted before the final minute, it is impossible to distinguish between the other two quantities, but it turns out that it takes less energy to increase the temperature of ice than of liquid water.

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Question: `q012. Suppose you are in the center of a long, narrow swimming pool (e.g., a lap pool). Two friends with kickboards are using them to push waves in your direction. Their pushes are synchronized, and the crests of the waves are six feet apart as they travel toward you, with a 'valley' between each pair of crests. Since your friends are at equal distances from you the crests from both directions always reach you at the same instant, so every time the crests reach you the waves combine to create a larger crest. Similarly when the valleys meet you experience a larger valley, and as a result you bob up and down further than you would if just one person was pushing waves at you.

Now if you move a bit closer to one end of the pool the peak from that end will reach you a bit earlier, and the peak from the other end will reach you a little later. So the peaks won't quite be reaching you simultaneously, nor will the valleys, and you won't bob up and down as much. If you move far enough, in fact, the peak from one end will reach you at the same time as the valley from the other end and the peak will 'fll in' the valley, with the result that you won't bob up and down very much.

If the peaks of the approaching waves are each 6 inches high, how far would you expect to bob up and down when you are at the center point?

How far would you have to move toward one end or the other in order for peaks to meet valleys, placing you in relatively calm water?

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Your solution:

2 friends and the waves are 6inches so you will bob 12 inches or 1 foot

You are 6 feet from the friends and if you moved 1 foot then you would be 5 ft from 1 and 7 from the other if you go 2 feet then it woud be 4 and 8 Im not sure what to do to solve this one

confidence rating:

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Given Solution:

`aIf the two 6-inch peaks meet and reinforce one another completely, the height of the 'combined' peak will be 6 in + 6 in = 12 in.

If for example you move 3 ft closer to one end you move 3 ft further from the other and peaks, which are 6 ft apart, will still be meeting peaks. [ Think of it this way: If you move 3 ft closer to one end you move 3 ft further from the other. This shifts your relative position to the two waves by 6 feet (3 feet closer to the one you're moving toward, 3 feet further from the other). So if you were meeting peaks at the original position, someone at your new position would at the same time be meeting valleys, with two peaks closing in from opposite directions. A short time later the two peaks would meet at that point. ]

However if you move 1.5 ft the net 'shift' will be 3 ft and peaks will be meeting valleys so you will be in the calmest water.

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Self-critique (if necessary):

So I need to find half of the distance and then take half again half of 6 is 3 and half of 3 is 1.5?

That's what it comes down to.

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Self-critique Rating:

f your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

001. Rates

Note that there are 10 questions in this assignment. The questions are of increasing difficulty--the first questions are fairly easy but later questions are very tricky. The main purposes of these exercises are to refine your thinking about rates, and to see how you process challenging information. Most students in most courses would not be expected to answer all these questions correctly; all that's required is that you do your best and follows the recommended procedures for answering and self-critiquing your work.

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Question: If you make $50 in 5 hr, then at what rate are you earning money?

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Your solution:

50/5= $10 a hour

confidence rating:

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Given Solution:

The rate at which you are earning money is the number of dollars per hour you are earning. You are earning money at the rate of 50 dollars / (5 hours) = 10 dollars / hour. It is very likely that you immediately came up with the $10 / hour because almosteveryone is familiar with the concept of the pay rate, the number of dollars per hour. Note carefully that the pay rate is found by dividing the quantity earned by the time required to earn it. Time rates in general are found by dividing an accumulated quantity by the time required to accumulate it.

You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.

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Question: `q003.If you make $60,000 per year then how much do you make per month?

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Your solution:

60000/12 months= $5000 a month

confidence rating:

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Given Solution:

Most people will very quickly see that we need to divide $60,000 by 12 months, giving us 60,000 dollars / (12 months) = 5000 dollars / month. Note that again we have found a time rate, dividing the accumulated quantity by the time required to accumulate it.

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Question: `q004. Suppose that the $60,000 is made in a year by a small business. Would be more appropriate to say that the business makes $5000 per month, or that the business makes an average of $5000 per month?

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Your solution:

It would be better to say an average of $5000 because it states that it was 60000 a year and some months may be more or less but on average it is $5000

confidence rating:

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Given Solution:

Small businesses do not usually make the same amount of money every month. The amount made depends on the demand for the services or commodities provided by the business, and there are often seasonal fluctuations in addition to other market fluctuations. It is almost certain that a small business making $60,000 per year will make more than $5000 in some months and less than $5000 in others. Therefore it is much more appropriate to say that the business makes and average of $5000 per month.

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Question: `q005. If you travel 300 miles in 6 hours, at what average rate are you covering distance, and why do we say average rate instead of just plain rate?

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Your solution:

300/6= 50 miles a hour

some times you may go faster or slower depending on traffic or other factors

confidence rating:

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Given Solution:

The average rate is 50 miles per hour, or 50 miles / hour. This is obtained by dividing the accumulated quantity, the 300 miles, by the time required to accumulate it, obtaining ave rate = 300 miles / ( 6 hours) = 50 miles / hour. Note that the rate at which distance is covered is called speed. The car has an average speed of 50 miles/hour. We say 'average rate' in this case because it is almost certain that slight changes in pressure on the accelerator, traffic conditions and other factors ensure that the speed will sometimes be greater than 50 miles/hour and sometimes less than 50 miles/hour; the 50 miles/hour we obtain from the given information is clearly and overall average of the velocities.

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Question: `q006. If you use 60 gallons of gasoline on a 1200 mile trip, then at what average rate are you using gasoline, with respect to miles traveled?

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Your solution:

60 gallons/1200 mile= .05 gallons per mile

confidence rating:

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Given Solution:

The rate of change of one quantity with respect to another is the change in the first quantity, divided by the change in the second. As in previous examples, we found the rate at which money was made with respect to time by dividing the amount of money made by the time required to make it.

By analogy, the rate at which we use fuel with respect to miles traveled is the change in the amount of fuel divided by the number of miles traveled. In this case we use 60 gallons of fuel in 1200 miles, so the average rate it 60 gal / (1200 miles) = .05 gallons / mile.

Note that this question didn't ask for miles per gallon. Miles per gallon is an appropriate and common calculation, but it measures the rate at which miles are covered with respect to the amount of fuel used. Be sure you see the difference.

Note that in this problem we again have here an example of a rate, but unlike previous instances this rate is not calculated with respect to time. This rate is calculated with respect to the amount of fuel used. We divide the accumulated quantity, in this case miles, by the amount of fuel required to cover t miles. Note that again we call the result of this problem an average rate because there are always at least subtle differences in driving conditions that result in more or fewer miles covered with a certain amount of fuel.

It's very important to understand the phrase 'with respect to'. Whether the calculation makes sense or not, it is defined by the order of the terms.

In this case gallons / mile tells you how many gallons you are burning, on the average, per mile. This concept is not as familiar as miles / gallon, but except for familiarity it's technically no more difficult.

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Question: `q007. The word 'average' generally connotes something like adding two quantities and dividing by 2, or adding several quantities and dividing by the number of quantities we added. Why is it that we are calculating average rates but we aren't adding anything?

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Your solution:

We are not adding anything because we are using one factor of multiple quantities and finding out what one would be. Everything has been added for us when it says we went so many miles or used so much gas etc

confidence rating:

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Given Solution:

The word 'average' in the context of the dollars / month, miles / gallon types of questions we have been answering was used because we expect that in different months different amounts were earned, or that over different parts of the trip the gas mileage might have varied, but that if we knew all the individual quantities (e.g., the dollars earned each month, the number of gallons used with each mile) and averaged them in the usual manner, we would get the .05 gallons / mile, or the $5000 / month. In a sense we have already added up all the dollars earned in each month, or the miles traveled on each gallon, and we have obtained the total $60,000 or 1200 miles. Thus when we divide by the number of months or the number of gallons, we are in fact calculating an average rate.

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Question: `q008. In a study of how lifting strength is influenced by various ways of training, a study group was divided into 2 subgroups of equally matched individuals. The first group did 10 pushups per day for a year and the second group did 50 pushups per day for year. At the end of the year to lifting strength of the first group averaged 147 pounds, while that of the second group averaged 162 pounds. At what average rate did lifting strength increase per daily pushup?

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Your solution:

first group 10 per day per year second group 50 per day per year

average 147 pounds average 162 pounds

second group did 40 more push ups per day per year and lifted 15 more pounds

15/40 = .38 pounds per push up

confidence rating:

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Given Solution:

The second group had 15 pounds more lifting strength as a result of doing 40 more daily pushups than the first. The desired rate is therefore 15 pounds / 40 pushups = .375 pounds / pushup.

STUDENT COMMENT:

I have a question with respect as to how the question is interpreted. I used the interpretation given in the solution

to question 008 to rephrase the question in 009, but I do not see how this is the correct interpretation of the question as

stated.

INSTRUCTOR RESPONSE:

This exercise is designed to both see what you understand about rates, and to challenge your understanding a bit with concepts that aren't always familiar to students, despite their having completed the necessary prerequisite courses.

The meaning of the rate of change of one quantity with respect to another is of central importance in the application of mathematics. This might well be your first encounter with this particular phrasing, so it might well be unfamiliar to you, but it is important, unambiguous and universal.

You've taken the first step, which is to correctly apply the wordking of the preceding example to the present question.

You'll have ample opportunity in your course to get used to this terminology, and plenty of reinforcement.

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Question: `q009. In another part of the study, participants all did 30 pushups per day, but one group did pushups with a 10-pound weight on their shoulders while the other used a 30-pound weight. At the end of the study, the first group had an average lifting strength of 171 pounds, while the second had an average lifting strength of 188 pounds. At what average rate did lifting strength increase with respect to the added shoulder weight?

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Your solution:

first group 30 push up with 10 pounds second group 30 push up with 30 pounds

average lift 171 pounds average lift of 188 pounds

lifted 17 more pounds with 20 pounds of weight 17/20=.85 lift pounds

confidence rating:

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Given Solution:

The difference in lifting strength was 17 pounds, as a result of a 20 pound difference in added weight. The average rate at which strength increases with respect added weight would therefore be 17 lifting pounds / (20 added pounds) = .85 lifting pounds / added pound. The strength advantage was .85 lifting pounds per pound of added weight, on the average.

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Question: `q010. During a race, a runner passes the 100-meter mark 12 seconds after the start and the 200-meter mark 22 seconds after the start. At what average rate was the runner covering distance between those two positions?

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Your solution:

100 meter 12 seconds 200 meter at 22 seconds

100 meters/ 10 seconds = 10 meters per second

confidence rating:

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Given Solution:

The runner traveled 100 meters between the two positions, and required 10 seconds to do so. The average rate at which the runner was covering distance was therefore 100 meters / (10 seconds) = 10 meters / second. Again this is an average rate; at different positions in his stride the runner would clearly be traveling at slightly different speeds.

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Question: `q011. During a race, a runner passes the 100-meter mark moving at 10 meters / second, and the 200-meter mark moving at 9 meters / second. What is your best estimate of how long it takes the runner to cover the 100 meter distance?

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Your solution:

It says estimate so we must take the time for the 100 and the time for the 200 meter and get an idea of how fast. 10+9=19 19/2= 9.5 seconds so if we go 100 meters/ 9.5 seconds pr meter = 10.5 seconds

confidence rating:

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Given Solution:

At 10 meters/sec, the runner would require 10 seconds to travel 100 meters. However the runner seems to be slowing, and will therefore require more than 10 seconds to travel the 100 meters. We don't know what the runner's average speed is, we only know that it goes from 10 m/s to 9 m/s. The simplest estimate we could make would be that the average speed is the average of 10 m/s and 9 m/s, or (10 m/s + 9 m/s ) / 2 = 9.5 m/s. Taking this approximation as the average rate, the time required to travel 100 meters will be (100 meters) / (9.5 m/s) = 10.5 sec, approx.. Note that simply averaging the 10 m/s and the 9 m/s might not be the best way to approximate the average rate--for example we if we knew enough about the situation we might expect that this runner would maintain the 10 m/s for most of the remaining 100 meters, and simply tire during the last few seconds. However we were not given this information, and we don't add extraneous assumptions without good cause. So the approximation we used here is pretty close to the best we can do with the given information.

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Question: `q012. We just averaged two quantities, adding them and dividing by 2, to find an average rate. We didn't do that before. Why we do it now?

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Your solution:

We did it before sort of. We were given quantities that were already figured for us likes miles, gallons, seconds. This time we were given 2 different quantities and asked to make an estimate so we had to add and then divide to get an average because we did not have one.

confidence rating:

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Given Solution:

In previous examples the quantities weren't rates. We were given the amount of change of some accumulating quantity, and the change in time or in some other quantity on which the first was dependent (e.g., dollars and months, miles and gallons). Here we are given 2 rates, 10 m/s and 9 m/s, in a situation where we need an average rate in order to answer a question. Within this context, averaging the 2 rates was an appropriate tactic.

STUDENT QUESTION:

I thought the change of an accumulating quantity was the rate?

INSTRUCTOR RESPONSE:

Quick response: The rate is not just the change in the accumulating quantity; if we're talking about a 'time rate' it's the change in the accumulating quantity divided by the time interval (or in calculus the limiting value of this ratio as the time interval approaches zero).

More detailed response: If quantity A changes with respect to quantity B, then the average rate of change of A with respect to B (i.e., change in A / change in B) is 'the rate'. If the B quantity is clock time, then 'the rate' tells you 'how fast' the A quantity accumulates. However the rate is not just the change in the quantity A (i.e., the change in the accumulating quantity), but change in A / change in B.

For students having had at least a semester of calculus at some level: Of course the above generalizes into the definition of the derivative. y ' (x) is the instantaneous rate at which the y quantity changes with respect to x. y ' (x) is the rate at which y accumulates with respect to x.

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If your solution to stated problem does not match the given solution, you should self-critique per instructions at

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qa areas etc

001. Areas

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Question: `q001. There are 11 questions and 7 summary questions in this assignment.

What is the area of a rectangle whose dimensions are 4 m by 3 meters.

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Your solution:

4*3= 12m^2

confidence rating:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`aA 4 m by 3 m rectangle can be divided into 3 rows of 4 squares, each 1 meter on a side. This makes 3 * 4 = 12 such squares. Each 1 meter square has an area of 1 square meter, or 1 m^2. The total area of the rectangle is therefore 12 square meters, or 12 m^2.

The formula for the area of a rectangle is A = L * W, where L is the length and W the width of the rectangle. Applying this formula to the present problem we obtain area A = L * W = 4 m * 3 m = (4 * 3) ( m * m ) = 12 m^2.

Note the use of the unit m, standing for meters, in the entire calculation. Note that m * m = m^2.

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Question: `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?

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Your solution:

a=1/2 b*h a=.5(4*3)= .5*12= 6m^2

confidence rating:

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Given Solution:

`aA right triangle can be joined along its hypotenuse with another identical right triangle to form a rectangle. In this case the rectangle would have dimensions 4.0 meters by 3.0 meters, and would be divided by any diagonal into two identical right triangles with legs of 4.0 meters and 3.0 meters.

The rectangle will have area A = L * W = 4.0 m * 3.0 m = 12 m^2, as explained in the preceding problem. Each of the two right triangles, since they are identical, will therefore have half this area, or 1/2 * 12 m^2 = 6.0 m^2.

The formula for the area of a right triangle with base b and altitude h is A = 1/2 * b * h.

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Question: `q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters?

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Your solution:

5*2=10m^2

confidence rating:

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Given Solution:

`aA parallelogram is easily rearranged into a rectangle by 'cutting off' the protruding end, turning that portion upside down and joining it to the other end. Hopefully you are familiar with this construction. In any case the resulting rectangle has sides equal to the base and the altitude so its area is A = b * h.

The present rectangle has area A = 5.0 m * 2.0 m = 10 m^2.

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Question: `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm?

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Your solution:

.5*5*2= .5*10= 5 cm^2

confidence rating:

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Given Solution:

`aIt is possible to join any triangle with an identical copy of itself to construct a parallelogram whose base and altitude are equal to the base and altitude of the triangle. The area of the parallelogram is A = b * h, so the area of each of the two identical triangles formed by 'cutting' the parallelogram about the approriate diagonal is A = 1/2 * b * h. The area of the present triangle is therefore A = 1/2 * 5.0 cm * 2.0 cm = 1/2 * 10 cm^2 = 5.0 cm^2.

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Question: `q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km?

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Your solution:

4*5=20 km^2

confidence rating:

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Given Solution:

`aAny trapezoid can be reconstructed to form a rectangle whose width is equal to that of the trapezoid and whose altitude is equal to the average of the two altitudes of the trapezoid. The area of the rectangle, and therefore the trapezoid, is therefore A = base * average altitude. In the present case this area is A = 4.0 km * 5.0 km = 20 km^2.

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Question: `q006. What is the area of a trapezoid whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm?

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Your solution:

2 different altitudes so take an average

3+8=11/2= 5.5 so the altitude is 5.5 5.5*4= 22cm^2

confidence rating:

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Given Solution:

`aThe area is equal to the product of the width and the average altitude. Average altitude is (3 cm + 8 cm) / 2 = 5.5 cm so the area of the trapezoid is A = 4 cm * 5.5 cm = 22 cm^2.

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Question: `q007. What is the area of a circle whose radius is 3.00 cm?

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Your solution:

a= pi*r^2 a= 3.14*9= 28.26cm^2

confidence rating:

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Given Solution:

`aThe area of a circle is A = pi * r^2, where r is the radius. Thus

A = pi * (3 cm)^2 = 9 pi cm^2.

Note that the units are cm^2, since the cm unit is part r, which is squared.

The expression 9 pi cm^2 is exact. Any decimal equivalent is an approximation. Using the 3-significant-figure approximation pi = 3.14 we find that the approximate area is A = 9 pi cm^2 = 9 * 3.14 cm^2 = 28.26 cm^2, which we round to 28.3 cm^2 to match the number of significant figures in the given radius.

Be careful not to confuse the formula A = pi r^2, which gives area in square units, with the formula C = 2 pi r for the circumference. The latter gives a result which is in units of radius, rather than square units. Area is measured in square units; if you get an answer which is not in square units this tips you off to the fact that you've made an error somewhere.

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Question: `q008. What is the circumference of a circle whose radius is exactly 3 cm?

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Your solution:

2*pi*r= C 2*3.14*3= 18.84 cm

confidence rating:

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Given Solution:

`aThe circumference of this circle is

C = 2 pi r = 2 pi * 3 cm = 6 pi cm.

This is the exact area. An approximation to 3 significant figures is 6 * 3.14 cm = 18.84 cm.

Note that circumference is measured in the same units as radius, in this case cm, and not in cm^2. If your calculation gives you cm^2 then you know you've done something wrong.

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Question: `q009. What is the area of a circle whose diameter is exactly 12 meters?

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Your solution:

a radius is 1/2 diameter so 12/2=6 3.14*6^2= 3.14*36=113.04m^2

confidence rating:

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Given Solution:

`aThe area of a circle is A = pi r^2, where r is the radius. The radius of this circle is half the 12 m diameter, or 6 m. So the area is

A = pi ( 6 m )^2 = 36 pi m^2.

This result can be approximated to any desired accuracy by using a sufficient number of significant figures in our approximation of pi. For example using the 5-significant-figure approximation pi = 3.1416 we obtain A = 36 m^2 * 3.1416 = 113.09 m^2.

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Question: `q010. What is the area of a circle whose circumference is 14 `pi meters?

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Your solution:

C= 2*3.14*r 14 pi m= 2*pi r divide by 2 pi = 7 m = r

r= 7m 3.14* 49= 153.86 m^2

confidence rating:

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Given Solution:

`aWe know that A = pi r^2. We can find the area if we know the radius r. We therefore attempt to use the given information to find r.

We know that circumference and radius are related by C = 2 pi r. Solving for r we obtain r = C / (2 pi). In this case we find that

r = 14 pi m / (2 pi) = (14/2) * (pi/pi) m = 7 * 1 m = 7 m.

We use this to find the area

A = pi * (7 m)^2 = 49 pi m^2.

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Self-critique (if necessary):

I figured pi as 3.14 then multiplied instead of leaving it 49 pi

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Question: `q011. What is the radius of circle whose area is 78 square meters?

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Your solution:

78= area area= 3.14 * r^2 78= 3.14 *r^2 78/3.14=24.84 r^2= 24.84 square of 24.84= 4.98 m

confidence rating:

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Given Solution:

`aKnowing that A = pi r^2 we solve for r. We first divide both sides by pi to obtain A / pi = r^2. We then reverse the sides and take the square root of both sides, obtaining r = sqrt( A / pi ).

Note that strictly speaking the solution to r^2 = A / pi is r = +-sqrt( A / pi ), meaning + sqrt( A / pi) or - sqrt(A / pi). However knowing that r and A are both positive quantities, we can reject the negative solution.

Now we substitute A = 78 m^2 to obtain

r = sqrt( 78 m^2 / pi) = sqrt(78 / pi) m.{}

Approximating this quantity to 2 significant figures we obtain r = 5.0 m.

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Self-critique (if necessary):

Did my way make sense?

Your solution was completely equivalent to the given solution. Very good.

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Question: `q012. Summary Question 1: How do we visualize the area of a rectangle?

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Your solution:

length times width

confidence rating:

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Given Solution:

`aWe visualize the rectangle being covered by rows of 1-unit squares. We multiply the number of squares in a row by the number of rows. So the area is A = L * W.

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Question: `q013. Summary Question 2: How do we visualize the area of a right triangle?

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Your solution:

area of a triangle is 1/2* base* height

confidence rating:

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Given Solution:

`aWe visualize two identical right triangles being joined along their common hypotenuse to form a rectangle whose length is equal to the base of the triangle and whose width is equal to the altitude of the triangle. The area of the rectangle is b * h, so the area of each triangle is 1/2 * b * h.

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Question: `q014. Summary Question 3: How do we calculate the area of a parallelogram?

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Your solution:

a parallelogram area is base * height height is altitude

confidence rating:

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Given Solution:

`aThe area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base.

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Question: `q015. Summary Question 4: How do we calculate the area of a trapezoid?

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Your solution:

area = altitude * width you need to get an average altitude when you do this because the figure may have different altitudes

confidence rating:

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Given Solution:

`aWe think of the trapezoid being oriented so that its two parallel sides are vertical, and we multiply the average altitude by the width.

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Question: `q016. Summary Question 5: How do we calculate the area of a circle?

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Your solution:

area of circle = pi* r^2 pi =3.14 r= radius

confidence rating:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`aWe use the formula A = pi r^2, where r is the radius of the circle.

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Question: `q017. Summary Question 6: How do we calculate the circumference of a circle? How can we easily avoid confusing this formula with that for the area of the circle?

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Your solution:

C= 2*pi*r pi=3.14 r= radius When you get circumference it is not units ^2

confidence rating:

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Given Solution:

`aWe use the formula C = 2 pi r. The formula for the area involves r^2, which will give us squared units of the radius. Circumference is not measured in squared units.

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Question: `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.

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Self-critique (if necessary):

I am not sure what is being asked in this question. I have the formulas in my notebook if that is what is being asked.

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If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

qa areas etc

002. Volumes

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Question: `q001. There are 9 questions and 4 summary questions in this assignment.

What is the volume of a rectangular solid whose dimensions are exactly 3 cm by 5 cm by 7 cm?

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Your solution:

V= area * height we have to find the area of 5*7 = 35 cm^2 and multiply by the hegth of 3 cm

35*3= 105 cm^3

confidence rating:

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Given Solution:

`aIf we orient this object so that its 3 cm dimension is its 'height', then it will be 'resting' on a rectangular base whose dimension are 5 cm by 7 cm. This base can be divided into 5 rows each consisting of 7 squares, each 1 meter by 1 meter. There will therefore be 5 * 7 = 35 such squares, showing us that the area of the base is 35 m^2.

Above each of these base squares the object rises to a distance of 3 meters, forming a small rectangular tower. Each such tower can be divided into 3 cubical blocks, each having dimension 1 meter by 1 meter by 1 meter. The volume of each 1-meter cube is 1 m * 1 m * 1 m = 1 m^3, also expressed as 1 cubic meter. So each small 'tower' has volume 3 m^3.

The object can be divided into 35 such 'towers'. So the total volume is 35 * 3 m^3 = 105 m^3.

This construction shows us why the volume of a rectangular solid is equal to the area of the base (in this example the 35 m^2 of the base) and the altitude (in this case 3 meters). The volume of any rectangular solid is therefore

V = A * h,

where A is the area of the base and h the altitude.

This is sometimes expressed as V = L * W * h, where L and W are the length and width of the base. However the relationship V = A * h applies to a much broader class of objects than just rectangular solids, and V = A * h is a more powerful idea than V = L * W * h. Remember both, but remember also that V = A * h is the more important.

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Question: `q002. What is the volume of a rectangular solid whose base area is 48 square meters and whose altitude is 2 meters?

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Your solution:

48m^2* 2m = 96m^3

confidence rating:

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Given Solution:

`aUsing the idea that V = A * h we find that the volume of this solid is

V = A * h = 48 m^2 * 2 m = 96 m^3.

Note that m * m^2 means m * (m * m) = m * m * m = m^2.

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Question: `q003. What is the volume of a uniform cylinder whose base area is 20 square meters and whose altitude is 40 meters?

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Your solution:

20 m^2* 40m= 800m^3

confidence rating:

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Given Solution:

`aV = A * h applies to uniform cylinders as well as to rectangular solids. We are given the altitude h and the base area A so we conclude that

V = A * h = 20 m^2 * 40 m = 800 m^3.

The relationship V = A * h applies to any solid object whose cross-sectional area A is constant. This is the case for uniform cylinders and uniform prisms.

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Question: `q004. What is the volume of a uniform cylinder whose base has radius 5 cm and whose altitude is 30 cm?

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Your solution:

5 ^2 cm* 3.14= 25*3.14=78.5 cm^2 78.5 cm^2 * 30 cm= 2355 cm^3

confidence rating:

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Given Solution:

`aThe cylinder is uniform, which means that its cross-sectional area is constant. So the relationship V = A * h applies.

The cross-sectional area A is the area of a circle of radius 5 cm, so we see that A = pi r^2 = pi ( 5 cm)^2 = 25 pi cm^2.

Since the altitude is 30 cm the volume is therefore

V = A * h = 25 pi cm^2 * 30 cm = 750 pi cm^3.

Note that the common formula for the volume of a uniform cylinder is V = pi r^2 h. However this is just an instance of the formula V = A * h, since the cross-sectional area A of the uniform cylinder is pi r^2. Rather than having to carry around the formula V = pi r^2 h, it's more efficient to remember V = A * h and to apply the well-known formula A = pi r^2 for the area of a circle.

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Self-critique (if necessary):

Do I need to quit solving for pi and just write pi in my answer or is it ok to keep solving?

In some situations it's OK to approximate; but only answers expressed as multiples of pi are exact. Multiples of pi are in some situations easier to express and make sense of.

As long as you can do it both ways you'll be OK.

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Question: `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates?

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Your solution:

the can has a diameter of 3 in and the height of 6 in

the radius is 1.5 a= 3.14* 1.5^2= 7.07 in^2 7.07 in^2* 6 in = 42.42 in ^3

confidence rating:

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Given Solution:

`aPeople will commonly estimate the dimensions of a can of food in centimeters or in inches, though other units of measure are possible (e.g., millimeters, feet, meters, miles, km). Different cans have different dimensions, and your estimate will depend a lot on what can you are using.

A typical can might have a circular cross-section with diameter 3 inches and altitude 5 inches. This can would have volume V = A * h, where A is the area of the cross-section. The diameter of the cross-section is 3 inches so its radius will be 3/2 in.. The cross-sectional area is therefore A = pi r^2 = pi * (3/2 in)^2 = 9 pi / 4 in^2 and its volume is

V = A * h = (9 pi / 4) in^2 * 5 in = 45 pi / 4 in^3.

Approximating, this comes out to around 35 in^3.

Another can around the same size might have diameter 8 cm and height 14 cm, giving it cross-sectional area A = pi ( 4 cm)^2 = 16 pi cm^2 and volume V = A * h = 16 pi cm^2 * 14 cm = 224 pi cm^2.

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Question: `q006. What is the volume of a pyramid whose base area is 50 square cm and whose altitude is 60 cm?

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Your solution:

v= 1/3 a*h

50 cm^2*60 cm= 3000 cm^3/3= 1000 cm^3

confidence rating:

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Given Solution:

`aWe can't use the V = A * h idea for a pyramid because the thing doesn't have a constant cross-sectional area--from base to apex the cross-sections get smaller and smaller. It turns out that there is a way to cut up and reassemble a pyramid to show that its volume is exactly 1/3 that of a rectangular solid with base area A and altitude h. Think of putting the pyramid in a box having the same altitude as the pyramid, with the base of the pyramid just covering the bottom of the box. The apex (the point) of the pyramid will just touch the top of the box. The pyramid occupies exactly 1/3 the volume of that box.

So the volume of the pyramid is V = 1/3 * A * h. The base area A is 30 cm^2 and the altitude is 60 cm so we have

V = 1/3 * 50 cm^2 * 60 cm = 1000 cm^3.

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Question: `q007. What is the volume of a cone whose base area is 20 square meters and whose altitude is 9 meters?

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Your solution:

20m ^2* 9 m= 180 m^3/ 3 = 60 m^3

confidence rating:

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Given Solution:

`aJust as the volume of a pyramid is 1/3 the volume of the 'box' that contains it, the volume of a cone is 1/3 the volume of the cylinder that contains it. Specifically, the cylinder that contains the cone has the base of the cone as its base and matches the altitude of the cone. So the volume of the cone is 1/3 A * h, where A is the area of the base and h is the altitude of the cone.

In this case the base area and altitude are given, so the volume of the cone is

V = 1/3 A * h = 1/3 * 20 m^2 * 9 m = 60 m^3.

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Question: `q008. What is a volume of a sphere whose radius is 4 meters?

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Your solution:

V= 4/3* 3.14* r^3 1.33*3.14*64= 267.28 m^3

confidence rating:

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Given Solution:

`aThe volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere. In this case r = 4 m so

V = 4/3 pi * (4 m)^3 = 4/3 pi * 4^3 m^3 = 256/3 pi m^3.

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Question: `q009. What is the volume of a planet whose diameter is 14,000 km?

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Your solution:

V= 4/3 pi*r^3 4/3* 3.14* 7000^3= 1.3333*3.14*343,000,000,000 km^3= 1,436,141,000,000 km^3

confidence rating:

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Given Solution:

`aThe planet is presumably a sphere, so to the extent that this is so the volume of this planet is V = 4/3 pi r^3, where r is the radius of the planet. The diameter of the planet is 14,000 km so the radius is half this, or 7,000 km. It follows that the volume of the planet is

V = 4/3 pi r^3 = 4/3 pi * (7,000 km)^3 = 4/3 pi * 343,000,000,000 km^3 = 1,372,000,000,000 / 3 * pi km^3.

This result can be approximated to an appropriate number of significant figures.

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Question: `q010. Summary Question 1: What basic principle do we apply to find the volume of a uniform cylinder of known dimensions?

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Your solution:

Volume is the area *height area is base*height

confidence rating:

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Given Solution:

`aThe principle is that when the cross-section of an object is constant, its volume is V = A * h, where A is the cross-sectional area and h the altitude. Altitude is measure perpendicular to the cross-section.

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Question: `q011. Summary Question 2: What basic principle do we apply to find the volume of a pyramid or a cone?

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Your solution:

V= 1/3*a*h area is b*h

confidence rating:

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Given Solution:

`aThe volumes of these solids are each 1/3 the volume of the enclosing figure. Each volume can be expressed as V = 1/3 A * h, where A is the area of the base and h the altitude as measured perpendicular to the base.

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Question: `q012. Summary Question 3: What is the formula for the volume of a sphere?

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Your solution:

V= 4/3*pi* r^3 pi= 3.14 r= radius

confidence rating:

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Given Solution:

`aThe volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere.

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Question: `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.

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Self-critique (if necessary):

I have the formulas in my notebook

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Self-critique Rating:

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

003. Misc: Surface Area, Pythagorean Theorem, Density

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Question: `q001. What is surface area of a rectangular solid whose dimensions are 3 meters by 4 meters by 6 meters?

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Your solution:

there are 6 surfaces so it would be 3x4, 3x4, 4x6, 4x6, 3x6, 3x6

3*4= 12 3*4=12 4*6=24 4*6=24 3*6=18 3*6=18 12+12+24+24+18+18=108 m^2

confidence rating:

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Given Solution:

`aA rectangular solid has six faces (top, bottom, front, back, left side, right side if you're facing it). The pairs top and bottom, right and left sides, and front-back have identical areas. This solid therefore has two faces with each of the following dimensions: 3 m by 4 m, 3 m by 6 m and 4 m by 6 m, areas 12 m^2, 18 m^2 and 24 m^2. Total area is 2 * 12 m^2 + 2 * 18 m^2 + 2 * 24 m^2 = 108 m^2.

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Question: `q002. What is the surface area of the curved sides of a cylinder whose radius is five meters and whose altitude is 12 meters? If the cylinder is closed what is its total surface area?

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Your solution:

surface area= C* altitude 2*pi*r 2*3.14*5= 31.4m

12* 31.4= 376.8m^2

if closed then we need to add top and bottom and sides

area of top and bottom 3.14* 5^2= 78.5m^2 78.5+78.5=157 m^2

total= 157m ^2+ 376.8 m^2= 533.8 m^2

confidence rating:

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Given Solution:

The circumference of this cylinder is 2 pi r = 2 pi * 5 m = 10 pi m. If the cylinder was cut by a straight line running up its curved face then unrolled it would form a rectangle whose length and width would be the altitude and the circumference. The area of the curved side is therefore

A = circumference * altitude = 10 pi m * 12 m = 120 pi m^2.

If the cylinder is closed then it has a top and a bottom, each a circle of radius 5 m with resulting area A = pi r^2 = pi * (5 m)^2 = 25 pi m^2. The total area would then be

total area = area of sides + 2 * area of base = 120 pi m^2 + 2 * 25 pi m^2 = 170 pi m^2.

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Question: `q003. What is surface area of a sphere of diameter three cm?

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Your solution:

surface area= 4*pi*r^2 4*3.14*1.5^2= 12.56*2.25=28.26 cm^3

confidence rating:

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Given Solution:

`aThe surface area of a sphere of radius r is A = 4 pi r^2. This sphere has radius 3 cm / 2, and therefore has surface area

A = 4 pi r^2 = 4 pi * (3/2 cm)^2 = 9 pi cm^2.

NOTE TO STUDENT:

While your work on most problems has been good, you left this problem blank and didn't self-critique.

You should self-critique here.

For example you should acknowledge having made note of the formula for the surface area of the sphere, which I expect you didn't know before.

I expect from your previous answers that you are very capable of applying the formula once you have it, and based on this history you probably wouldn't need to self-critique that aspect of the process.

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Question: `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?

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Your solution:

a^2+b^2= c^2 5^2+9^2= 25+81= 106 c= square root of 106 or c= 10.3 m

confidence rating:

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Given Solution:

`aThe Pythagorean Theorem says that the hypotenuse c of a right triangle with legs a and b satisfies the equation c^2 = a^2 + b^2. So, since all lengths are positive, we know that

c = sqrt(a^2 + b^2) = sqrt( (5 m)^2 + (9 m)^2 ) = sqrt( 25 m^2 + 81 m^2) = sqrt( 106 m^2 ) = 10.3 m, approx..

Note that this is not what we would get if we made the common error of assuming that sqrt(a^2 + b^2) = a + b; this would tell us that the hypotenuse is 14 m, which is emphatically not so. There is no justification whatsoever for applying a distributive law (like x * ( y + z) = x * y + x * z ) to the square root operator.

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Question: `q005. If the hypotenuse of a right triangle has length 6 meters and one of its legs has length 4 meters what is the length of the other leg?

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Your solution:

c^2=B^2+a^2 6^2= 4^2 + b^2 36= 16+ b ^2 20= b^2 b= 4.5 m

confidence rating:

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Given Solution:

`aIf c is the hypotenuse and a and b the legs, we know by the Pythagorean Theorem that c^2 = a^2 + b^2, so that a^2 = c^2 - b^2. Knowing the hypotenuse c = 6 m and the side b = 4 m we therefore find the unknown leg:

a = sqrt( c^2 - b^2) = sqrt( (6 m)^2 - (4 m)^2 ) = sqrt(36 m^2 - 16 m^2) = sqrt(20 m^2) = sqrt(20) * sqrt(m^2) = 2 sqrt(5) m,

or approximately 4.4 m.

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Question: `q006. If a rectangular solid made of a uniform, homogeneous material has dimensions 4 cm by 7 cm by 12 cm and if its mass is 700 grams then what is its density in grams per cubic cm?

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Your solution:

V= 4*7*12= 336 cm^3

density = 700 total each cm would be 700/336= 2.08 g per cm^3

confidence rating:

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Given Solution:

`aThe volume of this solid is 4 cm * 7 cm * 12 cm = 336 cm^3.

Its density in grams per cm^3 is the number of grams in each cm^3. We find this quantity by dividing the number of grams by the number of cm^3. We find that

density = 700 grams / (336 cm^3) = 2.06 grams / cm^3.

Note that the solid was said to be uniform and homogeneous, meaning that it's all made of the same material, which is uniformly distributed. So each cm^3 does indeed have a mass of 2.06 grams.

Had we not known that the material was uniform and homogeneous we could have said that the average density is 2.06 grams / cm^3, but not that the density is 2.06 grams / cm^3 (for example the object could be made of two separate substances, one with density less than 2.06 grams / cm^3 and the other with density greater than 2.06 g / cm^3, in appropriate proportions; neither substance would have density 2.06 g / cm^3, but the average density could be 2.06 g / cm^3).

NOTE TO STUDENT: (in this note the instructor attempts to clarify the idea of 'demonstrating what you do and do not understand about the statement of the problem' and 'giving a phrase-by-phrase analysis of the given solution')

You did not respond to the question and did not self-critique.

You would be expected to address the question, stating what you do and do not understand.

For example you should understand what a rectangular solid with dimensions 4 cm by 7 cm by 12 cm is, and how to find its volume and surface area. You might not know what to do with this information (for example you might well not understand that it's the volume and not the surface area that's related to density), but from previous work you should understand this much, and should at least mention something along the lines of 'well, I do know that I can find the volume and/or surface area of that solid' in a partial solution.

The word 'density' is clearly very important. Even if you don't know what density is, you could note from the statement of the problem that its units here are said to be 'grams per cubic centimeter'.

Having noted these things, you will be much better prepared to understand the information in the given solution.

Then you need to address the information in the given solution. A 'phrase-by-phrase' analysis is generally very beneficial:

I expect you understand the first statement from previous knowledge (you should have this understanding from prerequisite courses, and if not you encountered it in the preceding 'volumes' exercise): 'The volume of this solid is 4 cm * 7 cm * 12 cm = 336 cm^3.' It would of course be appropriate to ask a question here if necessary.

It is likely that, as is the case with many students, the concept of density is not that familiar to you. However if this wasn't addressed specifically in prerequisite courses, those courses would be expected to prepare you to understand this concept. The statement 'Its density in grams per cm^3 is the number of grams in each cm^3.' serves as a definition of density. In your self-critique you should have addressed what what this phrase means to you, and what you do or do not understand about it

The next phrase is 'We find this quantity by dividing the number of grams by the number of cm^3.' You would be expected to understand that this phrase is related to the preceding, and as best you can to address the connection. At this point many students would need to ask a question, and it would be perfectly appropriate to do so (or to have done so regarding previous statements).

The subsequent phrase 'density = 700 grams / (336 cm^3) = 2.06 grams / cm^3' is an illustration of the ideas and definitions in the preceding statements. A reasonable self-critique would demonstrate your attempt to understand this statement and its connection to the preceding. Once again questions would also be appropriate and welcome.

The above addresses sufficient information to solve the problem. If you get to this point, you're probably doing OK and you wouldn't necessarily be expected to address the rest of the given solution, which expands on the finer details of the problem and provides additional information. The basic prerequisite courses should have prepared you to understand the information, but students entering Liberal Arts Mathematics, College Algebra and even Precalculus or Applied Calculus (or Physics 121-122) courses probably don't need to address anything beyond the basic solution at this point. Though Precalculus and Applied Calculus students could benefit from doing so, and if time permits would certainly be encouraged to do so, time is also a factor and it would be understandable if these students chose to move on.

Students entering the Mth 173-4 sequence or the Phy 201-202 or 231-232 sequence would be expected to either completely understand all the details of the given solution, or address them in your self-critique.

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Question: `q007. What is the mass of a sphere of radius 4 meters if its average density is 3,000 kg/cubic meter?

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Your solution:

V= 4/3*3.14* (4)^3 4/3*200.96 = 267.95 m^3 mass= density*volume 267.95* 3000= 803,850 kg

confidence rating:

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Given Solution:

`aA average density of 3000 kg / cubic meter implies that, at least on the average, every cubic meter has a mass of 3000 kg. So to find the mass of the sphere we multiply the number of cubic meters by 3000 kg.

The volume of a sphere of radius 4 meters is 4/3 pi r^3 = 4/3 * pi (4m)^3 = 256/3 * pi m^3. So the mass of this sphere is

mass = density * volume = 256 / 3 * pi m^3 * 3000 kg / m^3 = 256,000 * pi kg.

This result can be approximated to an appropriate number of significant figures.

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Question: `q008. If we build a an object out of two pieces of material, one having a volume of 6 cm^3 at a density of 4 grams per cm^3 and another with a volume of 10 cm^3 at a density of 2 grams per cm^3 then what is the average density of this object?

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Your solution:

mass of first= 6 cm^3* 4g per cm^3 = 24g second 10cm^3*2 g= 20 g total 24+20=44g

density= total mass/total volume 44g/6+10 cm^3 44g/16cm^3 = 2.75g per cm^3

confidence rating:

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Given Solution:

`aThe first piece has a mass of 4 grams / cm^3 * 6 cm^3 = 24 grams. The second has a mass of 2 grams / cm^3 * 10 cm^3 = 20 grams. So the total mass is 24 grams + 20 grams = 44 grams.

The average density of this object is

average density = total mass / total volume = (24 grams + 20 grams) / (6 cm^3 + 10 cm^3) = 44 grams / (16 cm^3) = 2.75 grams / cm^3.

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Question: `q009. In a large box of dimension 2 meters by 3 meters by 5 meters we place 27 cubic meters of sand whose density is 2100 kg/cubic meter, surrounding a total of three cubic meters of cannon balls whose density is 8,000 kg per cubic meter. What is the average density of the material in the box?

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Your solution:

density= mass/volume 27 m^3 * 2100= 56700 kg 3*8000= 24000 kg

56700+24000kg= 80700 kg 27m^3+3m^3= 30 m^3

80700/30m^3= 2690kg per m^3

confidence rating:

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Given Solution:

`aWe find the average density from the total mass and the total volume. The mass of the sand is 27 m^3 * 2100 kg / m^3 = 56,700 kg. The mass of the cannonballs is 3 m^3 * 8,000 kg / m^3 = 24,000 kg.

The average density is therefore

average density = total mass / total volume = (56,700 kg + 24,000 kg) / (27 m^3 + 3 m^3) = 80,700 kg / (30 m^3) = 2,700 kg / m^3, approx..

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Question: `q010. How many cubic meters of oil are there in an oil slick which covers 1,700,000 square meters (between 1/2 and 1 square mile) to an average depth of .015 meters? If the density of the oil is 860 kg/cubic meter the what is the mass of the oil slick?

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Your solution:

mass= density*volume 1,700,000*.015= 25,500m ^3

860*25,500= 21,930,000 kg

confidence rating:

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Given Solution:

`aThe volume of the slick is V = A * h, where A is the area of the slick and h the thickness. This is the same principle used to find the volume of a cylinder or a rectangular solid. We see that the volume is

V = A * h = 1,700,000 m^2 * .015 m = 25,500 m^3.

The mass of the slick is therefore

mass = density * volume = 860 kg / m^3 * 24,400 m^3 = 2,193,000 kg.

This result should be rounded according to the number of significant figures in the given information.

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Question: `q011. Part 1 Summary Question 1: How do we find the surface area of a cylinder?

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Your solution:

surface area= circumference*altitude C= 2*pi*r if closed then you have a top and bottom so you have to find the area of these by using a=pi*r^2 and multiply this answer by 2 because of a top and bottom to get total add 1st answer to the total from the top and bottom

confidence rating:

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Given Solution:

`aThe curved surface of the cylinder can be 'unrolled' to form a rectangle whose dimensions are equal to the circumference and the altitude of the cylinder, so the curved surface has volume

Acurved = circumference * altitude = 2 pi r * h, where r is the radius and h the altitude.

The top and bottom of the cylinder are both circles of radius r, each with resulting area pi r^2.

{]The total surface area is therefore

Acylinder = 2 pi r h + 2 pi r^2.

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Question: `q012. Part 1 Summary Question 2: What is the formula for the surface area of a sphere?

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Your solution:

surface area= 4pi*r^2

confidence rating:

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Given Solution:

`aThe surface area of a sphere is

A = 4 pi r^2,

where r is the radius of the sphere.

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Question: `q013. Part 1 Summary Question 3: What is the meaning of the term 'density'.

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Your solution:

density is the mass or weight per unit of volume

confidence rating:

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Given Solution:

`aThe average density of an object is its mass per unit of volume, calculated by dividing its total mass by its total volume. If the object is uniform and homogeneous then its density is constant and we can speak of its 'density' as opposed to its 'average density'

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Question: `q014. Part 1 Summary Question 4: If we know average density and mass, how can we find volume?

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Your solution:

volume= mass/ density

confidence rating:

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Given Solution:

`aSince mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.

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Question: `q015. Part 1 Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.

I have the formulas and examples in my notebook

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

004. Units of volume measure

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Question: `q001. There are 10 questions and 5 summary questions in this assignment.

How many cubic centimeters of fluid would require to fill a cubic container 10 cm on a side?

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Your solution:

10*10*10= 100 cm^3

confidence rating:

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Given Solution:

`aThe volume of the container is 10 cm * 10 cm * 10 cm = 1000 cm^3. So it would take 1000 cubic centimeters of fluid to fill the container.

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Question: `q002. How many cubes each 10 cm on a side would it take to build a solid cube one meter on a side?

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Your solution:

it takes 10 to make a row so 10*10*10= 1000

confidence rating:

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Given Solution:

`aIt takes ten 10 cm cubes laid side by side to make a row 1 meter long or a tower 1 meter high. It should therefore be clear that the large cube could be built using 10 layers, each consisting of 10 rows of 10 small cubes. This would require 10 * 10 * 10 = 1000 of the smaller cubes.

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Question: `q003. How many square tiles each one meter on each side would it take to cover a square one km on the side?

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Your solution:

it takes 1000 tiles to make km 1000*1000= 1000000

confidence rating:

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Given Solution:

`aIt takes 1000 meters to make a kilometer (km). To cover a square 1 km on a side would take 1000 rows each with 1000 such tiles to cover 1 square km. It therefore would take 1000 * 1000 = 1,000,000 squares each 1 m on a side to cover a square one km on a side.

We can also calculate this formally. Since 1 km = 1000 meters, a square km is (1 km)^2 = (1000 m)^2 = 1,000,000 m^2.

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Question: `q004. How many cubic centimeters are there in a liter?

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Your solution:

cubic cm=10 10*10*10 1000cm^3 = 1 liter

confidence rating:

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Given Solution:

`aA liter is the volume of a cube 10 cm on a side. Such a cube has volume 10 cm * 10 cm * 10 cm = 1000 cm^3. There are thus 1000 cubic centimeters in a liter.

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Question: `q005. How many liters are there in a cubic meter?

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Your solution:

im not sure

confidence rating:

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Given Solution:

`aA liter is the volume of a cube 10 cm on a side. It would take 10 layers each of 10 rows each of 10 such cubes to fill a cube 1 meter on a side. There are thus 10 * 10 * 10 = 1000 liters in a cubic meter.

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Self-critique (if necessary):

1000 liters= 1m^3

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Question: `q006. How many cm^3 are there in a cubic meter?

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Your solution:

there is 100cm=m so (100cm)^3= (1m)^3

1000000cm^3= m^3

confidence rating:

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Given Solution:

`aThere are 1000 cm^3 in a liter and 1000 liters in a m^3, so there are 1000 * 1000 = 1,000,000 cm^3 in a m^3.

It's important to understand the 'chain' of units in the previous problem, from cm^3 to liters to m^3. However another way to get the desired result is also important:

There are 100 cm in a meter, so 1 m^3 = (1 m)^3 = (100 cm)^3 = 1,000,000 cm^3.

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Question: `q007. If a liter of water has a mass of 1 kg then what is the mass of a cubic meter of water?

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Your solution:cubic meter is 1000 liters so 1000 liters would weigh 1000kg

confidence rating:

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Given Solution:

`aSince there are 1000 liters in a cubic meter, the mass of a cubic meter of water will be 1000 kg. This is a little over a ton.

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Question: `q008. What is the mass of a cubic km of water?

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Your solution:

cubic m=1000kg (1000m)^3= (1km)^3

1000*1000000000= 1,000,000,000,000kg

confidence rating:

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Given Solution:

`aA cubic meter of water has a mass of 1000 kg. A cubic km is (1000 m)^3 = 1,000,000,000 m^3, so a cubic km will have a mass of 1,000,000,000 m^3 * 1000 kg / m^3 = 1,000,000,000,000 kg.

In scientific notation we would say that 1 m^3 has a mass of 10^3 kg, a cubic km is (10^3 m)^3 = 10^9 m^3, so a cubic km has mass (10^9 m^3) * 1000 kg / m^3 = 10^12 kg.

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Question: `q009. If each of 5 billion people drink two liters of water per day then how long would it take these people to drink a cubic km of water?

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Your solution:

cubic km= km^3= (1000m)^3 = 1,000,000,000 m^3 1000 liter = 1m^3 1,000,000,000*1000=

1,000,000,000,000 liters 1,000,000,000,000/ 10,000,000,000= 100 days

confidence rating:

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Given Solution:

`a5 billion people drinking 2 liters per day would consume 10 billion, or 10,000,000,000, or 10^10 liters per day.

A cubic km is (10^3 m)^3 = 10^9 m^3 and each m^3 is 1000 liters, so a cubic km is 10^9 m^3 * 10^3 liters / m^3 = 10^12 liters, or 1,000,000,000,000 liters.

At 10^10 liters per day the time required to consume a cubic km would be

time to consume 1 km^3 = 10^12 liters / (10^10 liters / day) = 10^2 days, or 100 days.

This calculation could also be written out:

1,000,000,000,000 liters / (10,000,000,000 liters / day) = 100 days.

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Question: `q010. The radius of the Earth is approximately 6400 kilometers. What is the surface area of the Earth? If the surface of the Earth was covered to a depth of 2 km with water that what would be the approximate volume of all this water?

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Your solution:

surface area= 4*pi*r^2 4*3.14*(6400)^2= 12.56* 40,960,000= 514,457,600 km^2

V=a*h 514,457,600*2 km= 1,028,915,200 km^3

confidence rating:

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Given Solution:

`aThe surface area would be

A = 4 pi r^2 = 4 pi ( 6400 km)^2 = 510,000,000 km^2.

A flat area of 510,000,000 km^2 covered to a depth of 2 km would indicate a volume of

V = A * h = 510,000,000 km^2 * 2 km = 1,020,000,000 km^3.

However the Earth's surface is curved, not flat. The outside of the 2 km covering of water would have a radius 2 km greater than that of the Earth, and therefore a greater surface area. However a difference of 2 km in 6400 km will change the area by only a fraction of one percent, so the rounded result 1,020,000,000,000 km^3 would still be accurate.

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Question: `q011. Summary Question 1: How can we visualize the number of cubic centimeters in a liter?

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Your solution:

there are 1000cubic cm in a liter

confidence rating:

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Given Solution:

`aSince a liter is a cube 10 cm on a side, we visualize 10 layers each of 10 rows each of 10 one-centimeter cubes, for a total of 1000 1-cm cubes. There are 1000 cubic cm in a liter.

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Question: `q012. Summary Question 2: How can we visualize the number of liters in a cubic meter?

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Your solution:

there are 1000 liters in m^3

confidence rating:

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Given Solution:

`aSince a liter is a cube 10 cm on a side, we need 10 such cubes to span 1 meter. So we visualize 10 layers each of 10 rows each of 10 ten-centimeter cubes, for a total of 1000 10-cm cubes. Again each 10-cm cube is a liter, so there are 1000 liters in a cubic meter.

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Question: `q013. Summary Question 3: How can we calculate the number of cubic centimeters in a cubic meter?

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Your solution:

there are 100cm in a m so (100cm)^3 = (m)^3

100*100*100= 1,000,000 cm^3= 1m^3

confidence rating:

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Given Solution:

`aOne way is to know that there are 1000 liters in a cubic meters, and 1000 cubic centimeters in a cubic meter, giving us 1000 * 1000 = 1,000,000 cubic centimeters in a cubic meter. Another is to know that it takes 100 cm to make a meter, so that a cubic meter is (100 cm)^3 = 1,000,000 cm^3.

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Question: `q014. Summary Question 4: There are 1000 meters in a kilometer. So why aren't there 1000 cubic meters in a cubic kilometer? Or are there?

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Your solution:

1000m = km (1000m)^3= (km)^3 1,000,000,000 m^3= km^3

confidence rating:

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Given Solution:

`aA cubic kilometer is a cube 1000 meters on a side, which would require 1000 layers each of 1000 rows each of 1000 one-meter cubes to fill. So there are 1000 * 1000 * 1000 = 1,000,000,000 cubic meters in a cubic kilometer.

Alternatively, (1 km)^3 = (10^3 m)^3 = 10^9 m^3, not 1000 m^3.

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Question: `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.

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Self-critique (if necessary):

I have the information in my notebook.

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Self-critique Rating:

physic orientation

Right. Very good.

Your answer made perfect sense. It didn't include all the detail of the given solution but everything you said was right, and on the right track. Much better than the answers I usually get to this question.

That's what it comes down to.

Your solution was completely equivalent to the given solution. Very good.

In some situations it's OK to approximate; but only answers expressed as multiples of pi are exact. Multiples of pi are in some situations easier to express and make sense of. As long as you can do it both ways you'll be OK.

&#Your work looks good. Let me know if you have any questions. &#

In some situations it's OK to approximate; but only answers expressed as multiples of pi are exact. Multiples of pi are in some situations easier to express and make sense of.

As long as you can do it both ways you'll be OK.