Asst_2_Query2

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course PHY 201

2/4 515am

002. `ph1 query 2#$&* delim

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Question: Explain how velocity is defined in terms of rates of change.

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Your solution:

Velocity is defined as the rate of change in position with respect to clock time, or change in

clock time.

The formula for this is vAve = 'ds (displacement) / 'dt (change in clock time)

confidence rating #$&*:3

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Given Solution: Average velocity is defined as the average rate of change of position with

respect to clock time.

The average rate of change of A with respect to B is (change in A) / (change in B).

Thus the average rate of change of position with respect to clock time is

ave rate = (change in position) / (change in clock time).

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Self-critique (if necessary):

ok

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Self-critique Rating:

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Question: Why can it not be said that average velocity = position / clock time?

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Your solution:

Average velocity is a rate that is calculated by a change in position divided by a change in

clock time. If I roll a ball down an incline, and started the clock at the time the ball commenced

its travel, the velocity would be less than what I would measure at the finish. Therefore the

average velocity is the average rate the ball moved throughout its travel, and not just at the

start or finish.

confidence rating #$&*:3

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Given Solution: The definition of average rate involves the change in one quantity, and the

change in another.

Both position and clock time are measured with respect to some reference value. For example,

position might be measured relative to the starting line for a race, or it might be measured

relative to the entrance to the stadium. Clock time might be measure relative to the sound of the

starting gun, or it might be measured relative to noon.

So position / clock time might, at some point of a short race, be 500 meters / 4 hours (e.g.,

500 meters from the entrance to the stadium and 4 hours past noon). The quantity (position /

clock time) tells you nothing about the race.

There is a big difference between (position) / (clock time) and (change in position) / (change in

clock time).

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Self-critique (if necessary):

ok

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Self-critique Rating:

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Question: Explain in your own words the process of fitting a straight line to a graph of y vs. x

data, and briefly discuss the nature of the uncertainties encountered in the process. For

example, you might address the question of how two different people, given the same graph,

might obtain different results for the slope and the vertical intercept.

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Your solution:

Using a string or ruler to fit a straight line between different data points is done by estimating a

point toward the extreme left data point and the extreme right data point that when a line is

drawn between the two would best represent all the data points of the graphed data, without

necessarily intersecting any given point (coodinate pair).

The issues that arise with this method are that it is not a precise method of obtaining the best fit

line of the graphed data. An individual that uses this method may come close to a second

person's straight line estimation, but they would most likely guess slightly varying points and as

such would calculating different points from each other. This was shown in the Fitting a Straight

Line lab by the instructor coming up with a line that was often more than .01 units different than

my best guessed line.

confidence rating #$&*:3

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Question:

(Principles of Physics and General College Physics students) What is the range of speeds of a

car whose speedometer has an uncertainty of 5%, if it reads 90 km / hour? What is the range

of speeds in miles / hour?

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Your solution:

An uncertainty of 5% means the speedometer could vary as much as 4.5km/hr; .05 * 90 =

4.5. This means the true speed could be as little as 90km/hr - 4.5km/hr = 85.5km/hr, or as

fast as 90km/hr +4.5km/hr = 94.5km/hr.

To convert to Imperial form, I used the conversion that 1 m = 3.28084 ft. There are 1000

meters in a km, so 1000 * 3.28084 = 3280.84 ft in a km. To find the number of ft in 94.5km,

I multiply 94.5 * 3280.84 = 310039.38 ft in 94.5km. There are 5280 ft in a mile, so

310,039.38 / 5280 = 58.7196 mph. 58.7196mph is the max speed of the the vehicle

travelling 90km/hr with a 5% uncertainty.

confidence rating #$&*:3

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Given Solution: 5% of 90 km / hour is .05 * 90 km / hour = 4.5 km / hour. So the actual

speed of the car might be as low as 90 km / hour - 4.5 km / hour = 85.5 km / hour, or as

great as 90 km / hour + 4.5 km / hour = 94.5 km / hour.

To convert 90 km / hour to miles / hour we use the fact, which you should always know, that 1

inch = 2.54 centimeters. This is easy to remember, and it is sufficient to convert between SI units

and British non-metric units.

Using this fact, we know that 90 km = 90 000 meters, and since 1 meter = 100 centimeter this

can be written as 90 000 * (100 cm) = 9 000 000 cm, or 9 * 10^6 cm.

Now since 1 inch = 2.54 cm, it follows that 1 cm = (1 / 2.54) inches so that 9 000 000 cm = 9

000 000 * (1/2.54) inches, or roughly 3 600 000 inches (it is left to you to provide the

accurate result; as you will see results in given solutions are understood to often be very

approximate, intended as guidelines rather than accurate solutions). In scientific notation, the

calculation would be 9 * 10^6 * (1/2.54) inches = 3.6 * 10^6 inches.

Since there are 12 inches in a foot, an inch is 1/12 foot so our result is now 3 600 000 *(1/12

foot) = 300 000 feet (3.6 * 10^6 * (1/12 foot) = 3 * 10^5 feet).

Since there are 5280 feet in a mile, a foot is 1/5280 mile so our result is 300 000 * (1/5280

mile) = 58 miles, again very approximately.

So 90 km is very roughly 58 miles (remember this is a rough approximation; you should have

found the accurate result).

Now 90 km / hour means 90 km in an hour, and since 90 km is roughly 58 miles our 90

km/hour is about 58 miles / hour.

A more formal way of doing the calculation uses 'conversion factors' rather than common sense.

Common sense can be misleading, and a formal calculation can provide a good check to a

commonsense solution:

We need to go from km to miles. We use the facts that 1 km = 1000 meters, 1 meter = 100

cm, 1 cm = 1 / 2.54 inches, 1 inch = 1/12 foot and 1 foot = 1 / 5280 mile to get the

conversion factors (1000 m / km), (100 cm / m), (1/2.54 in / cm), (1/12 foot / in) and

(1/5280 mile / ft) and string together our calculation:

90 km / hr * (1000 m / km) * (100 cm / m) * (1/2.54 in / cm) * (1/12 foot / in) * (1/5280

mile / ft) = 58 mi / hr (again not totally accurate).

Note how the km divides out in the first multiplication, the m in the second, the cm in the third, the

inches in the fourth, the feet in the fifth, leaving us with miles / hour.

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Self-critique (if necessary):

ok

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Self-critique Rating:

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Question: (Principles of Physics students are invited but not required to submit a solution) Give

your solution to the following: Find the approximate uncertainty in the area of a circle given

that its radius is 2.8 * 10^4 cm.

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Your solution:

The uncertainty of 2.8*10^4cm could be +- .05 of 2.8, so 2.8 + .05 = 2.85 * 10^4cm or 2.8

-.05 = 2.75*10^4cm.

Using these percents, the area could be as little as pi * (2.75 * 10^4)^2 = 7.5*10^8 pi cm^2

or as much as pi * (2.85 * 10^4)^2 = 7.8*10^8 pi cm^2.

confidence rating #$&*:3

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Given Solution:

** Radius 2.8 * 10^4 cm means that the radius is between 2.75 * 10^4 cm and 2.85 * 10^4

cm.

We regard 2.75 *10^4 cm as the lower bound and 2.85 *10^4 cm as the upper bound on

the radius. 2.75 is .05 less than 2.8, and 2.85 is .05 greater than 2.8, so we say that the

actual number is 2.8 +- .05.

Thus we express the actual radius as (2.8 +- .05) * 10^4 cm, and we call .05 * 10^4 cm the

uncertainty in the measurement.

The area of a circle is pi r^2, with which you should be familiar (if for no reason other than that

you used it and wrote it down in the orientation exercises)..

With this uncertainty estimate, we find that the area is between a lower area estimate of pi *

(2.75 * 10^4 cm)^2 = 2.376 * 10^9 cm^2 and and upper area estimate of pi * (2.85 *

10^4 cm)^2 = 2.552 * 10^9 cm^2.

The difference between the lower and upper estimate is .176 * 10^9 cm^2 = 1.76 * 10^8

cm^2.

The area we would get from the given radius is about halfway between these estimates, so the

uncertainty in the area is about half of the difference.

We therefore say that the uncertainty in area is about 1/2 * 1.76 * 10^8 cm^2, or about .88

* 10^8 cm^2.

Note that the .05 * 10^4 cm uncertainty in radius is about 2% of the radius, while the .88 *

10^8 cm uncertainty in area is about 4% of the area.

The area of a circle is proportional to the squared radius.

A small percent uncertainty in the radius gives very nearly double the percent uncertainty in the

squared radius. **

STUDENT COMMENT:

I don't recall seeing any problems like this in any of our readings or assignments to this point

INSTRUCTOR RESPONSE:

The idea of percent uncertainty is presented in Chapter 1 of your text.

The formula for the area of a circle should be familiar.

Of course it isn't a trivial matter to put these ideas together.

STUDENT COMMENT:

I don't understand the solution. How does .176 * 10^9 become 1.76 * 10^8? I understand that

there is a margin of error because of the significant figure difference, but don't see how this was

calculated.

INSTRUCTOR RESPONSE:

.176 = 1.76 * .1, or 1.76 * 10^-1.

So .176 * 10^9 = 1.76 * 10^-1 * 10^9. Since 10^-1 * 10^9 = 10^(9 - 1) =10^8, we have

.176 * 10^9 = 1.76 * 10^8.

The key thing to understand is the first statement of the given solution:

Radius 2.8 * 10^4 cm means that the radius is between 2.75 * 10^4 cm and 2.85 * 10^4 cm.

This is because any number between 2.75 and 2.85 rounds to 2.8. A number which rounds to

2.8 can therefore lie anywhere between 2.75 and 2.85.

The rest of the solution simply calculates the areas corresponding to these lower and upper

bounds on the number 2.8, then calculates the percent difference of the results.

STUDENT COMMENT: I understand how squaring the problem increases uncertainty and I

understand the concept of

a range of uncertainty but I am having trouble figuring out how the range of 2.75 * 10^4 and

2.85*10^4 were established

for the initial uncertainties in radius.

INSTRUCTOR RESPONSE:

The key is the first sentence of the given solution:

'Radius 2.8 * 10^4 cm means that the radius is between 2.75 * 10^4 cm and 2.85 * 10^4 cm.'

You know this because you know that any number which is at least 2.75, and less than 2.85,

rounds to 2.8.

Ignoring the 10^4 for the moment, and concentrating only on the 2.8:

Since the given number is 2.8, with only two significant figures, all you know is that when

rounded to two significant figures the quantity is 2.8. So all you know is that it's between 2.75

and 2.85.

STUDENT QUESTION

I honestly didn't consider the fact of uncertainty at all. I misread the problem and thought I

was simply solving for area. I'm still not really sure how to determine the degree of uncertainty.

INSTRUCTOR RESPONSE

Response to Physics 121 student:

This topic isn't something critical to your success in the course, but the topic will come up. You're

doing excellent work so far, and it might be worth a little time for you to try to reconcile this

idea.

Consider the given solution, the first part of which is repeated below, with some questions

(actually the same question repeated too many times). I'm sure you have limited time so don't

try to answer the question for every statement in the given solution, but try to answer at least a

few. Then submit a copy of this part of the document.

Note that a Physics 201 or 231 student should understand this solution very well, and should

seriously consider submitting the following if unsure. This is an example of how to break down a

solution phrase by phrase and self-critique in the prescribed manner.

##&*

** Radius 2.8 * 10^4 cm means that the radius is between 2.75 * 10^4 cm and 2.85 * 10^4

cm.

Do you understand what this is saying, and why it is so? If not, tell me what you think you

understand, what you are pretty sure you don't understand, and what you think you might

understand but aren't sure.

##&*

We regard 2.75 *10^4 cm as the lower bound and 2.85 *10^4 cm as the upper bound on

the radius. 2.75 is .05 less than 2.8,

and 2.85 is .05 greater than 2.8, so we say that the actual number is 2.8 +- .05.

Do you understand what this is saying, and why it is so? If not, tell me what you think you

understand, what you are pretty sure you don't understand, and what you think you might

understand but aren't sure.

##&*

Thus we express the actual radius as (2.8 +- .05) * 10^4 cm, and we call .05 * 10^4 cm the

uncertainty in the measurement.

Do you understand what this is saying, and why it is so? If not, tell me what you think you

understand, what you are pretty sure you don't understand, and what you think you might

understand but aren't sure.

##&*

The area of a circle is pi r^2, with which you should be familiar (if for no reason other than that

you used it and wrote it

down in the orientation exercises).

With this uncertainty estimate, we find that the area is between a lower area estimate of pi *

(2.75 * 10^4 cm)^2 = 2.376 *

10^9 cm^2 and and upper area estimate of pi * (2.85 * 10^4 cm)^2 = 2.552 * 10^9 cm^2.

Do you understand what this is saying, and why it is so? If not, tell me what you think you

understand, what you are pretty sure you don't understand, and what you think you might

understand but aren't sure.

##&*

The difference between the lower and upper estimate is .176 * 10^9 cm^2 = 1.76 * 10^8

cm^2.

Do you understand what this is saying, and why it is so? If not, tell me what you think you

understand, what you are pretty sure you don't understand, and what you think you might

understand but aren't sure.

##&*

The area we would get from the given radius is about halfway between these estimates, so the

uncertainty in the area is about

half of the difference.

We therefore say that the uncertainty in area is about 1/2 * 1.76 * 10^8 cm^2, or about .88

* 10^8 cm^2.

Note that the .05 * 10^4 cm uncertainty in radius is about 2% of the radius, while the .88 *

10^8 cm uncertainty in area is

about 4% of the area.

Do you understand what this is saying, and why it is so? If not, tell me what you think you

understand, what you are pretty sure you don't understand, and what you think you might

understand but aren't sure.

##&*

The area of a circle is proportional to the squared radius.

A small percent uncertainty in the radius gives very nearly double the percent uncertainty in the

squared radius.

Do you understand what this is saying, and why it is so? If not, tell me what you think you

understand, what you are pretty sure you don't understand, and what you think you might

understand but aren't sure.

##&*

If you wish you can submit the above series of questions in the usual manner.

STUDENT QUESTION

I said the uncertainty was .1, which gives me .1 / 2.8 = .4.

INSTRUCTOR RESPONSE

A measurement of 2.8 can be taken to imply a number between 2.75 and 2.85, which means

that the number is 2.8 +- .05 and the uncertainty is .05. This is the convention used in the given

solution.

(The alternative convention is that 2.8 means a number between 2.7 and 2.9; when in doubt

the alternative convention is usually the better choice. This is the convention used in the text.

It should be easy to adapt the solution given here to the alternative convention, which yields an

uncertainty in area of about 8% as opposed to the 4% obtained here).

Using the latter convention, where the uncertainty is estimated to be .1:

The uncertainty you calculated would indeed be .04 (.1 / 2.8 is .04, not .4), or 4%. However

this would be the percent uncertainty in the radius.

The question asked for the uncertainty in the area. Since the calculation of the area involves

squaring the radius, the percent uncertainty in area is double the percent uncertainty in radius.

This gives us a result of .08 or 8%. The reasons are explained in the given solution.

NOTE FOR UNIVERSITY PHYSICS STUDENTS (calculus-based answer):

Note the following:

A = pi r^2, so the derivative of area with respect to radius is

dA/dr = 2 pi r. The differential is therefore

dA = 2 pi r dr.

Thus an uncertainty `dr in r implies uncertainty

`dA = 2 pi r `dr, so that

`dA / `dr = 2 pi r `dr / (pi r^2) = 2 `dr / r.

`dr / r is the proportional uncertainty in r.

We conclude that the uncertainty in A is 2 `dr / r, i.e., double the uncertainty in r.

STUDENT QUESTION

I looked at this, and not sure if I calculated the uncertainty correctly, as the radius squared

yields double the uncertainty. I know where this is in the textbook, and do ok with uncertainty,

but this one had me confused a bit.

INSTRUCTOR RESPONSE:

In terms of calculus, since you are also enrolled in a second-semester calculus class:

A = pi r^2

The derivative r^2 with respect to r is 2 r, so the derivative of the area with respect to r is dA /

dr = pi * (2 r).

If you change r by a small amount `dr, the change in the area is dA / dr * `dr, i.e., rate of

change of area with respect to r multiplied by the change in r, which is a good commonsense

notion.

Thus the change in the area is pi * (2 r) `dr. As a proportion of the original area this is pi ( 2 r)

`dr / (pi r^2) = 2 `dr / r.

The change in the radius itself was just `dr. As a proportion of the initial radius this is `dr / r.

The proportional change in area is 2 `dr / r, compared to the proportional change in radius `dr

/ r.

That is the proportional change in area is double the proportional change in radius.

STUDENT COMMENT

I used +-.1 instead of using +-.05. I understand why your solution used .05 and will use this

method in the future.

INSTRUCTOR RESPONSE

Either way is OK, depending on your assumptions. When it's possible to assume accurate

rounding, then the given solution works. If you aren't sure the rounding is accurate, the method

you used is appropriate.

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Self-critique (if necessary):

Based on my understanding of uncertainty so far, I used +-.05 as you did, but left my answer

in terms of pi, should I have calculated the problem through, as I am assuming you did? If so

how can I tell what a reliable rounding of pi to be, or rather how many digits after the decimal

do I use for pi? 3.14 or 3.14159?

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Self-critique Rating:

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@&

Your uncertainty and the area of the circle would both be expressed as multiples of pi, so when you divide the uncertainty by the area the pi will divide out. So it makes no difference whether you use multiple-of-pi notation or a numerical approximation.

The approximation you use for pi therefore makes no difference.

The information in this problem was to two significant figures, so a two-significant-figure approximation to pi would be sufficient. However there would be no harm in using the more common three-figure approximation 3.14, and rounding the final result to 2 significant figures. There would be no point to using more than three significant figures with the given information.

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&#This looks good. See my notes. Let me know if you have any questions. &#