Revised Query 1

#$&*

course Phy 121

6/14 at 12:30 PM

Query 1#$&*

course Phy 121

6/12/12 at 12:36 PM

ph1 query 1delim #$&*

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

Question: `qExplain in your own words how the standard deviation of a set of numbers is calculated.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

First, you need to determine the mean. Then, subtract the mean from every number to get the list of deviations. Create a list of these numbers. Next, square the resulting list of numbers. Add up all of the resulting squares to get their total sum. Divide your result by one less than the total number of items in the list. To get the standard deviation, just take the square root of the resulting number

confidence rating #$&*: 3

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

#$&*

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

Question: State the given definition of the average rate of change of A with respect to B.

Briefly state what you think velocity is and how you think it is an example of a rate of change.

In reference to the definition of average rate of change, if you were to apply that definition to get an average velocity, what would you use for the A quantity and what would you use for the B quantity?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Average rate of change of A with respect to B = change in A / change in B. I think that velocity is similar to speed except it involves a difference between two distances in a specific direction in a certain time. This is an example of a rate of change because it provides an average rate of change between positions over changes in a time.

confidence rating #$&*: 3

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

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

Given Solution:

A rate is a change in something divided by a change in something else.

This question concerns velocity, which is the rate of change of position: change in position divided by change in clock time. **

NOTE ON NOTATION

Students often quote a formula like v = d / t. It's best to avoid this formula completely.

The average velocity on an interval is defined as the average rate of change of position with respect to clock time. By the definition of average rate, then, the average velocity on the interval is v_ave = (change in position / change in clock time).

• One reason we might not want to use v = d / t: The symbol d doesn't look like a change in anything, nor does the symbol t. Also it's very to read 'd' and 'distance' rather than 'displacement'.

• Another reason: The symbol v doesn't distinguish between initial velocity, final velocity, average velocity, change in velocity and instantaneous velocity, all of which are important concepts that need to be associated with distinct symbols.

In this course we use `d to stand for the capital Greek symbol Delta, which universally indicates the change in a quantity. If we use d for distance, then the 'change in distance' would be denoted `dd. It's potentially confusing to have two different d's, with two different meanings, in the same expression.

We generally use s or x to stand for position, so `ds or `dx would stand for change in position. Change in clock time would be `dt. Thus

v_Ave = `ds / `dt

(or alternatively, if we use x for position, v_Ave = `dx / `dt).

With this notation we can tell that we are dividing change in position by change in clock time.

For University Physics students (calculus-based note):

If x is the position then velocity is dx/dt, the derivative of position with respect to clock time. This is the limiting value of the rate of change of position with respect to clock time. You need to think in these terms.

v stands for instantaneous velocity. v_Ave stands for the average velocity on an interval.

If you used d for position then you would have the formula v = dd / dt. The dd in the numerator doesn't make a lot of sense; one d indicates the infinitesimal change in the other d.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I did not know the correct notation in order to describe what velocity is. Now I know that v_Ave = `ds / `dt the correct way to define it.

------------------------------------------------

Self-critique rating: 3

#$&*

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

Question: Given average speed and time interval how do you find distance moved?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Distance moved = average speed * time interval

confidence rating #$&*: 3

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

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

Given Solution:

** You multiply average speed * time interval to find distance moved.

For example, 50 miles / hour * 3 hours = 150 miles. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating: OK

#$&*

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

Question: Given average speed and distance moved how do you find the corresponding time interval?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Distance / average speed = time interval

confidence rating #$&*:: 3

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

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

Given Solution:

** time interval = distance / average speed. For example if we travel 100 miles at 50 mph it takes 2 hours--we divide the distance by the speed.

In symbols, if `ds = vAve * `dt then `dt = `ds/vAve.

Also note that (cm/s ) / s = cm/s^2, not sec, whereas cm / (cm/s) = cm * s / cm = s, as appropriate in a calculation of `dt. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating: OK

#$&*

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

Question: Given time interval and distance moved how do you get average speed?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Average speed = distance / time interval

confidence rating #$&*:: 3

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

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

Given Solution:

** Average speed = distance / change in clock time. This is the definition of average speed.

For example if we travel 300 miles in 5 hours we have been traveling at an average speed of 300 miles / 5 hours = 60 miles / hour. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): Instead of using “time interval” I should have been more specific saying “change in clock time”.

------------------------------------------------

Self-critique rating: 3

#$&*

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

Question: A ball rolls from rest down a book, off that book and onto another book, where it picks up speed before rolling off the end of that book. Consider the interval that begins when the ball first encounters the second book, and ends when it rolls of the end of the book.

For this interval, place in order the quantities initial velocity (which we denote v_0), and final velocity (which we denote v_f), average velocity (which we denote v_Ave).

During this interval, the ball's velocity changes. It is possible for the change in its velocity to exceed the three quantities you just listed? Is it possible for all three of these quantities to exceed the change in the ball's velocity? Explain.

Note that the change in the ball's velocity is denoted `dv.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

v_0 < v_Ave < v_f for the first interval. I believe that it is possible for the change in the ball’s velocity to exceed these three quantities. This is because the ball has more speed once it travels to the second book and it still has an incline. For example, with my previous experiment for query 0, we know that during the first interval the v_f is the fastest and it will only become faster as it reaches the continuing ramp.

confidence rating #$&*: 3

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

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

Question: If the velocity at the beginning of an interval is 4 m/s and at the end of the interval it is 10 m/s, then what is the average of these velocities, and what is the change in velocity?

List the four quantities initial velocity, final velocity, average of initial and final velocities, and change in velocity, in order from least to greatest.

Give an example of positive initial and final velocities for which the order of the four quantities would be different.

For positive initial and final velocities, is it possible for the change in velocity to exceed the other three quanities?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Average of these velocities is (4 m/s + 10 m/s ) / 2 = 7 m/s. The change in velocity is 10 m/s - 4 m/s = 6 m/s.

Initial velocity < change in velocity < average of initial and final velocities < final velocity.

This order would be different if the initial velocity was 10 m/s and the final velocity was 4 m/s. In this case, the average of the two would still be 7 m/s; however, the change in velocity would now be -6 m/s. Furthermore the new order would be change in velocity < final velocity < average of initial and final velocities < initial velocity.

For positive initial and final velocities it is possible for the change in velocity to exceed only the initial velocity; not the other two quantities. For example, if the initial velocity was 20 m/s and the final velocity was 50 m/s the change in velocity would be 30 m/s and the average would be 35 m/s. In this situation the order is initial velocity < change in velocity < average of initial and final velocities < final velocity. The change in velocity can only possibly be larger than the initial velocity.

confidence rating #$&*: 3

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

#$&*

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

Question: If the position of an object changes by 5.2 meters, with an uncertainty of +-4%, during a time interval of 1.3 seconds, with an uncertainty of +-2%, then

What is the uncertainty in the change in position in meters?

What is the uncertainty in the time interval in seconds?

What is the average velocity of the object, and what do you think is the uncertainty in the average velocity?

(this last question is required of University Physics students only, but other are welcome to answer): What is the percent uncertainty in the average velocity of the object, and what is the uncertainty as given in units of velocity?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The uncertainty in the change in position is 5.2 m * 0.04 = +-0.208 m.

The uncertainty in the time interval is 1.3 sec * 0.02 = +-0.026 sec.

The average velocity of the object is (5.2 meters / 1.3 seconds) = 4 m/s. The uncertainty in units of velocity is (0.208 m / 0.026 sec) = 8 m/s.

confidence rating #$&*: 3

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

#$&*

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

#$&*

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

Question: Given average speed and distance moved how do you find the corresponding time interval?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Distance / average speed = time interval

confidence rating #$&*:: 3

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

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

Given Solution:

** time interval = distance / average speed. For example if we travel 100 miles at 50 mph it takes 2 hours--we divide the distance by the speed.

In symbols, if `ds = vAve * `dt then `dt = `ds/vAve.

Also note that (cm/s ) / s = cm/s^2, not sec, whereas cm / (cm/s) = cm * s / cm = s, as appropriate in a calculation of `dt. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating: OK

#$&*

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

Question: Given time interval and distance moved how do you get average speed?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Average speed = distance / time interval

confidence rating #$&*:: 3

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

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

Given Solution:

** Average speed = distance / change in clock time. This is the definition of average speed.

For example if we travel 300 miles in 5 hours we have been traveling at an average speed of 300 miles / 5 hours = 60 miles / hour. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): Instead of using “time interval” I should have been more specific saying “change in clock time”.

------------------------------------------------

Self-critique rating: 3

#$&*

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

For this interval, place in order the quantities initial velocity (which we denote v_0), and final velocity (which we denote v_f), average velocity (which we denote v_Ave).

Note that the change in the ball's velocity is denoted `dv.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*: 3

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

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

Question: If the velocity at the beginning of an interval is 4 m/s and at the end of the interval it is 10 m/s, then what is the average of these velocities, and what is the change in velocity?

List the four quantities initial velocity, final velocity, average of initial and final velocities, and change in velocity, in order from least to greatest.

Give an example of positive initial and final velocities for which the order of the four quantities would be different.

For positive initial and final velocities, is it possible for the change in velocity to exceed the other three quanities?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Average of these velocities is (4 m/s + 10 m/s ) / 2 = 7 m/s. The change in velocity is 10 m/s - 4 m/s = 6 m/s.

Initial velocity < change in velocity < average of initial and final velocities < final velocity.

confidence rating #$&*: 3

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

#$&*

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

Question: If the position of an object changes by 5.2 meters, with an uncertainty of +-4%, during a time interval of 1.3 seconds, with an uncertainty of +-2%, then

What is the uncertainty in the change in position in meters?

What is the uncertainty in the time interval in seconds?

What is the average velocity of the object, and what do you think is the uncertainty in the average velocity?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The uncertainty in the change in position is 5.2 m * 0.04 = +-0.208 m.

The uncertainty in the time interval is 1.3 sec * 0.02 = +-0.026 sec.

The average velocity of the object is (5.2 meters / 1.3 seconds) = 4 m/s. The uncertainty in units of velocity is (0.208 m / 0.026 sec) = 8 m/s.

confidence rating #$&*: 3

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

#$&*

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

#*&!

The velocity is not uncertain by 8 m/s. This would mean that the given information would be consistent with any velocity from 4 m/s - 8 m/s = -4 m/s to 4 m/s + 8 m/s = 12 m/s.

The change in position would be between 5.2 m - .208 m and 5.2 m + .208 m; the change in clock time between 1.3 s - .026 s and 1.3 s + .026 s.

Using the minimum change in position and maximum change in clock time we get the minimum average velocity. What is this result?

Using the maximum change in position and maximum change in clock time we get the maximum average velocity. What is this result?

By what percent do these results differ from the 4 m/s result?

&&&&

Using minimum change in position and clock time we can find minimum average velocity. Minimum vAve = (5.2m - .208m) / (1.3s - .026s) = (4.992m) / (1.274s) = 3.92 m / s. Using maximum change in position and clock time we can find maximum average velocity. Maximum vAve = (5.2m + .208m) / (1.3s + .026s) = (5.408m) / (1.326s) = 4.08 m / s. These results differ from the 4 m/s result by +- .08 m/s. The percent that these results differ from the 4 m/s result is (.08 / 4) * 100 = 2%.

&&&&

#$&*

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

Question: Given average speed and distance moved how do you find the corresponding time interval?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Distance / average speed = time interval

confidence rating #$&*:: 3

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

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

Given Solution:

** time interval = distance / average speed. For example if we travel 100 miles at 50 mph it takes 2 hours--we divide the distance by the speed.

In symbols, if `ds = vAve * `dt then `dt = `ds/vAve.

Also note that (cm/s ) / s = cm/s^2, not sec, whereas cm / (cm/s) = cm * s / cm = s, as appropriate in a calculation of `dt. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating: OK

#$&*

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

Question: Given time interval and distance moved how do you get average speed?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Average speed = distance / time interval

confidence rating #$&*:: 3

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

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

Given Solution:

** Average speed = distance / change in clock time. This is the definition of average speed.

For example if we travel 300 miles in 5 hours we have been traveling at an average speed of 300 miles / 5 hours = 60 miles / hour. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): Instead of using “time interval” I should have been more specific saying “change in clock time”.

------------------------------------------------

Self-critique rating: 3

#$&*

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

For this interval, place in order the quantities initial velocity (which we denote v_0), and final velocity (which we denote v_f), average velocity (which we denote v_Ave).

Note that the change in the ball's velocity is denoted `dv.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*: 3

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

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

Question: If the velocity at the beginning of an interval is 4 m/s and at the end of the interval it is 10 m/s, then what is the average of these velocities, and what is the change in velocity?

List the four quantities initial velocity, final velocity, average of initial and final velocities, and change in velocity, in order from least to greatest.

Give an example of positive initial and final velocities for which the order of the four quantities would be different.

For positive initial and final velocities, is it possible for the change in velocity to exceed the other three quanities?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Average of these velocities is (4 m/s + 10 m/s ) / 2 = 7 m/s. The change in velocity is 10 m/s - 4 m/s = 6 m/s.

Initial velocity < change in velocity < average of initial and final velocities < final velocity.

confidence rating #$&*: 3

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

#$&*

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

Question: If the position of an object changes by 5.2 meters, with an uncertainty of +-4%, during a time interval of 1.3 seconds, with an uncertainty of +-2%, then

What is the uncertainty in the change in position in meters?

What is the uncertainty in the time interval in seconds?

What is the average velocity of the object, and what do you think is the uncertainty in the average velocity?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The uncertainty in the change in position is 5.2 m * 0.04 = +-0.208 m.

The uncertainty in the time interval is 1.3 sec * 0.02 = +-0.026 sec.

The average velocity of the object is (5.2 meters / 1.3 seconds) = 4 m/s. The uncertainty in units of velocity is (0.208 m / 0.026 sec) = 8 m/s.

confidence rating #$&*: 3

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

#$&*

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

#*&!

The velocity is not uncertain by 8 m/s. This would mean that the given information would be consistent with any velocity from 4 m/s - 8 m/s = -4 m/s to 4 m/s + 8 m/s = 12 m/s.

The change in position would be between 5.2 m - .208 m and 5.2 m + .208 m; the change in clock time between 1.3 s - .026 s and 1.3 s + .026 s.

Using the minimum change in position and maximum change in clock time we get the minimum average velocity. What is this result?

Using the maximum change in position and maximum change in clock time we get the maximum average velocity. What is this result?

By what percent do these results differ from the 4 m/s result?

#*&!

Only this last question needs to be revised.

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).

Be sure to include the entire document, including my notes.

Revised Query 1

@& self-critique

@&
self-critique