Ch2_ChungA

=** Labs **=

toc ** Lab: A Crash Course in Velocity (Part 1) **
Partner: Amanda Fava


 * Objective: ** What is the speed of a Constant Motion Vehicle (CMV)?

Constant Motion Vehicle, Tape measure and/or metersticks, spark timer and spark tape
 * Available Materials ** :


 * Interpreting Your Results: **
 * Be sure to provide concrete evidence for the success of your experiment.
 * Any graphs that you make should be done in Excel, not by hand.
 * You need to collect at least 10 data points.


 * Discussion questions / Analysis **
 * 1) Why is the slope of the position-time graph equivalent to average velocity?
 * 2) Because the slope of the position-time graph is represented by the equation Delta Y over Delta X, which is the change in position over the change in time (if you look at the graph). If you divide change in position by change in time you get average velocity as a result.
 * 3) Why is it average velocity and not instantaneous velocity? What assumptions are we making?
 * 4) The slope of the position-time graph would be average velocity and not instantaneous velocity. This is because instantaneous velocity is the velocity at a specific moment. Average velocity; on the other hand, take into account the entire run. Since we are measuring time and distance, the slope of the position-time graph would be equivalent to average velocity.
 * 5) Why was it okay to set the y-intercept equal to zero?
 * 6) Because the y-intercept is the starting position. Since we are starting at 0 cm, setting the y-intercept to anything other than zero would not make sense.
 * 7) What is the meaning of the R2 value?
 * 8) The R2 value basically gives you a percentage of how good the fit is on a graph of data. The R2 value on our graph is .99912, which signifies that the line connecting our data points fits very well.
 * 9) If you were to add the graph of another CMV that moved more slowly on the same axes as your current graph, how would you expect it to lie relative to yours?
 * 10) The graph of the slow CMV would lie below the graph I have. This is because while time would remain constant, the position would grow at a smaller pace. The graph of the slow CMV would also be more horizontal than the current graph.


 * Hypothesis **


 * 1) How fast does a CMV move?
 * 2) A CMV possibly moves at 1 to 2 miles per hour. I can walk faster than a CMV and I walk at around 5 miles per hour. Thus, the speed of the CMV would be less than my walking speed.
 * 3) How precisely can distance be measured?
 * 4) Distance can be measured up to a centimeter.
 * 5) What does a position-time graph tell you?
 * 6) A position-time graph graphs time and position. Time is represented on the X Axis while Position is represented on the Y Axis. It can tell us how far something moves from a starting point relative to time.


 * Data **

Position Time Data Table:


 * Conclusion **

The results of our lab were fairly successful. The data table and graph show that the slope (average velocity) of the CMV was 64.629 centimeters per second. In the hypothesis, I stated that I believed that the CMV moved at around a mile per hour. The result of converting one mile per hour to centimeters per second yielded the result that one-mile per hour is the same as 44.704 centimeters per second. 2 miles per hour is 89.408 centimeters per second. Thus, we can safely say that the CMV moved faster than one mile per hour but less than two miles per hour, proving our first hypothesis. In the second hypothesis, I stated that distance could be measured up to a centimeter. This proved to be false. The meter sticks that were used in this experiment measured things up to the tenth of a centimeter. By correcting for significant figures, we could actually measure things up to a hundredth of a centimeter (the hundredth place being the estimation sig-fig). As for the third hypothesis, our idea of a position-time graph was accurate. It indeed did tell us how far something moved relative to time This experiment was not entirely error proof. For example, the ruler could have shifted causing an inaccurate measure of the distance between the dots on the spark tape. The spark timer could have also failed to indicate dots correctly, which would throw off the entire experiment. Also, if the CMV were moving slower or faster than normal, due to a variety of reasons, the experiment would have been inaccurate. If I had to redo this lab, I would use other measurement devices than a meter stick. A meter stick relies too heavily on human involvement, which is sure to complicate things. Measuring tape would have been very useful to use as it is flat, thus allowing for a more accurate reading of measurement. It is also to tape down measuring tape which would eliminate the problem of a shifting meter stick. In terms of the CMV, I would compare the speed of the CMV my group used to other CMVs, to make sure that it is not faster or slower than normal.

Graphical Representations of Equilibrium
Graphs:


 * This graph shows the velocity, position, and acceleration graphs of an object at rest in front of a motion detector. As such, there is no change in the graph.**


 * This graph shows the position, velocity, and acceleration of an object moving at constant speed. The run labeled #5 is the run in which our object moved away from the detector. Run #6 is the run in which our object moved closer to the motion detector. The graph shows that an object's graph while moving at constant motion is fairly symmetric.**




 * The below graphs show the position, velocity, and acceleration of an object moving at a faster constant speed. Run #1 is moving away from the motion detector while run #2 is moving towards it. The position graph for the faster speed rises and falls at a steeper level than the graph above.**




 * This graph shows the position, velocity, and acceleration of an object moving at slow constant speed.The position graph is not as steep as the two graphs above. Run #3 is the run in which our object moved away from the motion detector and run #4 is the run in which our object moved towards the motion detector.**




 * Questions**

1. How can you tell that there is no motion on a
 * Position Time Graph
 * The Graph will simply be a straight line
 * Velocity vs Time Graph
 * The Graph will simply be a straight line
 * Acceleration vs Time Graph
 * The graph will simply be a straight line

2. How can you tell that your motion is steady on a
 * Position vs Time Graph
 * The graph will have a constant slope
 * Velocity vs Time Graph
 * There will be a horizontal line that is not x = 0
 * Acceleration vs Time Graph
 * There will be a horizontal line that is not x = 0

3. How can you tell that your motion is fast vs. slow on a
 * Position vs Time Graph
 * There will be a steeper slope
 * Velocity vs Time Graph
 * The slope will still be greater
 * Acceleration vs Time Graph
 * Will be 0. Acceleration refers to the rate at which something goes faster. This does not happen with constant speed.

4. How can you tell that an object changed direction on a
 * Position vs Time Graph
 * The slope of the graph will change. A positive slope indicates that the object is moving away from the motion detector while a negative slope indicates the opposite.
 * Velocity vs Time Graph
 * Velocity is arbitrarily designed a positive and negative value. Thus, if the graph dips below or above x =0 the object is changing direction.
 * Acceleration vs Time Graph
 * This cannot be done. An object with a positive velocity can still have a negative acceleration. It's very hard if not impossible to tell.

5. What are the advantages of representing motion using a
 * Position vs Time Graph
 * You can see where something is at a certain time. This graph is also an accurate representation of speed. The steeper the slope, the greater the speed.
 * Velocity vs Time Graph
 * We can know the rate at which the position is changing.
 * Acceleration vs Time Graph
 * This graph very accurately shows the acceleration of an object at a particular time. We can see when exactly something speeds up or slows down.

6. What are the disadvantages of representing motion using a
 * Position vs Time Graph
 * We cannot tell the direction in which something is moving.
 * Velocity vs Time Graph
 * We cannot tell the direction in which something is moving.
 * Acceleration vs Time Graph
 * This graph does not show whether something is speeding up or slowing down. Speeding up or Slowing Down happens when, respectively, velocity and acceleration have the same sign and when they do not have the same sign. As velocity is not part of this graph, you cannot see when this is happening.

7. Define the following
 * No Motion - this object is simply at rest. It's acceleration and velocity are equal to zero.
 * Constant speed - this means that an object moves at a steady rate. Since it isn't increasing it's speed, acceleration doesn't change, which means that it is zero.

**LAB: Acceleration Graphs**
Partner: Amanda Fava

__Objectives__:
 * What does a position time graph for increasing speed look like?
 * What information can be found from the graph?

__Available Materials:__
 * Spark tape, spark timer, track, dynamics cart, ruler/meter stick/ measuring tape

__Hypothesis:__
 * The slope will get steeper as the cart gains speed
 * We can learn position, velocity, and acceleration from the graph as well as direction.

__Procedure:__
 * 1) Place the track on the textbook to create a ramp.
 * 2) Place the spark timer on the end of the ramp that is higher up.
 * 3) Thread the spark tape through the spark timer and attach it to the cart.
 * 4) Place the cart on top of the ramp.
 * 5) Turn on the spark timer.
 * 6) Let the cart roll down the ramp.
 * 7) Turn off the spark timer.
 * 8) Pull out the spark tape and measure the distance between each dot.


 * 1) Place the track on the textbook to create a ramp.
 * 2) Place the spark timer on the end of the ramp that is lower.
 * 3) Thread the spark tape through the spark timer and attach it to the cart.
 * 4) Place the cart on the bottom of the ramp.
 * 5) Turn on the spark timer.
 * 6) Give the cart a push to get it going up the ramp.
 * 7) Have someone catch the cart before it rolls back down.
 * 8) Turn off the spark timer.
 * 9) Pull out the spark tape and measure the distance between each dot.

__Data:__





__Graph:__



__Analysis:__


 * 1) There are several things about the graph above that need to be mentioned. Each equation above has both an A value and a B value. The A value of the equations is equal to twice the acceleration. Furthermore, B is the initial velocity. This equation helps us a great deal with graph analysis. For instance, since B is supposed to be the initial velocity it should be as close to zero as possible. A nonzero B value indicates that measurement might have gone awry. The R2 Value for the cart moving down the ramp is .99996 which indicates that the line of best fit we used is accurate. Similarly, the R2 value for the cart going up the ramp is .9982 which also indicates that the line of best fit we used is accurate. We chose to use a polynomial line because it yielded better results than the other lines. Since the A value is equal to 1/2 of the acceleration, we know that the acceleration for the cart moving down the ramp is 21.46 cm/s2. Subsequently, the acceleration for the cart moving up the ramp is 3.9822 cm/s2. The Y intercept would be zero when the time is equal to zero.

2.





To solve Analysis #2, I found the derivative of the original equation. Plugging in the time values for the halfway point and the end point into the derivatives subsequently gave me the instantaneous speed at those points. The instantaneous speeds for going down the ramp are 27.14 cm/s (halfway) and 40.014 (endpoint). The instantaneous speeds for going up the ramp are 40.70 cm/s (halfway) and 38.91 (endpoint).

3.



Down the Ramp = 27.138 cm/s Up the Ramp = 40.69 cm/s

__Discussion Questions:__

1. What would your graph look like if the incline had been steeper?
 * If the incline had been steeper, the slope would be steeper. This is because the cart would accelerate quicker causing the position to be greater within each time interval. This would result in the graph being very steep.

2. What would your graph look like if the cart had been decreasing up the incline?
 * We actually completed this experiment. As shown in our graph above, the graph starts out normally but stops increasing in position near the end of the run. This results in the graph getting flat at the top.

3. Compare the instantaneous speed at the halfway point with the average speed for the entire trip.
 * Down the Ramp - The instantaneous speed for the halfway point and the average speed for the entire trip are the same. The instantaneous speed is 27.138 cm/s and the average speed is also 27.138 cm/s.
 * Up the Ramp - The instantaneous speed is 40.70 cm/s while the average speed is 40.69 cm/s. These values are very close to each other as well.
 * It appears that the average speed is equal to roughly the middle of the graph. We know this because the instantaneous speed at the halfway point is in the middle of the graph and the average speed and instantaneous speed are close (or equal to) each other.

4. Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense?
 * The instantaneous speed is the slope of the tangent line. This can be best explained using calculus. Basically, the slope of the tangent line is the derivative of the graph. Derivatives measure rate of change. So, the slope of the tangent line at a particular point (the derivative at that point) would be the rate of change. Instantaneous speed is defined as the speed at that point. Thus, these values are the same. It is also worth noting that velocity is the derivative of a position graph. And since speed is the absolute value of velocity we can find the instantaneous speed by finding the derivative at a particular point (which ends up being the slope of the tangent line) on the position graph.

5. Draw a v-t graph of the motion of the cart. Be as quantitative as possible.



The velocity time graph constantly increases. This is because the experiment measures a cart that is increasing in speed. I graphed this easily by simply plugging in time values into the derivative function of the position graph. This works because the derivative of a position graph is velocity.

__Conclusion:__

The results of our lab proved our hypothesis. Our graphs show that as the cart increased in speed, the position-time graph got steeper. The reverse is also true. As the cart slowed down, the position-time graph got less steep. A visual representation of this idea can be found above. If you look at the graph with the blue line, it is clear that the graph is getting steeper (look at time = 0 to time = .8 and from time = .8 to time = 1.2. The latter is clearly more steep.) We also found that the instantaneous speed at the halfway point is equal to the average speed for the entire trip. The R2 value for the cart moving down the ramp is .99996 while the R2 value for the cart moving up the ramp is .9982. We used a polynomial line for the best fit. Since the R2 values are so close to one, we knew that choosing a polynomial line was the line of best fit. We also found that the velocity-time graph constantly increases. This makes perfect sense seeing as how the cart increased in speed as time went on. Our equations had high B values because we started measuring a bit later than we were supposed to. There are several sources of error for this experiment. For instance, the ramp could have been steeper from person to person. The spark timer also could have failed to mark dots properly. If I had to redo the lab, I would request that each ramp (from group to group) be set at the same level of inclination. I would also measure each dot with different measuring tools to ensure results that are consistent with each other. I would also redo the experiment with a different spark timer and compare results to see if one was particularly faulty.

Lab: A Crash Course in Velocity (Part 2) __- 9/21/11__

 * Lab Partners: Sammy Caspert, Amanda Fava, and John Chiavelli**


 * __Objectives:__**




 * __Procedure:__**

Crash:

media type="file" key="good vid crash (iPhone & iPod).m4v" width="300" height="300"

Catch Up:

media type="file" key="good catch up (iPhone & iPod).m4v" width="300" height="300"


 * __Sample Calculations:__**

A) My calculations show that the two CMV's will crash after 6.4 seconds (see picture below). If you look at the graph on top, after 6.4 seconds the slow CMV will have moved 186.374 centimeters while the fast CMV will have moved 413.626 centimeters. These values can be found by simply plugging in 6.4 seconds into the equations as such:

64.629 cm/s * (6.4 s) = 413.6256

And x can be found easily by doing:

600 - x = 413.6256 x = 186.374

CRASH PICTURE:

B) My calculations show that the x value is 81.7 cm. If you look at the graph, you will see that you must add 100 cm to this value which will give you the spot at which the fast CMV passes the slow CMV.

CATCHING UP PICTURE:




 * __Data Tables:__**





__**Analysis:**__

__Percent Error:__

The experimental values used in the calculations shown below are simply the averages of the data shown in the tables above. In the picture below, A is crash and B is catching up.



The percent errors for both of the experiments are very low. In A, the percent error of 2.39% indicates that our experiment ran fairly true to our hypothesis. The percent error of 0 in B shows that the experiment ran EXACTLY as predicted.

__Percent Difference:__

Crash:



Catching Up:




 * __Discussion Questions:__**

1. Where would the cars meet if their speeds were exactly equal?

The cars would meet at the 300 cm mark. This is rather intuitive and doesn't require much calculation to understand. If the speeds of both CMVs are exactly equal and they are 600 centimeters apart, logic dictates that they will meet in the middle. Since the middle of 600 is 300 (600/2), that is where they will meet.

2. Sketch position-time graphs to represent the catching up and crashing situations. Show the point where they are at the same place at the same time.

A)



B)



3. Sketch velocity-time graphs to represent the catching up situation. Is there any way to find the points when they are at the same place at the same time?



The velocity of the blue CMV is 64.629 cm/s.

The velocity of the yellow CMV is 29.06 cm/s.

There is no way to find the point when they are in the same place at the same time. A velocity-time graph only shows the velocity of each vehicle. Since these are constant motion vehicles, the velocities will remain steady throughout the experiment. This is why the graph above has two horizontal lines. Unfortunately, we cannot find out when the blue CMV passes the yellow (slower) one. This is because a V-T graph doesn't show position, it only shows how fast each vehicle is moving.


 * __Conclusion:__**

The results of our experiment were very successful. We calculated (in our hypothesis) that the two CMV's would crash at 186.374 centimeters. The trials that we conducted were very similar to our estimated value, so close in fact that we calculated a percent error of only 2.39%. To calculate the percent error, we simply used the percent error formula, plugging in the average of all the data values for the crash lab which came out to 181.914 centimeters. The results of our catching-up lab were even more successful. In our hypothesis, we said that the faster CMV would catch up with the slower one at 181.7 centimeters. The average of our trials for this part of the lab came out to 181.7 centimeters as well. Thus, our percent error was 0%. Percent difference was easy to calculate as well. We took the average of all the pieces of data and used that for the percent difference formula as shown in the pictures above. It is worth noting that the faster CMV (the blue one) moved at a speed of 64.629 cm/s while the slower one (the yellow one) moved at a speed of 29.06 cm/s.

This experiment has some possible sources of error. For instance, if you had to replace the batteries in the CMV it could move faster than normal, which would change the data. The CMVs also didn't move in a straight line. They would start turning which resulted in many complications. Due to the fact that the CMVs kept turning, we had to use the wall to steady the CMVs, which probably slowed down the CMVs due to friction.

If I had to redo this lab and change it to address the errors, I would use CMVs that didn't turn while they were moving. This would allow us to get a more accurate time for the crash and the catch-up. I would also run the experiment several times to determine whether the batteries were having a significant impact on the CMVs. I would also see if there were a way to measure more accurately the distances traveled by the CMVs, perhaps by attaching some small computerized gadget that measures distance traveled.

Egg Drop Project


Our project was rather simple. To start, we created a fairly long cone. The both of us had also read that if you fold up paper many times, it becomes almost impossible to tear or break. Thus, we folded up 4 separate pieces of paper into extremely compact little squares that we attached to the cone. We also had a cover inside of the cone that we placed over the egg (once the egg was inside the cone). Finally, we threaded several pieces of string through the cone and attached it to tin foil, which acted like a parachute. The parachute worked well because enough air was able to lift it.


 * Calculations**




 * Results**

We honestly couldn't have asked for more perfect results. The parachute worked exactly like we designed it to. Furthermore, there was no damage whatsoever to the egg. As the calculations show, our egg dropped extremely slowly due to the effective use of both the cone and the parachute. Because the cone doesn't take up much space, enough air is able to hit the parachute making it effective. The tip of the cone cushioned the fall and the egg stayed inside due to a combination of the cone's structure and our cover.


 * Conclusion**

Our egg drop project went as well as I could have hoped for. The parachute worked perfectly and our use of the cone turned out to be the right choice. If I were to go back and change the project, I would probably increase the size of the parachute. I might have also removed the folded up squares. While they appeared to have helped, I'm sure the egg would have survived without them. If we removed them completely, our cone would be very light. As for sources of error, if the wind was blowing really hard it could have caused our project to hit against the wall or fly really far out, which would have damaged our results.

Lab: Free Fall
Partner: Amanda Fava

__Purpose:__

In this experiment, my partner and I are attempting to find the acceleration of a falling body due to gravity.

__Hypothesis with Rationale:__

I believe that the acceleration of the falling body will be close to 9.8 m/s/s. This number is known as the acceleration due to gravity (g), and is the acceleration of any object falling in a vacuum. Analysis of the ticker tape should prove that the falling body did indeed accelerate and fall at a rate of 9.8 m/s/s.

__Data:__



__Graphs:__

__Analysis:__

The velocity-time graph equation is y=851.07x+2.525. This equation is based off of the Big 5 equation: Vf = Vi + at. Thus, 851.07 would be the acceleration, x would be the time, and 2.525 would be the initial velocity. The last piece of data raises a few questions. If it is supposed to be the initial velocity, why isn't it at zero? The reason that the value is a non-zero number is because the timer could have placed a dot on the ticker tape while it was already moving (due to the dropping of the weight). The acceleration also raises a few questions. If it is supposed to be acceleration, and acceleration is 9.81 m/s/s or 981 cm/s/s, then why is the value 851.07? This is simply due to the fact that the way a ticker tape works relies on an indicator that causes friction. This friction slows down the rate at which the object's acceleration is recorded. Thus, the value would be smaller than the expected value. My partner and I chose to use a linear trend line that resulted in a R^2 value of .97937, a value that clearly indicates that we could have had better data.

The position-time graph equation is y = 418.61x^2 + 1.6964x + 1.5025. This equation is based off of the Big 5 equation: y = 1/2 (at^2) + (Vi*t). Thus, we could find the acceleration by multiplying 418.61 by 2 which would equal 837.22 cm/s/s. This raises a few question. Why is the acceleration here 837.22 while the acceleration above is 851.07? The answer is pretty simple. On the velocity-time graph, we don't use a Y-intercept of zero. However, on the position-time graph, we do. This changes the value that we get for acceleration. Our initial velocity in this equation is 1.6964. While it should be zero, it is not due to the same reasons the initial velocity isn't zero in the paragraph above. My partner and I chose to use a polynomial trend line for this graph and we got a R^2 value of .99939, which shows that it is a great fit.

__% Error, % Difference:__





__Discussion Questions:__

//1. Does the shape of your v-t graph agree with the expected graph? Why or why not?//

Yes, to a degree. Coming into this experiment, I expected that the v-t graph would be a straight diagonal line. This is because I know that the acceleration of an object in free fall is 9.8 m/s/s. It makes sense that the graph would be a straight diagonal line. However, due to several sources of error present in this experiment, the graph wasn't as steep as I would have liked. This is because the slope was around 8.51 cm/s/s when it should have been 9.81 cm/s/s.

//2. Does the shape of your x-t graph agree with the expected graph? Why or why not?//

Yes, the shape of my x-t graph agrees with the expected graph. Since the acceleration is constant, the object should steadily be increasing with speed. This is what my graph showed. As time went on, the object covered a greater distance. The increase in distance was fluid and smooth because the acceleration is constant. This is why the graph is curved uwpards.

//3. How do your results compare to that of the class?//

My results were very similar to the rest of the class. I know this because when I calculated the percent difference, I found it to be a very small value (.26%). I knew that it would be small going in because the class average was 839.417 cm/s/s and my value was 837.22 cm/s/s which is pretty close to the class value. Thus, I knew that my values and results were very close to what my classmates got.

//4. Did the object accelerate uniformly? How do you know?//

In an ideal world, the object would have accelerated uniformly at a rate of 9 m/s/s. Ours, however, did not. I know this because the trend line for velocity-time doesn't match up with the last point. This is due to a multitude of different factors that could have messed with the way the spark timer was recording data.

//5. What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?//

There are many factors that could cause acceleration due to gravity to be higher or lower than it should be. If it were lower, it could be because friction exists between the spark timer and the ticker tape, which slows down the rate at which the spark timer makes dots. The spark timer could have also created a dot prematurely. The acceleration could be higher if the person dropping the weight put some force behind it by accident.

__Conclusion:__

Our hypothesis turned out to be correct. The acceleration due to gravity that we calculated was 837.22 cm/s/s which is 8.37 m/s/s. In my hypothesis, I stated that the value of G would be close to 9.8 m/s/s. This is true as shown by the data we collected.

We had a 14.66% error, which I find to be fairly high. The error most likely occurred in the collection of the data. For example, the spark timer probably created friction between itself and the ticker tape. Thus, the dots that were placed on the tape were created too slowly, causing the acceleration due to gravity to be a smaller value than it should be. It is also very possible that the measuring tape moved during measurement. This would cause incorrect data to be calculated for the acceleration due to gravity.

There are several ways in which I could remove the sources of error in our experiment. For one, I would make sure that the measuring tape is perfectly steady and that it cannot move even a millimeter during the course of measurement. I would also see if there exists a spark timer that reduces the friction between itself and the ticker tape. Doing this would yield more accurate values.

=Homework=


 * 9/8/11 - Reading at the Physics Classroom: 1D Kinematics, Lesson 1 (Method 2a) **


 * 1.What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.**

I felt that I truly understood the ideas behind displacement and distance. During class, Ms. Burns demonstrated that while you can travel a fair distance, your displacement can still be zero. She did this by moving back and forth across the classroom. This demonstration really cemented the idea of displacement and distance in my mind. I thus felt very comfortable reading about it on The Physics Classroom.


 * 2.What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.**

During class, we hadn't really talked about the difference between scalar and vector values pertaining to displacement and distance. Although I did learn about scalars and vectors in Honors Precalculus last year, we never really talked about what it means in terms of displacement. The Physics Classroom did a wonderful job of explaining the ideas and concepts to me. Now it seems very simple. Distance is a scalar. It's value is independent of direction. This differs from displacement, which is a vector, and thus depends on direction. These concepts also pertain to the idea of speed and velocity. Speed is a scalar because it isn't affected by direction. Velocity, on the other hand, IS a vector because it totally depends on direction. Velocity is measured by dividing displacement over time and, since displacement requires direction, is a vector.


 * 3.What (specifically) did you read that you still don’t understand? Please word these in the form of a question.**

I understood everything from the reading.


 * 4.What (specifically) did you read that was not gone over during class today?**

Two things that I read on The Physics Classroom that were not discussed in class today were scalars and vectors, and how they relate to displacement, distance, speed, and velocity.


 * 9/9/11 - Reading at the Physics Classroom: 1D Kinematics, Lesson 2 (Method 2a) **


 * 1. What (specifically) did your read that you already understood well from our class discussion? Describe at least 2 items fully.**

I understood the concept of a ticker tape very well from our class discussion. In short, a ticker tape places a dot on a piece of tape at regular intervals. The distance between the dots change if the speed of the thing the tape is attached to changes, ie. there is a change in acceleration. This concept was clear to me after the constant motion vehicle experiment. I also understood vector diagrams quite well from our last class discussion. A vector has both a direction and a magnitude. A vector diagram is a visual representation of a vector using an arrow (that could have varying lengths). I also understand vectors quite well because we discussed them at length in my honors precalculus course last year.


 * 2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.**

The idea of velocity and acceleration has been annoying me for a while now. We are going over the exact same thing in my calculus class and I could not for the life of me understand how something could speed up or slow down based on how velocity and acceleration were acting. From this lesson I now understand a bit more. Velocity is simply the act of moving towards a positive or negative side (arbitrarily designated of course). Acceleration (positive or negative as well) is simply the rate at which velocity happens. In calculus terms, acceleration is the slope (since slope measures rate of change) of velocity, also known as the derivative of velocity. I now feel as if I completely understand how position, acceleration, and velocity are interconnected.


 * 3. What (specifically) did you read that you still don't understand? Please word these in the form of a question.**

I understood everything from the reading.


 * 4. What (specifically) did you read that was not gone over during class today?**

Everything that I read was gone over in class.

**9/13/11 - Reading at the Physics Classroom: 1D Kinematics, Lesson 3 (Method 2a)**


 * 1. What (specifically) did your read that you already understood well from our class discussion? Describe at least 2 items fully.**

I already understood that acceleration was the change in the rate of velocity. I learned this fact both in my physics and calculus classes. In calculus, we learned a more mathematical approach and found that acceleration was the derivative of velocity. I also understood the fact that acceleration can be zero despite the fact that the object can be moving. This concept is tied back in with the first concept I mentioned. If something is moving fast at a constant rate (velocity is staying the same), the rate of change of velocity is not changing. This means that acceleration is also not changing. I also understood the positive and negative directional assignments from class. In class, we learned that these positions are arbitrarily defined but are still helpful when defining a vector quantity.


 * 2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.**

I was a bit confused with the ideas behind velocity and acceleration. What I mean is this. I was confused with the idea that you could be moving forward but have a negative acceleration. This idea was cleared up for me by the reading. Come to think about it, it makes perfect sense. For example, if a car is slowing down at a light, it is technically still moving forward while its acceleration is negative. The result is that the object slows down. I am really enjoying the fact that physics concepts are so easily applicable to every day instances like the one I just mentioned. When compared to my previous courses in science, I feel like I am really learning the rules of how the world works.


 * 3. What (specifically) did you read that you still don't understand? Please word these in the form of a question.**

I understood everything.


 * 4. What (specifically) did you read that was not gone over during class today?**

We did not go over free falling objects. For example, we didn't go over the idea that you could square the time and multiply that by the distance to get the total distance traveled.

**9/15/11**- Reading at the Physics Classroom: 1D Kinematics, Lesson 4 (Method 2a)


 * 1. What (specifically) did your read that you already understood well from our class discussion? Describe at least 2 items fully.**

I already understood that the graph of an object moving at constant speed would be a straight diagonal line. I learned this from the various laboratory experiments that we completed as a class. I also understood that an object increasing in velocity would curve upwards. I learned this from the accelerating cart lab that we completed.


 * 2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.**

I had trouble understanding the graphical representation of what negative velocity looked like. I had assumed that if something was negative it had to drop below the X axis. This turned out to not be the case. It is now clear to me that if a position-time graph displays a negative slope, you know that the velocity is negative. The idea seems very simple when you think about it. However, for some reason I just could not wrap my head around the idea. The graphs that the reading used to back up its points cemented the idea in my head. I now have a greater understanding of how velocity is related to slope.


 * 3. What (specifically) did you read that you still don't understand? Please word these in the form of a question.**

I understood everything.


 * 4. What (specifically) did you read that was not gone over during class today?**

I suppose the idea that velocity is the slope of a position-time graph was not gone over during class today. I did, however, have experience with this fact through my calculus course.


 * DESCRIBING MOTION WITH VELOCITY VS TIME GRAPHS **


 * 1. What (specifically) did your read that you already understood well from our class discussion? Describe at least 2 items fully.**

I already understood that velocity with zero acceleration would be a horizontal line. This makes perfect sense when you think about it. Since acceleration can be found by the slope of velocity, if there is no acceleration the slope would be zero. A line with a slope of zero is horizontal. I also understood the idea that something with positive velocity would be on the positive side of a graph. Similarly, something with negative velocity would be on the negative side of a graph. These values are arbitrary and serve to denote direction.


 * 2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.**

I was a little shaky on the idea of how speeding up and slowing down could be represented on a graph of velocity. From the various labs that we did, I knew how the position vs time graph of something speeding up should look. I found myself wondering, however, how the same thing could be represented on velocity. The reading cleared this entire thing up for me. Apparently, if the magnitude of a velocity graph increases, the object is speeding up. Similarly, if the magnitude of a velocity graph decreases, the object is slowing down.


 * 3. What (specifically) did you read that you still don't understand? Please word these in the form of a question.**

I understood everything.


 * 4. What (specifically) did you read that was not gone over during class today?**

We did not go over the idea that acceleration was the slope of velocity. We also did not go over the fact that the area bound by the line and axis of a velocity-time graph is also the total displacement of the object. More specifically, we did not go over the various areas that could be bound by the line and axis (triangle, rectangle, trapezoid).

**1D Kinematics - Lesson 5, Summarizing Method #1**

__Introduction__

Free fall occurs when only the force of gravity acts upon an object

Objects in free fall do not encounter air resistance. They also fall at a rate of 9.8/ms^2.

You can use a ticker tape diagram because objects in free fall fall at a constant rate.

__The Acceleration of Gravity__

The acceleration of gravity is simply 9.8m/s^2 which is sometimes rounded to 10m/s^2. The symbol "g" is often used for this number.

__Representing Free Fall__



This position time graph shows that the object in free fall starts slow and speeds up in a downward fashion.



When you have a diagonal line, you know that the object is accelerating. In this case, the object is accelerating downward and reaching a velocity of -9.8 m/s/s.

__How Fast? And How Far?__

Vf = g * t

This makes sense because the velocity is the same value (9.8 m/s/s) as it is a constant and it changes by that rate every second. We can thus multiply g by t to get the total velocity.

__The Big Misconception__

Heavier things fall quicker than lighter things ONLY IF YOU FACTOR IN AIR RESISTANCE.

In free fall, things accelerate downward at an equal rate, 9.8 m/s/s.

=**Class Notes**=

Constant Speed Notes
Types of Motion
 * At rest
 * Constant speed
 * Increasing speed (acceleration)
 * Decreasing speed (acceleration)

Motion Diagrams
 * At rest
 * V= 0
 * A=0
 * Constant Speed
 * V constant length
 * A = 0
 * Increasing speed
 * V arrows get bigger
 * Acceleration is positive
 * Decreasing speed
 * V arrows start big and get smaller
 * Acceleration is opposite direction (negative)

Motion Diagrams and Ticker Tape Diagrams
Ticker Tape
 * At Rest - single dot
 * Constant speed - even distance between each dot
 * increasing speed - more space between each dot
 * decreasing speed - less space between each dot
 * These are good because they have numerical values and are very exact
 * Unfortunately, you cannot see direction
 * A ticker tape is generally used when you need to get exact measurements

Signs are arbitrary

**Graph Shapes**
At Rest
 * Position Graph
 * Straight line
 * Velocity Graph
 * Straight line at zero
 * Acceleration Graph
 * Straight line at zero

Constant Speed Graph
 * Position
 * Would be a linear line with a positive or negative slope
 * Velocity
 * Would be a horizontal line at any y value
 * Acceleration
 * Horizontal line at zero

The Big 5
BIG 5:



Increasing and Decreasing Speed Graphs
Increasing Speed Position-Time Graph:

Increasing Speed Velocity-Time Graph:

Increasing Speed Acceleration-Time Graph:



Decreasing Speed Position-Time Graph:

Decreasing Speed Velocity-Time Graph:

Decreasing Speed Acceleration-Time Graph:



Freefall
Only gravity acts upon an object falling during free fall - NO OTHER FORCES!!

This is important because in free fall, you would ignore things such as air resistance











=Quantitative Graph Interpretation=