ph1 query 3

#$&*

course Phy 201

3/9 at 12:33pm

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.

003. `Query 3

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

Question: What do the coordinates of two points on a graph of position vs. clock time tell you about the motion of the object? What can you reason out once you have these coordinates?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The coordinates of two points on a graph of position vs. clock time tell us the object’s change in position over time, also revealing the object’s average velocity. The rise between two points reveals the change in position while the run reveals the change in time. The slope of the line gives the average velocity for the corresponding time interval.

confidence rating #$&*: 3

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

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

Given Solution: The coordinates a point on the graph include a position and a clock time, which tells you where the object whose motion is represented by the graph is at a given instant. If you have two points on the graph, you know the position and clock time at two instants.

Given two points on a graph you can find the rise between the points and the run.

On a graph of position vs. clock time, the position is on the 'vertical' axis and the clock time on the 'horizontal' axis.

• The rise between two points represents the change in the 'vertical' coordinate, so in this case the rise represents the change in position.

• The run between two points represents the change in the 'horizontal' coordinate, so in this case the run represents the change in clock time.

The slope between two points of a graph is the 'rise' from one point to the other, divided by the 'run' between the same two points.

• The slope of a position vs. clock time graph therefore represents rise / run = (change in position) / (change in clock time).

• By the definition of average velocity as the average rate of change of position with respect to clock time, we see that average velocity is vAve = (change in position) / (change in clock time).

• Thus the slope of the position vs. clock time graph represents the average velocity for the interval between the two graph points.

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

Self-critique (if necessary): OK

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

Self-critique Rating: 3

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

Question:

Pendulums of lengths 20 cm and 25 cm are counted for one minute. The counts are respectively 69 and 61. To how many significant figures do we know the difference between these counts?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

I think we can estimate to one more place. So if it was 69 and a little bit more of a cycle, we could estimate out to 69.3. But that is merely estimation and you are asking for what we KNOW, so I suppose it is what we are given, just significant digits.

confidence rating #$&*:

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

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

Question:

What are some possible units for position? What are some possible units for clock time? What therefore are some possible units for rate of change of position with respect to clock time?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

Possible units for position are cm, m, in, ft, yd, km. Possible units for clock time are m, s, h. So, possible units for rate of change of position are any of the units for position over any of the units of time: cm/h, mph, m/s, ft/m.

confidence rating #$&*:

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

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

Question: `qQuery Principles of Physics and General College Physics: Summarize your solution to Problem 1.19 (1.80 m + 142.5 cm + 5.34 * 10^5 `micro m to appropriate # of significant figures)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

1.80 m +142.5 cm + 5.34 * 10^5 ‘micro m

142.5 cm = 142.4 * 10^-2 m = 1.425 m

5.34 * 10^5 micro m = 5.34 * 10^-1 m = .534 m

So, 1.80 m + 1.425 m + .534 m

= 3.759 m

About 3.76 m (3 sig figs)

confidence rating #$&*: 3

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

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

Given Solution:

`a** 1.80 m has three significant figures (leading zeros don't count, neither to trailing zeros unless there is a decimal point; however zeros which are listed after the decimal point are significant; that's the only way we have of distinguishing, say, 1.80 meter (read to the nearest .01 m, i.e., nearest cm) and 1.000 meter (read to the nearest millimeter).

Therefore no measurement smaller than .01 m can be distinguished.

142.5 cm is 1.425 m, good to within .00001 m.

5.34 * `micro m means 5.34 * 10^-6 m, so 5.34 * 10^5 micro m means (5.34 * 10^5) * 10^-6 meters = 5.34 + 10^-1 meter, or .534 meter, accurate to within .001 m.

Then theses are added you get 3.759 m; however the 1.80 m is only good to within .01 m so the result is 3.76 m. **

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

Self-critique (if necessary): OK

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

Question: For University Physics students: Summarize your solution to Problem 1.31 (10th edition 1.34) (4 km on line then 3.1 km after 45 deg turn by components, verify by scaled sketch).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

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

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

Given Solution:

`a** THE FOLLOWING CORRECT SOLUTION WAS GIVEN BY A STUDENT:

The components of vectors A (2.6km in the y direction) and B (4.0km in the x direction) are known.

We find the components of vector C(of length 3.1km) by using the sin and cos functions.

Cx was 3.1 km * cos(45 deg) = 2.19. Adding the x component of the second vector, 4.0, we get 6.19km.

Cy was 2.19 and i added the 2.6 km y displacement of the first vector to get 4.79.

So Rx = 6.19 km and Ry = 4.79 km.

To get vector R, i used the pythagorean theorem to get the magnitude of vector R, which was sqrt( (6.29 km)^2 + (4.79 km)^2 ) = 7.9 km.

The angle is theta = arctan(Ry / Rx) = arctan(4.79 / 6.19) = 37.7 degrees. **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

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 determine the clock time at each of the three points, if we assume the clock started when the ball was released at the 'top' of the first book?

How would you use your information to sketch a graph of the ball's position vs. clock time?

(This question is more challenging that the others): How would you use your information to sketch a graph of the ball's speed vs. clock time, and how would this graph differ from the graph of the position?

confidence rating #$&*: 1

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

If we know the time intervals from one end of the book to the other, and how far the ball rolled along each book, we are able to find the average velocity over each interval (change in position over change in time). From there, we could estimate how far the ball would travel at clock times that we don’t have measured.

To sketch a graph, clock time would be along the x axis and position would be distributed over the y axis. We could plot the coordinate points on this graph to draw a line.

Speed vs. clock time would have how fast the ball was moving on the y axis and the change in time over the x. The position graph would consistently move in a positive direction, because the ball is always moving farther from the starting point. The speed graph may change because the speed of the ball may be changing over the course of the time.

"

Self-critique (if necessary):

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

Self-critique rating:

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

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 determine the clock time at each of the three points, if we assume the clock started when the ball was released at the 'top' of the first book?

How would you use your information to sketch a graph of the ball's position vs. clock time?

(This question is more challenging that the others): How would you use your information to sketch a graph of the ball's speed vs. clock time, and how would this graph differ from the graph of the position?

confidence rating #$&*: 1

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

If we know the time intervals from one end of the book to the other, and how far the ball rolled along each book, we are able to find the average velocity over each interval (change in position over change in time). From there, we could estimate how far the ball would travel at clock times that we don’t have measured.

To sketch a graph, clock time would be along the x axis and position would be distributed over the y axis. We could plot the coordinate points on this graph to draw a line.

Speed vs. clock time would have how fast the ball was moving on the y axis and the change in time over the x. The position graph would consistently move in a positive direction, because the ball is always moving farther from the starting point. The speed graph may change because the speed of the ball may be changing over the course of the time.

"

Self-critique (if necessary):

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

Self-critique rating:

#*&!

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