Rates

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course PHY 121

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 in 5 hours equals $10 per hour. You divide the total earned by the number of hours

to find the rate.

confidence rating #$&*: 3

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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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Self-critique (if necessary): OK

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Self-critique Rating: 3

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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:

There are 12 months in a year, so $60,000 per year is the same as $60,000 in 12 months.

This means that you divide the $60,000 by 12 to come up with the monthly rate of $5,000.

confidence rating #$&*: 3

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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.

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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Self-critique (if necessary): OK

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Self-critique Rating: 3

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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:

You do not know if the business makes the exact same amount per month. As a matter of

fact, it would be realistic to expect that to happen. Therefore, it would be more

appropriate to say that the business makes an average of $5000 per month.

confidence rating #$&*: 3

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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.

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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Self-critique (if necessary): OK

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Self-critique Rating: 3

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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:

The average rate is 300 divided by 6, or 50 miles per hour. You would say average rate

because you would probably not maintain a steady speed of 50 mph during the 6 hour

period.

confidence rating #$&*: 3

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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.

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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Self-critique (if necessary): OK

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Self-critique Rating: 3

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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:

If you use 60 gallons of gasoline on a 1200 mile trip, the rate at which you use the gas

is miles per gallon. This means that you divide 1200 by 60, which results in an average

of 20 miles per gallon.

confidence rating #$&*: 3

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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.

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.

STUDENT COMMENT

Very Tricky! I thought I had a rhythm going. I understand where I messed up. I am

comfortable with the calculations.

INSTRUCTOR RESPONSE

There's nothing wrong with your rhythm.

As I'm sure you understand, there is no intent here to trick, though I know most people

will (and do) tend to give the answer you did.

My intent is to make clear the important point that the definition of the terms is

unambiguous and must be read carefully, in the right order.

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Self-critique (if necessary):

I was overconficent and did it backwards. I need to make sure that I read the question

carefully and answer what is asked.

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Self-critique Rating: 3

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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:

When we add quantities together, we get a sum. We take this sum and divide it into

equal parts. For these questions, we start with the total amount. Therefore we don't

need to add to get the number to divide; we already have it.

confidence rating #$&*: 3

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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.

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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Self-critique (if necessary): OK

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Self-critique Rating: 3

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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:

The difference in the amount of daily push-ups is 40 push-ups per day (50-10); the

difference in strength is 15 pounds (162-147). Dividing 15 pounds by 40 push-ups should

give average rate of strength per push-up. This is .375 pounds per push-up.

confidence rating #$&*: 2

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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.

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.

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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Self-critique (if necessary): OK

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Self-critique Rating: 3

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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:

The difference in the amount of weight added to the shoulders was 30-10, or 20 pounds.

The difference in the result was 188-171, or 17 pounds.

I divided 17 by 20 and got .85. For every pound added to the shoulder weight, the

lifting strength was increased by .85 pounds.

confidence rating #$&*: 2

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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.

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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Self-critique (if necessary): OK

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Self-critique Rating: 3

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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:

The change in time was 22-12, or 10 seconds.

The change in distance was 200-100, or 100 meters.

Dividing 100 meters by 10 seconds yields 10 meters per second.

confidence rating #$&*: 3

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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.

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.

STUDENT QUESTION

Is there a formula for this is it d= r*t or distance equal rate times

time??????????????????

INSTRUCTOR RESPONSE

That formula would apply in this specific situation.

The goal is to learn to use the general concept of rate of change. The situation of this

problem, and the formula you quote, are just one instance of a general concept that

applies far beyond the context of distance and time.

It's fine if the formula helps you understand the general concept of rate. Just be sure

you work to understand the broader concept.

Note also that we try to avoid using d for the name of a variable. The letter d will

come to have a specific meaning in the context of rates, and to use d as the name of a

variable invite confusion.

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Self-critique (if necessary): OK

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Self-critique Rating: 3

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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 intervening 100 meter distance?

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Your solution:

The average speed would be 9.5 meters per second. It should take about 10.5263 seconds

for the runner to go from 100 to 200 meters at that average speed.

confidence rating #$&*: 2

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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.

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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Self-critique (if necessary): 3

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Self-critique Rating: OK

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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:

In this problem, we were given the parts to combine before dividing. In previous

problems, we were given the total to divide; the parts had already been put together.

confidence rating #$&*: 3

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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.

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.

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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Self-critique (if necessary):

I did not state, but did understand that this problem combined rates.

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Self-critique Rating: 3

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Question: `q013. The volume of water in a container increases from 1400 cm^3 to 1600

cm^3 as the depth of the water in the container changes from 10 cm to 14 cm. At what

average rate was the volume changing with respect to depth?

Optional question: What does this rate tell us about the container?

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Your solution:

The change in volume is 1600-1400=200 cubic centimeters

The change in depth is 14-10=4 cm.

The rate of change is 200 cubic centimeters/4 cm, or 50 cubic centimeters per cm.

confidence rating #$&*: 2

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Question: `q014. An athlete's rate of doing work increases more or less steadily from

340 Joules / second to 420 Joules / second during a 6-minute event. How many Joules of

work did she do during this time?

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Your solution:

The average Joules per second is (340+420)/2, or 380 Joules per second.

The number of seconds in 6 minutes is 6*60, or 360.

380 Joules per second times 360 seconds=136,800 Joules of work done during this time.

confidence rating #$&*: 2

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Self-critique Rating: ?Don't know. No answer given! I hope it's right."

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