Query 2

course Phy 121

002. `ph1 query 2*********************************************

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Velocity is the change of speed over the change of time

confidence rating #$&* 3

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

.............................................

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):

Not Necessary

&#Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the parts of the given solution on which your solution didn't agree, and if necessary asking specific questions (to which I will respond).

&#

Self-critique rating #$&*3

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Because velocity is the CHANGE in position and clock time, not just the current position

confidence rating #$&*: 3

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

.............................................

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):

Not Necessary

Self-critique rating #$&*3

*********************************************

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

A best fit line does not actually intercept any of the data point, but finds an “average” of the points in line format. Just eyeballing this, different people could get very different result, depending on the position, and slope of their best-fit line.

confidence rating #$&* 3

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

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Height in meters was 1.57 meters and mass in kg was 63.5. Just converted 62” to cm and then to m and converted 140 lbs to kg.

confidence rating #$&*: 3

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

.............................................

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%.

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%.

Self-critique (if necessary):

Not Necessary

Self-critique rating #$&*3

*********************************************

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 the ball requires 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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The average velocity non each book would be calculated by taking the distance traveled (length of each book) and dividing by the amount of time traveled (seconds to reach the end of the book). This would be done for each book. The changing speed could be calculated by finding how quickly the ball was covering the same distance3 ( top of book vs bottom of book) and then comparing times.

confidence rating #$&* 3

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

"

&#Good responses. See my notes and let me know if you have questions. &#

#$&*

Query 2

course Phy 121

002. `ph1 query 2*********************************************

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Velocity is the change of speed over the change of time

confidence rating #$&* 3

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

.............................................

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):

Not Necessary

&#Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the parts of the given solution on which your solution didn't agree, and if necessary asking specific questions (to which I will respond).

&#

Self-critique rating #$&*3

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Because velocity is the CHANGE in position and clock time, not just the current position

confidence rating #$&*: 3

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

.............................................

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):

Not Necessary

Self-critique rating #$&*3

*********************************************

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

A best fit line does not actually intercept any of the data point, but finds an “average” of the points in line format. Just eyeballing this, different people could get very different result, depending on the position, and slope of their best-fit line.

confidence rating #$&* 3

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

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Height in meters was 1.57 meters and mass in kg was 63.5. Just converted 62” to cm and then to m and converted 140 lbs to kg.

confidence rating #$&*: 3

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

.............................................

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%.

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%.

Self-critique (if necessary):

Not Necessary

Self-critique rating #$&*3

*********************************************

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 the ball requires 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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The average velocity non each book would be calculated by taking the distance traveled (length of each book) and dividing by the amount of time traveled (seconds to reach the end of the book). This would be done for each book. The changing speed could be calculated by finding how quickly the ball was covering the same distance3 ( top of book vs bottom of book) and then comparing times.

confidence rating #$&* 3

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

"

&#Good responses. See my notes and let me know if you have questions. &#

#$&*