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.

002.  `ph1 query 2

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NOTE:  If a question is labeled for your course (e.g., General College Physics or anything containing the abbreviation 'gen', Principle of Physics or anything labeled 'prin', University Physics or anything labeled 'univ') you should answer it.  If a question is not labeled for any course, everyone should do it.  If a question is labeled 'Openstax' you should do it if you are using the Openstax text and it has been assigned for your course.)

If you are in a more advanced course, you can still profit by thinking about the problems posed for less advanced courses and, when it is useful to do so, reading and thinking about the given solutions to those problems.

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

Your solution: 

Confidence Assessment:

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).

Self-critique (if necessary):

Self-critique Rating:

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

Your solution: 

Confidence Assessment:

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).

Self-critique (if necessary):

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.

Your solution: 

Confidence Assessment:

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?

Your solution: 

Confidence Assessment:

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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.

Self-critique (if necessary):

Self-critique Rating:

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Question:  Openstax:  An patient’s pulse rate is measured to be 110 ± 3 beats/min. What is the percent uncertainty in this measurement?

Your solution: 

Confidence Assessment:

Given Solution:  The uncertainty in this measurement is 3, and the measurement is 110.  So we calculate 3 as a percent of 110:

3/110 = .028, which is 2.8%.

However 3 has only one significant figure, while 2.8% has two, so the appropriate result is 3%.

Self-critique (if necessary):

Self-critique Rating:

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Question:  Openstax:  (a) If your speedometer has an uncertainty of 1.5 km/h at a speed of 80 km/h , what is the percent uncertainty?

(b) If it has the samepercent uncertainty when it reads 50 km/h , what is the range of speeds at which you could be moving?

Your solution: 

Confidence Assessment:

Given Solution: 

1.5 km/h is

1.5 km/h / 80 km/h = .019, or about 2%,

of 80 km/h.

So the uncertainty is about 2%.

An uncertainty of 2% at 50 km/h corresponds to

.02 * 50 km/h = 1 km/h.

Self-critique (if necessary):

Self-critique Rating:

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Question:  Openstax:  State how many significant figures are proper in the results of the following calculations:

(a) (106.7)(98.2) / (46.210)(1.01)

(b) (18.7)2

(c) 1.60×10^(-19) * (3712) .

Your solution: 

Confidence Assessment:

Given Solution: 

(106.7)(98.2) / (46.210)(1.01) would be appropriately calculated to three significant figures, since 98.2 and 1.01 each have three significant figures, while the other numbers involved in the calculation each have more than three.  So the number of significant figures in the result would be three.

(18.7)^2 would appropriately be calculated to three significant figures, since 18.7 has three significant figures and its square is just the product ot 18.7 with itself.

1.60×10^(-19) * (3712) would appropriately be calculated to three significant figures, since 1.60 has three significant figures and 3712 has more.

Self-critique (if necessary):

Self-critique Rating:

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Question:  Openstax:  If a marathon runner averages 9.5 mi/h, how long does it take him or her to run a 26.22-mi marathon?

Your solution: 

Confidence Assessment:

Given Solution: 

The calculation is straightforward:  divide the 26.22 mile distance by the 9.5 mi/h speed.  The 9.5 mi/h speed is known to only two significant figures, so only two significant figures are appropriate to the result.  We thus obtain

time required = 26.22 mi / (9.5 mi/h) = 2.8 hr (rounded to 2.8 from the exact result 2.76, which however includes too many significant figures).

Self-critique (if necessary):

Self-critique Rating:

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Question:  (Principles of Physics students are invited but not required to submit a solution; all others should)  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. 

Your solution: 

Confidence Assessment:

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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.</h3>

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.

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

 

`dr / r is the proportional uncertainty in r.

`dA / A is the proportional uncertainty in A.

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.

Self-critique (if necessary):

Self-critique Rating:

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Question:  What is your own height in meters and what is your own mass in kg (if you feel this question is too personal then estimate these quantities for someone you know)? 

Explain how you determined these. 

What are your uncertainty estimates for these quantities, and on what did you base these estimates?

Your solution: 

Confidence Assessment:

#$&*

Given Solution: 

Presumably you know your height in feet and inches, and have an idea of your ideal weight in pounds.  Presumably also, you can convert your height in feet and inches to inches. 

To get your height in meters, you would first convert your height in inches to cm, using the fact that 1 inch = 2.54 cm.  Dividing both sides of 1 in = 2.54 cm by either 1 in or 2.54 cm tells us that 1 = 1 in / 2.54 cm or that 1 = 2.54 cm / 1 in, so any quantity can be multiplied by 1 in / (2.54 cm) or by 2.54 cm / (1 in) without changing its value.

Thus if you multiply your height in inches by 2.54 cm / (1 in), you will get your height in cm.  For example if your height is 69 in, your height in cm will be 69 in * 2.54 cm / (1 in) = 175 in * cm / in.

in * cm / in = (in / in) * cm = 1 * cm = cm, so our calculation comes out 175 cm.

STUDENT SOLUTION

 5 feet times 12 inches in a feet plus six inches = 66 inches. 66inches * 2.54 cm/inch = 168.64 cm. 168.64 cm *
.01m/cm = 1.6764 meters.

INSTRUCTOR COMMENT: 

Good, but note that 66 inches indicates any height between 65.5 and 66.5 inches, with a resulting uncertainty of about .7%.

168.64 implies an uncertainty of about .007%.

It's not possible to increase precision by converting units.

STUDENT SOLUTION AND QUESTIONS

My height in meters is – 5’5” = 65inches* 2.54cm/1in = 165cm*1m/100cm = 1.7m. My weight is 140lbs*
1kg/2.2lbs = 63.6kg. Since 5’5” could be anything between 5’4.5 and 5’5.5, the uncertainty in height is ???? The
uncertainty in weight, since 140 can be between 139.5 and 140.5, is ??????

INSTRUCTOR RESPONSE

Your height would be 5' 5" +- .5"; this is the same as 65" +- .5".

.5" / 65" = .008, approximately, or .8%. So the uncertainty in your height is +-0.5", which is +-0.8%.

Similarly you report a weight of 140 lb +- .5 lb.

.5 lb is .5 lb / (140 lb) = .004, or 0.4%. So the uncertainty is +-0.5 lb, or +- 0.4%.</h3>

STUDENT QUESTION

I am a little confused. In the example from another student her height was 66 inches and you said that her height could be between 65.5 and 66.5 inches. but if you take the difference of those two number you get 1, so why do you divide by .5 when the difference
is 1

INSTRUCTOR RESPONSE

If you regard 66 inches as being a correct roundoff of the height, then the height is between 65.5 inches and 66.5 inches. This makes the height 66 inches, plus or minus .5 inches. This is written as 66 in +- .5 in and the percent uncertainty would be .5 / 66 = .007, about .7%.

If you regard 66 inches having been measured only accurately enough to ensure that the height is between 65 inches and 67 inches, then your result would be 66 in +- 1 in and the percent uncertainty would be 1 / 66 = .015 or about 1.5%.

STUDENT QUESTION

If a doctor were to say his inch marker measured to the nearest 1/4 inch, would that be the uncertainty?

Meaning, would I only have to multiply that by .0254 to find the uncertainty in meters, dividing that by my height to find the percent
uncertainty?

INSTRUCTOR RESPONSE

That's pretty much the case, though you do have to be a little bit careful about how the rounding and the uncertainty articulate.

For example I'm 72 inches tall. That comes out to 182.88 cm. It wouldn't make a lot of sense to say that I'm 182.88 cm tall, +- .64 cm. A number like 182.88 has a ridiculously high number of significant figures.

It wouldn't quite be correct to just round up and say that I'm 183 cm tall +- .64 cm. We might be able to say that I'm 183 cm tall, +- .76 cm, but that .76 cm again implies more precision than is present.

We would probably end up saying that I"m 183 cm tall, +- 1 cm.

Better to overestimate the uncertainty than to underestimate it.

As far as the percent uncertainty goes, we wouldn't need to convert the units at all.  In my case we would just divide 1/4 in. by 72 in., getting about .034 or 3.4%.

Self-critique (if necessary):

Self-critique Rating:

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Question:  A ball rolls from rest down a book, off that book and onto another book, where it picks up additional speed before rolling off the end of that book. 

Suppose you know all the following information:

• How far the ball rolled along each book.

• The time interval required by the ball  to roll from one end of each book to the other.

• How fast the ball is moving at each end of each book.

How would you use your information to calculate the ball's average velocity on each book?

How would you use your information to calculate how quickly the ball's speed was changing on each book?

Your solution: 

Confidence Assessment:

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Question:  Which changes more quickly, the speed of a spacecraft whose speed increases from 18000 mph to 19000 mph in an hour or the speed of a car whose speed increases by 10 mph in 20 seconds?

Compare those results to the similar result for a ball on a ramp whose speed increases by 90 cm/second in 2 seconds.

Put the three results in order from least to greatest.

Your solution: 

Confidence Assessment:

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