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= Constant Speed =

__CMV Lab 9/6/11__ Objective: What is the speed of a Constant Motion Vehicle (CMV)? Hypothesis: Based on observing the CMV and considering how fast it appears to be moving in relation to the space on the floor, the speed of the CMV is .5m/s. Distance can be measured pretty close to exact by hand, and exact using more accurate devices. A position-time graph tells the relation something is to its starting point at a certain time.

__Position-Time Data for CMV__

Discussion questions 1. Why is the slope of the position-time graph equivalent to average velocity? The slope is equal to the change in the y-axis/the change in the x-axis. The y-axis is position and the x-axis is time. The change in position/the change in time equals velocity. 2. Why is it average velocity and not instantaneous velocity? What assumptions are we making? It is average velocity because it takes into account multiple different measurements rather than just one. We are making the assumption that the velocity changed over time and was not completely constant. 3. Why was it okay to set the y-intercept equal to zero? It was okay to do this because when the time was zero the position was also zero since the time started when the vehicle began to move. 4. What is the meaning of the R2 value? The R2 value is the percentage of data collected that fits the slope. 5. 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? A graph of a slower moving CMV would lie beneath the graph of the one I measured and recorded data for.

Conclusion The slope and velocity of my CMV was 30.025 meters/seconds, and the percentage of data that fit this slope was 99.991%. My hypothesis was inaccurate as I predicted the CMV would move at 50cm/s when it actually moved at 30.025cm/s. Certain errors could have occurred as a result of the fact that we used tools that only went up to the tenth decimal place, and then we had to estimate to the hundredth. Another error could have occurred because the ruler wasn’t flat, making it difficult to line up the measurements with the dots on the strip. We also could have started measuring before the dots evened out to a constant speed. If I were to redo the lab I would suggest a flat measuring tape to reduce the possibility of misalignment of numbers to dots. A longer measuring tool would also be effective in helping to assure the ruler wouldn’t shift causing the results to be altered.

** Summary of Lesson 1 (a,b,c,d) September 8, 2011 **
What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. Distance is the total amount of ground that an object has covered. Displacement is how far an object is from its starting point.

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 not fully aware of the difference between speed and velocity. I now know that speed is a scalar quantity and that it refers to how fast an object moves. On the other hand, velocity is a vector quantity and it refers to the rate at which an object changes its position.

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

What (specifically) did you read that was not gone over during class today? I did not know that scalars and vectors were two categories of the mathematical quantities that are used to describe the motion of an object. Scalars are quantities that are described by magnitude alone, while vectors are described by magnitude as well as direction. We also didn't go over the difference between instantaneous speed and average speed. Instantaneous is the speed at any given instant, while average is the average of all the instantaneous speeds and equals distance/time. Constant speed is if you are going at a constant rate the whole time and your speed is not changing. Constant speed would be the same value as instantaneous speed.

class notes 9/9/11 __Types of Motion__ both increasing and decreasing speed are forms of acceleration (changing speed)
 * at rest
 * constant speed
 * increasing speed
 * decreasing speed

__Motion Diagrams__ --> --> --> a = 0
 * v = 0
 * a = 0

--> ---> > + --> + a

> ---> --> + <-- + a

__Ticker Tape__ at rest = one dot constant = evenly spaced dots increasing speed =increasing distance between dots decreasing speed = decreasing distance between dots

signs are arbitrary and subjective

**Summary of Lesson 2 (a,b,c) September 10, 2011**
What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. I understand that a ticker tape gives a visual representation of the speed of an object by making dots at regular intervals. The farther apart the dots are, the faster the object is moving. The ticker tape diagram also indicates whether an object is speeding up, slowing down, moving at a constant speed, or at rest. I understand that the arrows on a vector diagram change lengths depending on the rate at which an object is moving. A longer arrow indicates speeding up, while a smaller arrow means slowing down. If the arrows remain the same length throughout, the object is moving at a constant rate.

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 unaware that the motion diagram depicted with arrows is called a vector diagram. Now that I know that a vector diagram is the diagram with arrows used to depict speed, I understand it better.

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

What (specifically) did you read that was not gone over during class today? Everything I read was gone over previously.

**Graphical Representation of Equilibrium Lab 9/12/11**
Objectives:
 * What is the difference between static and dynamic equilibrium?
 * How is “at rest” represented on a position vs. time graph? On a velocity vs. time graph? On an acceleration vs. time graph?
 * How is constant speed represented on a position vs. time graph? On a velocity vs. time graph? On an acceleration vs. time graph?
 * How are changes in direction represented on a position vs. time graph? On a velocity vs. time graph? On an acceleration vs. time graph?

constant slow:

constant fast:

change in direction: constant normal:

at rest:

discussion questions:
 * 1) How can you tell that there is no motion on a…
 * 2) position vs. time graph : When this graph is at rest there is a horizontal line with a slope of zero.
 * 3) velocity vs. time graph : When this graph is at rest there is a horizontal line with a slope of zero.
 * 4) acceleration vs. time graph : When this graph is at rest there is a horizontal line with a slope of zero.
 * 5) How can you tell that your motion is steady on a…
 * 6) position vs. time graph : At constant motion this graph will be a diagonal line that rises up. It is linear.
 * 7) velocity vs. time graph : This graph will have a horizontal line slightly higher than the x-axis and will have a slope of 0.
 * 8) acceleration vs. time graph : In an acceleration vs. time graph, constant motion is represented with a line along the x-axis with a slope of 0.
 * 9) How can you tell that your motion is fast vs. slow on a…
 * 10) position vs. time graph : On a position vs. time graph, representing fast motion will be a steeper slope compared to that of slow motion.
 * 11) velocity vs. time graph : On a velocity vs. time graph it will be difficult to see the difference between fast vs. slow motion if the speed is constant because the line is virtually horizontal. The same length line will just get to a lower number faster or slower.
 * 12) acceleration vs. time graph : On an acceleration vs. time graph, if the motion is steady, there is no way to see a difference between fast vs. slow motion. The same length line will just get to a lower number faster or slower.
 * 13) How can you tell that you changed direction on a…
 * 14) position vs. time graph : On this kind of graph, there will be an equal slope rising, and then the same one falling as direction changes.
 * 15) velocity vs. time graph : On a velocity vs. time graph it is difficult to see a change in direction, besides that there might be a small falter in the graph as the rate of motion slightly changes when direction changes.
 * 16) acceleration vs. time graph : On an acceleration vs. time graph it is difficult to see a change in direction, besides that there might be a small falter in the graph as the rate of motion slightly changes when direction changes.
 * 17) What are the advantages of representing motion using a…
 * 18) position vs. time graph : A position vs. time graph is helpful in that it clearly and easily shows where something is in relation to the time.
 * 19) velocity vs. time graph : This graph is helpful in representing a change in speed
 * 20) acceleration vs. time graph : This graph shows small changes in speed.
 * 21) What are the disadvantages of representing motion using a…
 * 22) position vs. time graph : This type of graph is sensitive to slight changes in speed, therefore making falters often if the speed does not remain constant. It is also difficult to clearly depict a change in speed.
 * 23) velocity vs. time graph : Because of the fact that if acceleration is not taking place the slope is zero, this graph is not too helpful when there is no change in speed.
 * 24) acceleration vs. time graph : There is not much of a change unless acceleration occurs.
 * 25) Define the following:
 * 26) No motion : No motion is when something is at rest. There is a slope of zero.
 * 27) Constant speed : Constant speed is when something is moving at a steady rate. The intervals are equal and no acceleration is taking place.

**9/13/11 Kinematics Class Notes**




**Summary of Lesson 1e 9/13/11**
What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. Acceleration is defined as the rate at which an object changes its velocity. This is a vector quantity. The direction of acceleration depends on: whether the object is speeding up or slowing down, and whether the object is moving in the + or - direction. An object that has a negative acceleration is moving to the left or down. An object that has a positive acceleration is moving to the right or up. Negative and positive accelerations do not necessarily refer to negative and positive values.

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 total distance traveled is directly proportional to the sqare of the time. For example, if an object travels for three times the time, then it will cover nine times the distance; meaning that the distance traveled after three seconds is nine times the distance traveled after one second.

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

What (specifically) did you read that was not gone over during class today? Free-falling objects are constantly accelerating.

**Acceleration Graph Lab 9/14/11**
lab partner: George Souflis **Data and Graphs:**
 * Objectives:**
 * What does a position-time graph for increasing speeds look like?
 * What information can be found from the graph?
 * Hypothesis**: A position time graph will look like half of a parabola-like curved shape. We can tell the position the object is at a specific time, and how fast the acceleration increases or decreases.
 * Procedure**: Place the track on a textbook to create an incline. Run spark tape through spark timer and then attach to end of dynamics cart. Turn on spark timer and allow cart to move down incline. Measure distance between dots on spark timer. Create graph to interpre t results. To find a decrease in acceleration up the incline, set up the track and spark tape and timer the same way, but start cart at the bottom of the incline and push gently up track.



a) Interpret the equation of the line (slope, y-intercept) and the R2 value. The equation of the increasing line is y=10.398x2 + 2.7483x, and the R2 value is .99995. Because of the curve in the graph, we used polynomial trendline rather than a linear one, which gave us an R2 value of only .89863. The equation of the decreasing line is y=-16.515x2 + 64.35x, and the R2 value is .99956. Again we used a polynomial trendline because a linear one was less accurate with an R2 value of only .97211. b) Find the instantaneous speed at halfway point and at the end. (You may find this easier to do on a printed copy of the graph. Just remember to take a snapshot of it and upload to wiki when you are done.) Increasing acceleration graph: - halfway instantaneous speed = 8.82-0/.8-4.2 = 8.82/.38 = 23.21cm/s - end point instantaneous speed = 30.98-0/1.6-.82 = 30.98/.78 = 39.71cm/s decreasing acceleration graph: - halfway instantaneous speed = 32.82-6/.6-0 = 26.82/.6 = 44.7cm/s - end point instantaneous speed = 53.14-23/1.2-0 = 30.14/1.2 = 25.15cm/s c) Find the average speed for the entire trip. Increasing acceleration: ∆d/∆t = 30.98cm/1.6s = 19.3625m/s Decreasing acceleration: ∆d/∆t = 53.14/1.2 = 44.28m/s 1. What would your graph look like if the incline had been steeper? If the incline had been steeper, the graph would have been more curved. The curve would have been higher up, as the speed would have accelerated faster. 2. What would your graph look like if the cart had been decreasing up the incline? The graph of the cart decreasing up the incline curves the opposite way. The points start far apart and then get closer together as the acceleration decreases. There is a negative slope of this graph. 3. Compare the instantaneous speed at the halfway point with the average speed of the entire trip. For the accelerating down the incline, the instantaneous speed, 23.21cm/s, is greater than the average speed of 19.3625cm/s. This is true because as the cart moved down it gained momentum, but was not at its full speed at the halfway point, so there were still smaller values before it. For the accelerating up the incline, the instantaneous speed, 44.7cm/s, is slightly greater than the average speed of 44.28cm/s. This is true because around the halfway point, the cart began slowing down, causing the values after it to pull the average down. 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 because you want to find the slope of a particular point, and drawing a tangent line gives the most accurate representation of the slope. 5. Draw a v-t graph of the motion of the cart. Be as quantitative as possible.
 * Analysis:**
 * Discussion Questions:**


 * Conclusion**:

My hypothesis was accurate. I said that the shape would be a parabola-like curve, which it was. As the acceleration either decreased or increased, the shaped became steeper and more curved. There could have been a few sources of error contributing to inaccuracies. When comparing to other groups, if one incline was slightly steeper than the other, the results would have been different. Additionally, when measuring the distance between the dots on the spark tape, if one did not begin at the correct point, the results could have been thrown off. If I were to repeat this lab, I would make sure the track was as steep as everyone else’s for better comparative results.

**Summary of Lesson 3 9/15/11**
What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. The shape and slope of the lines on a position vs. time graph demonstrates specific features of the motion of objects. A small slope means a small velocity; a negative slope means a negative velocity; a constant slope (straight line) means a constant velocity; a changing slope (curved line) means a changing velocity.

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. On a position-time graph, whatever characteristics the velocity has, the slope will exhibit the same (and vice versa). The slope of the line on a position vs. time graph is equal to the velocity of the object

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

What (specifically) did you read that was not gone over during class today? Everything we read was gone over previously.

**Summary of Lesson 4 9/15/11**
What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. Positive velocity results in a line of zero slope. When there is a positive velocity, only the positive values are graphed. If acceleration is zero, slope is zero. If acceleration is positive, slope is positive. If acceleration is negative, slope is negative.

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 actual slope value of any straight line on a velocity-time graph is the acceleration of the graph. I now understand this better than I did before.

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

What (specifically) did you read that was not gone over during class today? Everything was already gone over in class.

A Crash Course in Velocity Part II Lab, 9/21/11
lab partners: George Souflis, Maddy Weinfeld, Ben Weiss



media type="file" key="New Project 3 - Mobile.m4v" width="270" height="270"
 * Part A Procedure:**

media type="file" key="New Project 4 - Mobile 1.m4v" width="270" height="270"
 * Part B Procedure:**

**Interpreting Your Results:**


 * Part A: Crashing**








 * Part B: **


 * [[image:formula_for_part_B.png width="344" height="195"]]﻿ **





**Discussion questions** If the speeds of the car were exactly equal and they are traveling in opposite directions, they would meet at 300 cm. Each car would travel half way, meeting in the center. If the cars are traveling in the same direction, they will never meet.
 * 1) Where would the cars meet if their speeds were exactly equal?
 * 1) 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.



Since v-t graphs do not show a change in direction, there is no way to tell when the two vehicles are at the same place at the same time.
 * 1) 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?

In both parts of the lab, the results we obtained were very close to what we thought they would theoretically be. For part A, we calculated that CMV 1, which has a velocity of 68.628 cm/s, would travel 417.26 cm to the right, and that CMV 2, which has a velocity of 30.025 would travel 182.55 cm to the left. They would then hit after 6.08 seconds. After performing the lab and gathering data, we averaged the trials together because they were so close, and discovered that we had a 0.274% error. The average of the percent difference is 0.534%. The trials for part A were very close to our theoretical calculation as well as to each other. For part B, we calculated that CMV 1 would travel 177.75 before catching and passing CMV 2, taking 2.59 seconds to do so. After gathering data, we averaged the trials together because they were so close, and figured out that we had a 1.63% error. The average percent difference is 1.61%. Again, while they were slightly more off, the results we obtained were still very close to what we had initially calculated. In this lab, some errors could have occurred, altering our data. One issue could have been that the both vehicles did not start at the exact same time. In addition, it was difficult to read the exact values due to the fact that the CMVs curved at some points. If we were to redo this lab, we could try to be even more careful in putting the vehicles down to start at the same time. We also did not use the tracks for part A, something that could be done and would potentially make a big difference.
 * Conclusion: **

**Interpreting Position Time Graphs**






Egg Drop Lab 10/1/11
partner: Caroline Braunstein

For our final model, we folded aluminum foil into a box shape. We then filled it with shredded newspaper as a way to cushion the fall. We attached a parachute made of paper with the strings to the top.
 * Picture:**

Our egg drop was unsuccessful. The egg, which was wrapped in newspaper, fell out of its carrier prior to hitting the floor, causing the egg to crack while wrapped up. The parachute, while successful in slowing the acceleration down, was unable to keep the box it was attached to upright.
 * Description of Results**


 * Calculations**



Before the final dropping, we had tested our idea out multiple times. For the most part, it did work. Because of this we decided to concentrate on making it weigh less, which led to us making less of a cover for it. This caused the egg to fall out and ultimately crack. Our design was an aluminum foil box, stuffed with rubber bands and shredded newspaper to lessen the impact. We also attached a parachute made of paper. One reason it was so unsteady was that the parachute was the same size as the box itself. It was unable to keep something the same size upright easily, and therefore, even if the box was dropped even slightly unbalanced, a flip could easily occur.
 * Analysis of Why it Didn't Work**

There are a couple of options on how we could redo this project in the future. If we were to stick with the box idea, we should have made it smaller and deeper, with a parachute that was clearly larger than the actual box. The shredded newspaper would have been more concentrated, making it deeper and more of a cushion for the egg. The parachute would also have had an easier time keeping the box upright. We also could have added a thin lid covering the entire box, which would have prevented the egg from falling out. We also could have tried to use less of each material to make everything lighter. If we were to go a different route, we could make a cone shape, which seemed to work for other groups. This would allow the egg some room to continue moving, even after the cone itself hit the ground.
 * What We Would Do Differently**

**Summary of Lesson 5: Free Fall and the Acceleration of Gravity 10/3/11**
A free falling object is an object that is falling under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. important motion characteristics that are true of free-falling objects: Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s, a ticker tape or dot diagram of its motion would depict an acceleration. acceleration of gravity - the acceleration for any object moving under the sole influence of gravity - the symbol g.
 * Free-falling objects do not encounter air resistance.
 * All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s (often approximated as 10 m/s/s for //back-of-the-envelope// calculations)
 * **Time (s)** || **Velocity (m/s)** ||
 * 0 || 0 ||
 * 1 || - 9.8 ||
 * 2 || - 19.6 ||
 * 3 || - 29.4 ||
 * 4 || - 39.2 ||
 * 5 || - 49.0 ||

A position versus time graph for a free-falling object is shown below.

A velocity versus time graph for a free-falling object is shown below.

vf = g * t

Free-fall is the motion of objects that move under the sole influence of gravity; free-falling objects do not encounter air resistance. More massive objects will only fall faster if there is an appreciable amount of air resistance present. the acceleration of an object is directly proportional to force and inversely proportional to mass. Increasing force tends to increase acceleration while increasing mass tends to decrease acceleration. the greater force on more massive objects is offset by the inverse influence of greater mass. all objects free fall at the same rate of acceleration, regardless of their mass.

Falling Body Lab 10/5/11
**Objective:** What is the acceleration of a falling body? **Hypothesis:** The acceleration of a falling body is 9.81 m/s/s. The v-t graph will have a negative slope and be below the origin. It will be linear from from left to right. Acceleration can be found from this graph by finding the instantaneous speed at any mid-time.

**Analysis:**

Data Tables:

Graphs:

In the equation for the velocity-time graph, the number 856.73 should equal 981, the value of acceleration due to gravity. The number -60.463 should equal 0, as the initial velocity should be 0. The R^2 value shows that this data is 98.882% of the data fits the linear slope. In the position-time graph, (429.87)2 should equal 981. Similarly to the first, 62.19 should equal zero for the same reasons. For this graph, 99.965% of the data fits the slope.





**Discussion Questions** 1. Does the shape of your v-t graph agree with the expected graph? Why or why not? I had expected the shape to be the opposite of what it is. I thought the slope would be negative. The reason it is not what I had anticipated is because we used positive values, rather than negative, even though we were dropping it down. I still thought it would be linear, because it would be constantly accelerating, just going the opposite way. 2. Does the shape of your x-t graph agree with the expected graph? Why or why not? The shape of this graph is also the opposite of what I thought it to be. Once again, I thought the slope would be negative, but is positive because we started it at a positive position even though it was falling. It is a curved shape as I had thought it would be. 3. How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.) Our results are 2.06% different than the class average. This is pretty close, meaning that the value that we obtained is very close to the mean. 4. Did the object accelerate uniformly? How do you know? The object did accelerate uniformly for the most part. This is evident in the r2 value. For the x-t graph, this value is .99965 and for the v-t graph this value is .98882. There was one point in the middle that slightly did not seem to follow the trendline, but even that point was not too far off. Also, just by looking at the graphs, it is evident that the linear shape for the v-t graph and the slight curve of the x-t graph indicate that the object accelerated uniformly. 5. What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be? Acceleration due to gravity could be higher than it should be if the data was read, interpreted, or gathered incorrectly. It could also be higher than it should be if there was some force pulling or pushing it down. One factor that would cause acceleration due to gravity to be lower than it should be would be the friction caused by both the spark tape rubbing against the railing, and that it had to pass through the spark timer. Failing to drop the object vertical to the ground is another source of unwanted friction.


 * Conclusion: **

The result of my experiment is that acceleration of a falling body due to gravity is 856.73cm/s/s. This result has a percent error of 12.66%, as the actual acceleration should be 981cm/s/s While this is not too far off, it is not accurate. It is clear that there were certain sources of error. Friction plays a large role in why the result was off. The spark tape rubbing against the railing, as well as needing to pass through the spark timer created another force acting upon the tape, skewing the results. Error could also have occurred in measuring the position of the points. It is difficult to get very precise measurements, and it becomes even more of a challenge when dealing with small dots and spark tape and measuring tape that both move very easily. Along the way, one of the two of these, or both, could have moved without us realizing it. This could explain why one of our points was further off than any other. No matter how careful one tries to be, it is difficult to keep everything in place when collecting the data. If I were to redo this lab, I would make sure the spark tape didn't touch the railing on its way down. I would also make sure the tape stayed as stable as possible when measuring the distances.