Ch2_RubensteinD

=__ Constant Speed __=

toc __**Class Notes 9/6/11**__ __ **Homework 9/8/11 - Kinematics, Lesson 1** __


 * After reading the material, answer the following questions:**
 * 1) **What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.** One thing I already understood from class while reading these sections was the difference between distance and displacement. Distance is how much ground one covered during its motion, while displacement shows how far an object is from its starting location. I understood that there was a difference between distance and displacement and that displacement was direction-aware while distance was not. Additionally, I already had a good understanding that speed did not require a direction like velocity does. I knew that speed could be determined without knowing what direction the object was going in and that it can constantly be changing while the object is in motion.
 * 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.** After class, I was slightly confused about the difference between speed and velocity. However, after reading, I am now more clear on on the topic. I did not fully understand that the two had such a significant difference, but now I know that velocity is a vector quality and requires direction while speed is a scalar quality that does not.
 * 3) **What (specifically) did you read that you still don’t understand? Please word these in the form of a question.** Why is the speed formula the same as I thought the formula was for velocity?
 * 4) **What (specifically) did you read that was not gone over during class today?** When I first began to read, I did not know what the terms scalar and vector were. At the time, I did not realize that they had to do with what I already knew about distance and displacement. However, now I understand them both and the difference between them. Scalars describe the motion of objects by solely using a numerical value, while vectors require both a numerical value and a direction. Scalars correspond with distance and speed, while vectors correspond with displacement and velocity.

__ **Lab: Constant Speed of CMV** __
**Date: 9/7/11** **Partner: Julia Sellman** **Objective- What is the speed of a CMV?** The speed of a CMV is 61.663 cm/s
 * Hypothesis-** I predict that the speed of the CMV is going to be 40 m/s because that is what I observed when watching one without calculating it.

__**Position Time Data for CMV**__ This graph of the CMV shows that it is moving at constant speed because of the slope of the line. Because it is straight instead of curved, it shows that there is no acceleration. Additionally, our R squared value tells us that our results were accurate and precise, as .993 is very close to 1.
 * Analysis:**

**Discussion questions** - The slope of the position-time graph and the average velocity are equivalent because the slope was created to take the data we put in and make a line to show where the average velocity should have been according to our information. - It is average velocity and not instantaneous velocity because we measured distance traveled in a given period of time rather than looking at how fast the CMV was going at a particular instance in time. We are assuming that it moved at a constant speed the entire time. - We were able to set the y-intercept equal to zero because we knew that the starting point was at zero. If it was possible that the starting point could have been less than zero, we would not have been able to do that. - The R2 value evaluates how good the data is. It shows a percentage of how close to accurate the results of the lab were. - I would expect it to lie below where my data is, because it would take more time for it to get as far as mine did. This means that the other CMV would lie below mine and would be more flat than mine is.
 * 1. Why is the slope of the position-time graph equivalent to average velocity?**
 * 2. Why is it average velocity and not instantaneous velocity? What assumptions are we making?**
 * 3. Why was it okay to set the y-intercept equal to zero?**
 * 4. What is the meaning of the R2 value?**
 * 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?**

**Conclusion** Our results showed that the average velocity of the blue car was 61.663 cm/s. My hypothesis was somewhat accurate, as after only watching the car I estimated that it would be around 50 cm/s. Although at 1.0 seconds we found that it was at 63.92 cm/s, which could have been due to different errors throughout the process. For example, the meterstick was a few centimeters thick on the sides, making it hard to see if it was exactly lined up with the dots. Also, we had to estimate the last digit of our measurements so that could have caused our results to be slightly off. To minimize these issues if I had to redo this lab, I would have requested to use measuring tape because it is flat and longer, making it easy to get accurate results. Additionally, I would have tried to find a measuring tool that was more exact, which would have allowed me to find the exact last digit instead of guessing.

- Constant speed means that it is not changing, speed is always the same value - Instantaneous is different at all different intervals - Equation is always the same for average speed, constant speed, and instantaneous speed - V = change in d / change in t
 * __ Notes 9/9/11 - Motion Diagrams and Ticker Tape Diagrams __**

1. at rest 2. at constant speed 3. increasing speed- acceleration 4. decreasing speed- acceleration
 * Types of Motion**
 * acceleration is whenever you are changing your speed

- Show direction of velocity - At rest, v = 0 and a = 0 - If the arrows are the same size, then it is at constant speed - If the arrows get bigger, then it is increasing its speed - Acceleration is pointed the same way as the arrows - If the arrows start bigger and get smaller, than it is decreasing its speed - Acceleration is pointed opposite ways of the arrows - If you are moving to the left then the direction if the arrows will change - Quick and easy, used during problem solving to talk about direction
 * Motion Diagrams**



- At rest, one dot - Constant speed, evenly spaced dots - Increasing speed, dots get farther apart - Decreasing speed, dots get closer together - Gernerally used when you need to take measurements
 * Ticker Tape Diagrams**



Signs are arbitrary and subjective

__ **Homework 9/9/11 - Kinematics Lesson 2** __


 * After reading the material, answer the following questions:**
 * 1) **What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.** Before I read, I already understood the ticker tape diagram. I knew that it created different dots on the paper to represent the speed of an object. Also, I understood that dots the same distance apart meant it was going at a constant speed, while dots increasing or decreasing length meant that it was either increasing its speed or decreasing it. Additionally, before the reading I was able to look at a ticker tape diagram and understand what it was showing about the motion of the object. Another topic I understand before reading was that vector diagrams use arrows to show the direction an object is going. Just like a vector quality and velocity, vector diagrams require a 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.** The reading helped clarify that vector diagrams related directly to vector qualities and velocity. Because we used a different name for it in class, I was not fully aware of the connection. However, I now know that vector diagrams show direction, which is required for velocity.
 * 3) **What (specifically) did you read that you still don’t understand? Please word these in the form of a question.** I now fully understand both the vector and ticker tape diagrams and therefore have no questions regarding the subject.
 * 4) **What (specifically) did you read that was not gone over during class today?** After class, I did not know that vector diagrams were also going to be used to show force and momentum. Also, I did not know exactly how the ticker tape diagram worked but now I know how it transfers how fast an object is going into different dots on the tape.

**__Activity: Graphical Representations of Equilibrium 9/12/11__**


 * At Rest-** these graphs show us when we were at rest, standing directly in front of the motion sensor


 * Constant Speed-** these graphs show us walking at constant speed towards and away from the motion sensor
 * Change in Direction-** these graphs show our change in direction in front of the motion sensor


 * Constant Speed: Fast-** these graphs also show us walking at constant speed, however, during these trials we were moving at a faster rate


 * Discussion questions**
 * 1. How can you tell that there is no motion on a...**
 * - Position vs. time graph:** it is in a straight, flat line along the x-axis
 * - Velocity vs. time graph:** also is a straight, flat line along the x-axis
 * - Acceleration vs. time graph:** also a straight, flat line along the x-axis


 * 2. How can you tell your motion is steady on a...**
 * - Position vs. time graph:** the line is linear, however, it is not flat and either starts high and gets lower or starts low and gets high
 * - Velocity vs. time graph:** the line is a straight, horizontal line along the x-axis
 * - Acceleration vs. time graph:** the line is also a straight, horizontal line along the x-axis


 * 3. How can you tell that your motion is fast vs. slow on a...**
 * - Position vs. time graph:** if the line is steeper, that means you are going faster; if it is less steep, then you are going slower
 * - Velocity vs. time graph:** if the motion was faster, the y value would be greater because you would be moving faster
 * - Acceleration vs. time graph:** if the motion was faster, the y value would also be greater because you would be moving faster and getting to a certain position in less amount of time


 * 4. How can you tell that you changed direction on a...**
 * - Position vs. time graph:** slope will change from negative to positive varying on whether you are going forwards or backwards
 * - Velocity vs. time graph:** the line goes above and below zero depending on which direction and how fast you are going
 * - Acceleration vs. time graph:** acceleration does not detect direction


 * 5. What are the advantages of representing motion using a...**
 * - Position vs. time graph:** shows the rate of how fast an object is moving at while at a certain position
 * - Velocity vs. time graph:** shows the increase and/or decrease in velocity, can get the most information because you can also figure out acceleration and displacement
 * - Acceleration vs. time graph:** shows the increase and/or decrease in velocity


 * 6. What are the disadvantages of representing motion using a...**
 * - Position vs. time graph:** changes in speed are not represented by this type of graph
 * - Velocity vs. time graph:** if the object goes back to its original point, there would be no velocity but this graph des not show that
 * - Acceleration vs. time graph:** does not show anything beside the change in speed while in motion


 * 7. Define the following**
 * - No motion:** an object is at rest, so the position, velocity, and acceleration all remain at zero
 * - Constant speed:** an object is changing its position at a constant rate, however, the acceleration and velocity both remain 0 because there is no change

**__Class Notes 9/13/11 - Graph Shapes (At Rest and Constant Speed)__**



**__Class Notes- Kinematics: The Big 5 Notes__** acceleration- rate that velocity is changing (a) m/s ^ 2 a = (vf-vi) / t 1/2(vi + vf) = change in d / change in t vi = initial velocity vf = final velocity vf = vi + a(t) **- has no change in d** change in d = 1/2(vi + vf)t **- has no a** change in d = vi(t) + 1/2(a)(t^2) **- has no vf** vf^2 = vi^2 + 2(a)(change in d) **- has no t**
 * v = d/t ONLY for average or constant speeds
 * v = (vi + vf) / 2 ONLY for average speed


 * __ Class Notes- Increasing and Decreasing Speed Graphs __**
 * [[image:Screen_shot_2011-09-13_at_9.32.51_AM.png width="480" height="439"]]'**

=**__ Acceleration __**=


 * __ Homework 9/13/11 - Lesson 1E Kinematics __**


 * After reading the material, answer the following questions:**
 * 1) **What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.** I already knew that acceleration was the rate at which an object changed it velocity. From class, I understood that acceleration did not simply mean an object was speeding up. If the acceleration is higher, that does not necessarily mean the object is moving faster. It simply is saying that the rate at which an object is changing its velocity is high. Additionally, I also knew that if an object has a constant velocity it is not accelerating, because the velocity is not changing. An object only has an acceleration if there is a change in the rate at which the velocity is changing.
 * 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.** Before reading, I understood that acceleration had to do somewhat with direction, however, I was not completely sure why. Now I know that the direction of the acceleration vector depends on whether the object is speeding up or slowing down and whether the object is moving in the positive or negative direction. I also now know the "rule of thumb", which is that if an object is slowing down, then its acceleration is in the opposite direction of its motion.
 * 3) **What (specifically) did you read that you still don’t understand? Please word these in the form of a question.** Where do the acceleration units come from? How do you understand and use these units?
 * 4) **What (specifically) did you read that was not gone over during class today?** While reading, I learned about free-falling objects and we did not speak about that in class today. Free-falling objects constantly accelerate and cover different distances in each second it is falling. Also, the total distance traveled is directly proportional to the square of the time.

**__ Lab: Acceleration on an Incline __**

 * Date: 9/14/11**
 * Partner:** Julia Sellman
 * Objectives:** What does a position-time graph for increasing speeds look like? What information cnan be foiund fromt he graph?
 * Available Materials:** Spark tape, spark timer, track, dynamics cart, measuring tape


 * Hypothesis:** Due to the increasing acceleration, the graph would gradually get steeper. We will be able to tell what position the cart was at during a certain time.

media type="file" key="Movie on 2011-09-14 at 09.19.mov" width="300" height="300" This video is a demonstration of the procedure we used while completing this lab. We used the ticker tape and the spark timer to calculate the distance and speed of the cart while it was going down the ramp. After this, we recorded the data and made a graph to organize our results.
 * Procedure:**


 * Data:**


 * Graph:**

Our equation of the line was y = 13.25x^2 + 8.10x. This implies that 8.10 is the initial velocity and 13.25 multiplied by 2 is the acceleration because the "A" column is equal to 1/2 acceleration. Our r squared value was originally .96 or 96%, however, we were using a linear trend line at first. Once we changed it to a polynomial trend line, our r squared value went up to .99 or 99%.
 * Analysis:**
 * a) Interpret the equation of the line (slope, y-intercept) and the r squared value**
 * b) Find the instantaneous speed at halfway point and at the end**


 * Halfway Point: v = 36.6 cm/s**
 * Final: v = 93.33 cm/s**

30.36 cm/s
 * c) Find the average speed for the entire trip. (average speed = d/t)**


 * Discussion Questions:**
 * 1. What would your graph look like if the incline had been steeper?** The curve would have been steeper and would have gone up higher sooner.
 * 2. What would your graph look like if the cart had been decreasing up the incline?** It would be a curve starting steeper that gets flatter as time increases because the acceleration is decreasing as it goes up the incline.
 * 3. Compare the instantaneous speed at the halfway point with the speed of the entire trip.** At the halfway point it was 36.6 cm/s while for the entire trip it was about 36 cm/s. This shows that it must have gotten increasingly faster because it was able to make such a far jump in a pretty short amount of time.
 * 4. Explain why the instantaneous speed is the slope for the tangent line. In other words, why does this make sense.** The instantaneous speed is the slope of the tangent line because the slope always represents the speed at that instant. Because it is a curve, the overall graph does not have a slope. However, if you draw a line tangent to a certain point you can use the slope of that tangent line to calculate the instantaneous speed at that particular point on the graph.
 * 5. Draw a v-t graph of the motion of the cart. Be as quantitative as possible.**


 * Conclusion:** Our final results were that the average speed of the cart was 36 cm/s, the initial velocity was 8.10 cm/s, and the acceleration was 13.23 cm/s. My hypothesis saying that the graph was going to be an upward curve that got steeper was correct, but I did not initially mention anything about being able to use the graph to accumulate so much data and information. We might have experienced some errors but estimating slightly off, accidentally pushing the cart a little to start it off, and starting at a ticker tape dot that we should not have. To eliminate these errors, I would have done another trial just in case and made sure it matched up with the results we got in this trial.


 * __ Homework 9/15/11 __**


 * Lesson 3 **
 * After reading the material, answer the following questions:**
 * 1) **What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.** From class discussions in both physics and math classes, I already knew a lot about slope. I already knew how to find the slope if given two points, as I have been required to use the formula a lot in the past. Additionally, from our class discussion, I already knew how to describe position, velocity, and acceleration time graphs by looking at the slope/curves. I knew how to do this because we were required to do so in our lab from this week, and I fully understood how to do it after that.
 * 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.** After class, I was aware of the definitions of both velocity and acceleration but I was slightly confused about their relationship. However, the graphs in the lesson helped to show and teach me more about this topic. Before, I did not realize that if an object had a changing velocity it is acceleration, but now the pictures show me that. [[image:Screen_shot_2011-09-14_at_10.26.20_PM.png width="434" height="147"]]
 * 3) **What (specifically) did you read that you still don’t understand? Please word these in the form of a question.** After reading and listening during class, I fully understand the material and have no questions.
 * 4) **What (specifically) did you read that was not gone over during class today?** I believe that everything in the lesson was gone over during class and I was pretty clear on most of the material before reading.

**Lesson 4**


 * After reading the material, answer the following questions:**
 * 1) **What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.** From the class discussion and completing the lab, I already fully understood the significance of slope on a v-t graph. I knew that if the velocity was constant, it would have a flat slope, while if the velocity was positive or negative, the slope would either be going up or going down. I knew that the slope of the v-t graph determined something about the objects acceleration.
 * 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.** Although discussed in class, I was slightly confused about determining whether the velocity was positive or negative by just looking at the v-t graph. However, the visuals from the lesson helped me to understand this much more clearly. [[image:U1L4a7.gif]]
 * 3) **What (specifically) did you read that you still don’t understand? Please word these in the form of a question.** I now fully understand everything from the reading and have no questions.
 * 4) **What (specifically) did you read that was not gone over during class today?** During class, we did not discuss anything about finding area on a v-t graph. When I first read this, I was completely unfamiliar with this concept, however, now I understand. To find the area, all you have to do is figure out what shape your graph makes and solve using the correct area formula that we already know from math class.

__Lab: A Crash Course In Velocity (Part II)__

 * Date:** 9/21/11
 * Partners:** Julia Sellman, Caroline Braunstein, Ali Cantor
 * Objectives:** Both algebraically and graphically, solve the following 2 problems. Then set up the situation and run trials to confirm your calculations.

Our calculations show that when the two cars meet, the blue should travel 372.321 cm and the yellow car should travel 229.214 cm. Below is a video of the actual situation of the two cars meeting up and crashing. media type="file" key="My First Project - Medium.m4v" width="300" height="300" Here, the two cars crash at a certain point. We then marked the place where they crashed, and saw that our results were approximately correct.
 * Calculations for Part A:**


 * Data Table Showing Each Trial:**

According to these calculations, the blue, faster car should travel 260.156 cm before catching up and becoming parallel with the yellow, slower car. Below is a video demonstrating this situation. media type="file" key="new - Medium.m4v" width="300" height="300" The blue car is catching up to the yellow car at a certain point. We then proceeded to mark the spot where they met and saw that our calculations from before were correct.
 * Calculations for Part B:**


 * Data Table Showing Each Trial:**

Based our percent error of .57%, we can conclude that the results we calculated in the beginning were almost exactly correct. The smaller the percent, the closer the calculations were compared to the actuality of the situation. Also, we chose to take an average of our experimental values because they were all very close to each other. This shows that our results were very accurate and precise. For percent difference, we took the individual experimental value and subtracted it from the average experimental value. Then we divided that by the average experimental value and multiplied it by 100. Because our percentages were very small, it showed the precision and accuracy in our results. By calculating that our results were extremely close to the average results, it proves that our results were accurate and precise.
 * Analysis:**
 * Part A: Percent Error**
 * Part A: Percent Difference**

One again, we used percent error to calculate how accurate our results were. Because we got .03%, it proves that our results were almost exact.
 * Part B: Percent Error**

Once again, our results were shown as very close to the average results, proving that our results for Part B were precise and accurate as well.
 * Part B: Percent Difference**

**Catching Up:** **Crashing:** **Catching Up:**
 * Discussion Questions:**
 * 1. Where would the cars meet if their speeds were exactly equal?** The cars would meet at the exact halfway point, which would be 300 cm. This is so because they are going the same exact speed so they would travel the same amount of distance in the same amount of time, causing them to meet halfway. For Part B, the cars would never meet because the first car would have still started ahead so the second car would not be fast enough to catch up with it.
 * 2. Sketch position-time graphs to represent the catching up and crashing situations. Show the points where they are at the same place at the same time.**
 * 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?**

**Crashing:** In conclusion, we came with two sets of results, one for part A and one for part B. For part A, our calculations showed that the two cars would meet and crash once the blue traveled 372. 321 cm and the yellow traveled 229.214 cm. For part B, our calculations showed that the faster, blue car would catch up with the slower, yellow car at the 260.156 cm mark. After recreating both of these situations, we learned that our results were almost exactly equal to our calculations. To prove our calculations correct even further, we found our percent error for the experiment, which was .57%. One of our sources of error was using significant figures, because we had to truncate, or cut off, the decimal at a certain point. Although it did not throw of our results drastically, it caused our results from part A to add up to about 601 cm instead of exactly 600. To fix this error in the future, we could have tried using a few more decimal places. Also, error could have been found while marking the exact point when the cars met up, due to human reaction time. Additionally, the measuring tape could have been altered while finding the distances of each car, as it easily could have been accidentally moved while trying to get the measurements. If I could change this lab to address the error, I would have made sure that the measuring tape was flat against the distance we were measuring and that nobody was altering it. Finally, the class average experimental values could have been off because everyone was using different cars with batteries that have been in use for different amounts of times. To fix this, we could have started each CMV with a set of fresh batteries to ensure that the batteries were not a factor that tampered with the final results.
 * Conclusion:**

=Egg Drop Project=
 * Partner: Danielle Bonnett**
 * Our Project:**


 * Description:** Our egg drop was a cylinder shaped cushioning around the egg that left enough room for the egg to get pulled slowly and softly to the ground after impact. It was made of six different cylinder shaped parts that each were filled with about ten straws. The straws were used so that the egg would not directly hit the ground after being dropped. Also, the tightness around the middle of the egg helped to strongly hold in in place while also lessening the likeliness of the egg being hit on the bottom and breaking.


 * Results:** Unfortunately, our egg completely cracked and leaked after being dropped.


 * Analysis:** For our first prototype, we used the same shape and idea that we used for our final one except it was much bigger. Although this prototype was successful in keeping the egg intact, we decided to make our second prototype similar but much smaller and lighter. However, the second prototype, which used less straws and was half the size, did not succeed in keeping the egg whole after being dropped. For our final one, we used the exact shape and idea of our first prototype but literally cut it in half to reduce the weight. However, our final project did not keep the egg intact either. After calculating, our results were that our project moved 9.8 m/s^2, which is very close to the average free-fall velocity. Here is a picture of the work we did to find this result.


 * What we would do to fix it?** To fix our project, we should have added some sort of parachute the slow down the rate at which it fell from the window. Also, we should have dropped it vertically instead of horizontally so the impact was on the flat part of the project instead of on the part directly surrounding the egg.

__** Quantitative Graph Interpretation **__
 * Graph E:**
 * Graph F:**
 * Graph G:**

= __Free Fall__ =

__** Homework 10/3/11 Summarizing: Method 1 Kinematics Lesson 5 **__ A free falling object is an object that is falling under the sole influence of gravity. A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth). It is known as the ** acceleration of gravity ** - the acceleration for any object moving under the sole influence of gravity. Physicists have a special symbol to denote it - the symbol ** g **. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude. The value of the acceleration of gravity (**g**) is different in different gravitational environments. Even on the surface of the Earth, there are local variations in the value of the acceleration of gravity (**g**). These variations are due to position and the geological structure of the region. One means of describing the motion of objects is through the use of graphs. Position-Time Graph: Observe that the line on the graph curves. Since the slope of any position vs. time graph is the velocity of the object the small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity. Finally, the negative slope of the line indicates a negative velocity. Velocity-Time Graph: Observe that the line on the graph is a straight, diagonal line. An object that is moving in the negative direction and speeding up is said to have a negative acceleration. Since the slope of any velocity versus time graph is the acceleration of the object, the constant, negative slope indicates a constant, negative acceleration. The velocity of a free-falling object that has been dropped from a position of rest is dependent upon the time that it has fallen. The formula for determining the velocity of a falling object after a time of t seconds is vf = g * t where g is the acceleration of gravity. The distance that a free-falling object has fallen from a position of rest is also dependent upon the time of fall. The distance fallen after a time of t seconds is given by the formula d = 0.5 * g * t2. More massive objects will only fall faster if there is an appreciable amount of air resistance present. The greater force on more massive objects is offset by the inverse influence of greater mass. Subsequently, all objects free fall at the same rate of acceleration, regardless of their mass
 * 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)

__Lab: Free Fall 10/5/11__

 * Date:** 10/5/11
 * Partner:** Julia Sellman
 * Objective:** What affect does gravity have on the acceleration of an object?
 * Hypothesis:** Gravity will cause the acceleration of the object to be 9.81 m/s, or 981 cm/s
 * Materials:** Ticker Tape Timer, Timer tape, Masking tape, Mass, clamp, meterstick

1. Put the ticker tape through the ticker tape timer 2. Take the ticker tape to a small 100 g circle weight 3. Hold the spark time and weight over the balcony in the lobby 4. Drop from the top of the balcony without pushing it downwards 5. Measure the distances from the beginning of the ticker tape's dots to each dot along the tape 6. Record data
 * Procedure:**


 * Data: (Left shows the position-time data and the right shows the Instantaneous Speed-Mid-time results)**


 * Sample Calculations for Instantaneous Speed (s) and Mid Time (s)**


 * Analysis:**

This graph is our velocity-time graph of the free falling object. Because it is almost exactly linear, it shows that our results were correct, because theoretically, the velocity-time graph of a free-falling object should be linear, as it is constantly accelerating at 981 cm/s/s. Additionally, by adding the trend line and displaying the y value and the r squared value we were able to conclude more about our results. The slope of the line was 710.65, which was equal to the experimental acceleration. Although this result was not too far off, it was not exact because the theoretical acceleration of a free-falling object should be 981 cm/s/s. Finally, because our r squared value was .996, it shows that our results were extremely precise and accurate.
 * Graphs:**
 * Velocity-Time Graph of the Free-Falling Object**

This graph is our position-time graph of the free-falling object. Just by looking at it, one could tell that our results were accurate because of the shape of the graph. Theoretically, a graph of a free-falling object would start with a flatter slope and would end up with a steeper one because of the increase in speed. We used a polynomial shape trend line to also show that our results fit the trend line almost perfectly. Additionally, our r squared value is .99 which shows that our results were precise and correct. Through substitution of the equation d = 1/2at^2 + vit we know that the number before the x^2 can be doubled to equal the acceleration of the object. If we do this, we get the result to be 711.34 cm/s, which is extremely close to our actual result which was 710.65 m/s.
 * Position-Time Graph of the Free-Falling Object**


 * Percent Difference = 15.34 %**


 * Percent Error = 27.56%**


 * Discussion Questions:**
 * 1. Does the shape of your v-t graph agree with the expected graph? Why or why not?** The shape of our v-t graph agrees with the expected graph in the sense that it is a linear line due to constant speed. However, because the object was being dropped, one would expect the line to be moving negatively away from the origin. However, because we omitted the negatives in our data, we received the graph that we did.
 * 2. Does the shape of your x-t graph agree with the expected graph? Why or why not?** Yes, the graph we received perfectly matches the graph that we expected. This is so, because we knew that the object would be gaining speed as it accelerated downward, and this is shown with the j-shaped curve of our graph.
 * 3. How do your results compare to that of the class? (Use Percent difference to discuss quantitatively)** Our final result was that our acceleration was 710.651 cm/s/s. However, the average acceleration of the class was 839.417 cm/s/s. After finding the percent difference, we calculated that we had a difference of 15.34%. Although this value still falls within the 20% difference range, it was not particularly close to equal to the class average, which could mean that our results were slightly off.
 * 4. Did the object accelerate uniformly? How do you know?** Yes, the object did accelerate uniformly. We know this because after looking at the v-t graph, we can see that it has a constant acceleration because the line is linear. Although it did not accelerate at exactly 981 cm/s/s, it accelerated at a constant rate pretty close to that, which shows that it did accelerate uniformly.
 * 5. What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?** Factors that could have caused gravity to be higher than it should be could have been the accidental force of the person who was letting the weight drop. Factors that would have caused acceleration due to gravity to be lower than it should be could have been friction of the ticker tape going through the spark timer.

Our final results showed that our weight fell with an acceleration of 710.651 cm/s/s. Although this was close to 981 cm/s/s, the theoretical result, it was not exact. Additionally, we found out that our percent error was 27.56%, which shows that our results were not as precise as they could have been but they were pretty accurate. We also found that the percent difference from the class average was 15.34%, which was fell within the 20% difference range, proving our results were close enough to the class average to be considered accurate. Our hypothesis was somewhat correct, because the acceleration of the mass would have been 981 cm/s/s, however, there could have been sources of error that could have contributed to this inaccuracy. For example, the friction from the ticker tape going through the spark timer could have caused the acceleration to be lower. To fix this for further labs, one could figure out the amount that the friction lowered the acceleration and add it to the result that they calculated. Additionally, the fact that a human had to drop the mass off the balcony could have been a source of error because he/she could have slightly pushed the mass downwards, adding another factor besides gravity. Finally, there could have been error in the lab while measuring the spaces between the dots on the ticker tape. The measuring tape could have slightly moved during the process of recording data, which would evidently alter the results. To fix this error, one could have taped down the measuring tape to make sure that it didn't move.
 * Conclusion:**


 * __ Class Notes- Freefall __**