Query 3

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course Phy 201

6/7 around 7:30

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: 

 If you know the coordinates of two points on a graph you can use this information to find the slope. After you find the slope you know what the average rate of change is since slope is a measurement of the average velocity on a position vs. clock time graph. The coordinates will be able to tell you if the object is accelerating or decelerating as well if the graph is increasing or decreasing at an increasing or decreasing rate.

confidence rating #$&*: 3

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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.

 

 

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Self-critique (if necessary):

 OK

 

 

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Self-critique Rating: 3

 

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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?

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Your Solution: 

 We use 2 significant figures. Therefore, 69-61=8.0

 

confidence rating #$&*:

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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?

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Your Solution: 

 Hertz=frequency of oscillations

 Minutes or seconds just depending

Therefore: Hz/min or hz/sec

confidence rating #$&*:

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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) 

 

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Your solution: 

 1.80 m has three significant figures

142.5 cm is 1.425 m

(5.34 * 10^5) * 10^-6 meters = 5.34 + 10^-1 meter, or .534 meter

1.80 m+1.425 m+.534 m=3.759 m--rounded to 3 sig figs=3.76 m 

 

 

confidence rating #$&*: 3

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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.  **

 

 

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Self-critique (if necessary):

 

 OK

 

 

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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).   

 

 

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Your solution: 

 

 

confidence rating #$&*:

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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. **

 

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Self-critique (if necessary):

 

 

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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 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 #$&*: 3

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I would use the following information to determine the clock time at each of the three points: initial, midpoint, and endpoint.

For example, initial point would read 0 m/0 sec. midpoint : 10 m / 4 s and endpoint: 20 m/ 3 s.

When graphing these results of ball’s speed vs. clock time, as clock time decreased between the midpoint and endpoint, the graph would show an increase in velocity/ball’s speed. It would show a graph of decrease at an increasing rate, as opposed to position vs. clock time, which would be normally increasing at an increasing rate."

Self-critique (if necessary):

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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 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 #$&*: 3

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I would use the following information to determine the clock time at each of the three points: initial, midpoint, and endpoint.

For example, initial point would read 0 m/0 sec. midpoint : 10 m / 4 s and endpoint: 20 m/ 3 s.

When graphing these results of ball’s speed vs. clock time, as clock time decreased between the midpoint and endpoint, the graph would show an increase in velocity/ball’s speed. It would show a graph of decrease at an increasing rate, as opposed to position vs. clock time, which would be normally increasing at an increasing rate."

Self-critique (if necessary):

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Self-critique rating:

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&#Very good responses. Let me know if you have questions. &#