Ch2_CaspertS

__Constant Speedtoc__

__**Lab: A Crash Course in Velocity (Part I)**__

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

** Hypothesis :** I conclude that the CMV will move at 2 ft/s, which is 60.96 cm/s. One can precisely measure the distance to the nearest 1/10 of a centimeter. Also, the position-time graph gives you the distance traveled by an object over a period of time.

**Discussion questions** The slope, which is 29.058, and the average velocity, which is 28.13, are almost equivalent. That is because slope is the “change in Y”/”change in X”, which is the “change in position”/”change in time”, which is the equation for velocity.
 * Position-Time Data for CMV**
 * 1. Why is the slope of the position-time graph equivalent to average velocity?**

Instantaneous velocity is the velocity at one point. Therefore it doesn’t take into account the difference in speed from the beginning to end, which may occur. By using the average velocity we are moreprecise it answering the speed of the CMV (cm/s).
 * 2. Why is it average velocity and not instantaneous velocity? What assumptions are we making?**

It is okay because the position of the CMV has to have a starting point, which is 0 cm. The Y-intercept tells us at one point the CMV started so that would be 0 cm.
 * 3. Why was it okay to set the y-intercept equal to zero?**

The R2 value tells us the percentage of values described by the tandem line. In this situation the R2 is 0.9975, which is very high. So the points fit very well on a linear line telling us that the speed of the CMV is at a consistent pace.
 * 4. What is the meaning of the R2 value?**

The CMV would start at the same point. But since it is moving slower, at each time interval it would have traveled a less distance. This means this new graph would have a more horizontal tandem line making it lie under mine.
 * 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:** When it comes to how fast the CMV move I was wrong; I hypothesized it to be at 2 ft/s or 60.96 cm/s, but it travled much slower 209.06 cm/s. I thought we measured the distances to a 1/10 of a cm; however, a distance can be measured to the 1/100 of a cm with the last decimal being an educated guess. I was correct in stating that the position-time graph tells us the velocity of the CMV. As for sources of error goes, both ripping off the tape at an inconsistent point and the fact that the meter stick is raised above the table contribute to inaccuracies. To reduce the amount of issues we can measure the distances between the dots until we get a relatively similar answer throughout. And for the meter stick, it was hard for me to measure the lines with the points made by the CMV because the stick wasn’t flat on the table. So if we were able to use a flat meter stick/tape that would reduce much of my inaccuracies.
 * In essay format, answer the following questions: **
 * What results did you get? Was your hypothesis accurate? (Be specific, using data from the lab to support claims.) **
 * What sources of error may have contributed to inaccuracies? **
 * What could you do to minimize these issues if you had to redo this lab? **

**__Homework: 1-D Kinematics - Lesson 1: (Motion w/ Words) - 9/8__**
 * 1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.**
 * While reading this, I understood the main idea of distance . I knew that distance is how much an object has traveled from the starting point. Therefore, distance can technically be seen within an absolute value because no matter what direction an object travels, the distance will always be positive. In the picture of the coach, you add up A to B, etc., and you get a total distance traveled of 95 yards.
 * I was also fully comprehensive on the idea of displacement. This explains the distance and direction an object is from an original position. In the coach picture, it made sense that he was only displaced 55 yards to the left. So technially an object can run up and down a football field 20 times and will have a displacement of 0.




 * 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 bit confused on the idea of speed versus velocity. Coincidentally, I was also just learning this in Calculus and didn't truly understand it there as well. But the reading helped to show the idea that "speed equals distance over time" while "velocity equals displacement over time". This is because velocity, just like displacement, takes direction into account. On the flip side, neither speed nor distance take direction into account. So, you can have a negative velocity, but you will not have a negative speed.


 * 3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question.**
 * I was able to understand all of the material in the assigned sections.


 * 4.What (specifically) did you read that was not gone over during class today?**
 * In class we did not talk about scalars and vectors. However, I found it pretty easy to comprehend considering I learned it last year in math. I knew that a scalar is only measured by magnitude, which is a numerical value. Vector, however, is not only measured by magnitude, but also direction. Therefore, as I read in the next section, scalar is related to distance and vector is related to displacement.

__**Speed Notes - 9/9/11**__
 * Distance: how far travleed
 * Position: where you are located, relative to a known position
 * Displacement: requires direction and distance of travel from original point
 * Speed: how fast you're going; rate of change of position
 * Velocity: deals with displacement and direction
 * Speed:
 * **Average Speed:**Speed of an object over time
 * = Change in Position / Change in Time
 * **Constant Speed:**Speed is always the same value
 * By this, instantaneous speed is always the same
 * Types of Motion:
 * At rest
 * Velocity = 0
 * Acceleration = 0
 * At constant speed
 * -v-> -v-> -v-> -v-> or <-v- <-v- <-v-- <-v-
 * A = 0 for btoh
 * Ticker Tape: [[image:Constant_Speed.png]]
 * Acceleration
 * Increasing speed
 * Decreasing Speed
 * [[image:Decrease_Accel_NOtes.png]]
 * Ticker Tape vs Motion Diagrams
 * Ticker Tape is more precise
 * Motion Diagrams give direction

__**Graph Shapes: Kinematics Practice**__ Position Graphs --> Velocity Graphs

Velocity Graphs --> Position Graphs

**__Homework: 1-D Kinematics - Lesson 2: (Motion w/ Diagrams) - 9/9__**


 * 1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.**
 * Ticker tapes are a way to show the motion of an object. From class, I was able to understand the idea about the distance of dots. If the dots are at a constant distance, then the object is at constant speed. If the dots gradually increase their distance, there is a positive acceleration and there is a negative acceleration when the dots gradually decrease their distance.
 * Vector diagrams show the speed and direction of an object. If the length of the arrows stays consistent, the object is at constant speed. If the length of the arrows grows with each arrow, the object is increasing acceleration. And if the length of the arrows decreases with each arrow, the object is decreasing in 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.**
 * When it comes to the ticker tapes, I was not completely sure what the distance between two dots meant. But now I understand that if while at constant speed the dots are very close to each other, the object is moving slow. The farther the dots are from each other, the quicker the object is moving. This is because if an object is moving quickly, the ticker tape goes through the machine quickly making the dots spread out.


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


 * 4.What (specifically) did you read that was not gone over during class today?**
 * We went over both ticker taps and vector diagrams in class.

__**The Big 5 Equations **__ 

**__Class Activity: Graphical Acceleration__**
**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?

No Motion

In these graphs, there is no motion. So, there is a horizontal line in the position-time graph meaning the velcoity and acceleration graph are both horizontal at y=0.

Constant Speed Fast --vs.--Constant Speed Slow

In the constant speed fast there is no acceleration because there is constant speed. The position graphs both have a positive slope, with the fast one having a steeper slope than the slow one because it is moving farther per second. As for the velocity, the fast graph has a higher velocity than the slow one, both supposed to be horizontal lines.

Changing Directions For "changing directions" the position graph has an upward slope, a climax, and then a negative slope. As for the velocity, it should start off positive and then turn negative when the position slope turns negative.

Walking Towards Motion Detector For the position graph, there is a negative slope. The velocity graph should generally be a straight line because its constant motion. And the acceleration graph should mainly be around 0 because there shouldn't be a change in velocity.

Discussion Questions
 * 1) **How can you tell that there is no motion on a…**
 * position vs. time graph
 * There is a strait line because the position does not change over time.
 * velocity vs. time graph
 * There is a straight line on “velocity = 0”.
 * acceleration vs. time graph
 * There is a straight line on “acceleration = 0”.
 * 1) **How can you tell that your motion is steady on a…**
 * position vs. time graph
 * There will be a constant slope.
 * velocity vs. time graph
 * There will be a straight, horizontal line above “velocity=0”.
 * acceleration vs. time graph
 * There will be a straight, horizontal line.
 * 1) **How can you tell that your motion is fast vs. slow on a…**
 * position vs. time graph
 * The slope is steeper in a fast graph because it takes less time to travel farther. In a slow graph, the slope is more horizontal.
 * velocity vs. time graph
 * The velocity will be higher in a fast graph than slow graph.
 * acceleration vs. time graph
 * You cant tell because there shouldn’t be any acceleration in these 2 graphs
 * 1) **How can you tell that you changed direction on a…**
 * position vs. time graph
 * There will be a peak or valley in the graph.
 * velocity vs. time graph
 * The graph will consistently have a positive velocity and then it will consistently have a negative velocity.
 * acceleration vs. time graph
 * You cant tell on this type of graph because no matter which way the object is moving, there can be a negative or positive acceleration.
 * 1) **What are the advantages of representing motion using a…**
 * position vs. time graph
 * The slope tells you the distance traveled over time, which is speed.
 * velocity vs. time graph
 * It is precise in telling you the exact velocity of an object over a period of time.
 * acceleration vs. time graph
 * It is precise in telling you the exact acceleration of an object over a period of time.
 * 1) **What are the disadvantages of representing motion using a…**
 * position vs. time graph
 * It doesn’t tell you the direction of movement, whether it be north, south, east, or west.
 * velocity vs. time graph
 * This graph does not show change in position of an object.
 * acceleration vs. time graph
 * Unless you are trying to show any sort of acceleration, this graph will have a horizontal line at “acceleration=0”, which is pretty useless when used for comparison
 * 1) **Define the following:**
 * 2) __No motion:__ an object at rest with no velocity, acceleration, or change in position
 * 3) __Constant speed:__ an object that moves with the same velocity over a period of time

__**Increasing and Decresaing Speed Graphs**__ Increasing Speed - Decreasing Speed -

__**Homework: 1-D Kinematics - Lesson 1 (Acceleration) - 9/13**__
 * **What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.**
 * I understand the true meaning of acceleration and how you can be going really fast & have an acceleration of 0. Acceleration is the change in velocity, whether it be positive or negative. At constant speed, acceleration is negative.
 * It makes sense that the average acceleration is: ("initial velocity" - "final velocity")/("change in time"). Being that acceleration is the change in velocity, this equation works.


 * **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.**
 * There wasn't anything in class that about acceleration that I was confused up. The reading just did a good job of giving the material in a different way.


 * **What (specifically) did you read that you still don’t understand? Please word these in the form of a question.**
 * I believe I understand all of the material about acceleration.


 * **What (specifically) did you read that was not gone over during class today?**
 * I read and understood there "rule of thumbs" because I learned this in Calculus last week. It is the idea that when velocity and acceleration are alike (+ or -), the object speeds up. But when the velocity and acceleration are unlike (one is + and one is -), the object slows down.

__**Homework: 1-D Kinematics - Lesson 3: (Describing Motion w/ Position-Time Graphs) - 9/15**__
 * **What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.**
 * Position-time graphs can have both a constant slope and a changing slope. When it is a constant slope, that means the velocity does not change. For example, an object that moves consistently at 10 m/s will have a slope of 10 throughout the graph. When there is a changing slope it means the velocity is changing; hence, there is a change in acceleration. This graph speeds up over time so, as the graph moves to the right the slope gets steeper. Therefore, it can be shown that the slope of the position-time graph equals the velocity in a v-t graph.
 * When a position-time graph has a negative slope, that means there is a negative velocity. This makes sense because an object neads to have a negative velocity to move back to the origin. On the flip side, an object has a positive velocity when the position-time graph has a positive slope.
 * **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.**
 * There was nothing I was confused about going into this reading.
 * **What (specifically) did you read that you still don’t understand? Please word these in the form of a question.**
 * I understand everything.
 * **What (specifically) did you read that was not gone over during class today?**
 * Everything here was pretty much gone over in class the past few days.

__**Homework: 1-D Kinematics - Lesson 4: (Describing Motion w/ Velocity-Time Graphs) - 9/15**__
 * **What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.**
 * The relationship velocity has with position-time graphs is the same relationship acceleration has with velocity-time graphs. When a v-t graph has no slope, object is moving at constant pace, there is no accleration. But when a v-t graph has a slope, there is a positive or negative acceleration depending on which one the slope is.
 * Speed is equal to the absolute value of velocity. So the farther away a point is from the origin on a v-t graph, the faster it is going. For example, an object moving at -9 m/s is moving faster than an object as 4 m/s.
 * **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.**
 * There were a few graphs in this section that really helped me with the understanding of speeding up versus slowing down. I know when velocity and acceleration are alike (+ or -) the object speeds up; visa-versa. But, I didn't really understand why. Although it is hard to explain, I have come to grasp with why this is the case.
 * **What (specifically) did you read that you still don’t understand? Please word these in the form of a question.**
 * I believe I understand everything.
 * **What (specifically) did you read that was not gone over during class today?**
 * I found it quite interesting determining the area on a v-t graph. For whatever shape the line and axis might make, you complete the area equation for it and the answer in the total displacement of the object over the allotted time.



**__Lab: Acceleration on an Incline__**
(with John Chiavelli) **Objectives/Hypotheses:**
 * What does a position-time graph for increasing speeds look like?
 * For this graph, there will not be a linear line but one that curves by getting a steeper slope as time goes on.
 * What information can be found from the graph?
 * From this graph, we can see position, acceleration, and velocity versus time.

**Materials:** Spark tape, spark timer, track, dynamics cart, ruler/meterstick/measuring tape

**Procedure:**
 * 1) Obtain a track, a dynamics cart, a textbook, ticker tape, and a spark timer.
 * 2) Align the track with on end on the physics textbook to form an incline, forming the ramp
 * 3) Put the spark timer on the textbook right in front of the beginning of the track.
 * 4) Pull the spark tape through spark timer and attach the end of it to the dynamic cart with a piece of tape.
 * 5) Make sure 10 hz is selected.
 * 6) Place dynamics cart at top of track and turn on the spark time.
 * 7) Led the dynamics cart go down the ramp.
 * 8) Hold the ticker tape and “feed” to machine for best results
 * 9) Measure the distance in cm from each point on the spark tape from zero.
 * 10) Graph the data by position vs. time.

**Data and Graph:**

**Analysis:** As for the "increasing down incline", there is a R2 value of 0.9995 for a polynomial line, which is very high. The line of best fit describes over 99% percent of the points meaning that the points are in a tight polynomial line.The Y-intercept is 0 because at zero seconds, the distance traveled by the cart is zero centimeters. The slope of it is 24.14x+2.56. This equation is the derivative of the first equation (position), which is in terms of velocity. Because this line is velocity vs. time and it has a positive slope, that means that over time acceleration increases.
 * a) Interpret the equation of the line (slope, y-intercept) and the R****2** **value.**

I used the derivate (slope) of the polynomial to find the instantaneous speed. At the halfway point, 0.75 s, the instantaneous speed is 20.67 cm/s. At the end, 1.5 s, the instantaneous speed is 38.77 cm/s.
 * 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.)**

The average speed is 20.33 cm/s.
 * c) Find the average speed for the entire trip.**Average Speed = 20.33 cm/s

**Discussion Questions:**
 * 1) **What would your graph look like if the incline had been steeper?**
 * 2) If the incline had been steeper, there would be a greater acceleration giving the graph a steeper slope.


 * 1) **What would your graph look like if the cart had been decreasing up the incline?**
 * 2) The graph would be the opposite of graphing the cart increasing down the incline. The acceleration would start fast and gradually slow down. So, the line would be heading up and there would be a curve to the right to show that as time goe on the position doesn't change as much.


 * 1) **Compare the instantaneous speed at the halfway point with the average speed of the entire trip.**
 * 2) The average speed is 20.33 cm/s while the instantaneous speed at the halfwway point 20.67 cm/s, which is clearly very similar. The total average speed is the average of the beginning & end, 2nd point & 2nd-to-last point, and so on. Therefore, it ends up being that the average speed is near the middle of the graph, around half. Similarly instantaneous speed at the halfway point is obviously at the half way mark.


 * 1) **Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense?**
 * 2) Instantaneous speed is the slope at one particular moment in time. Due to local linearity, as you zoom in and in on an individual point, the tangent line to that point starts to act more and more like the original graph. So the slope of the tangent line has the same slope as that individual point.

Due to the fact that the cart is always moving forward, as shown in the position-time graph, I knew there was always going to be a positive velocity. But also, there is a constant positive acceleration in the position graph, which is why the velocity graph has a slope of about 12.5.
 * 1) **Draw a v-t graph of the motion of the cart. Be as quantitative as possible.**

**Conclusion:** In this lab, my partner and I obtained successful results. We did an "increasing down" graph and a "decreasing up" graph, both of which were supposed to have polynomial looks due to the increase/decrease in acceleration. For the "increasing" graph we got a R2 of 0.9995 and for the "decreasing" graph we got a R2 of 0.9997. In both cases, there is a strong R2 meaning the points fit the polynomial line very well. For the "increasing" graph, there was an acceleration of 6.04 (A=2a=.5*12.071=6.04) and an initial velocity of 2.56 cm/s (initial velocity equals b). Both of these numbers, the 2.56 and 12.071, are constant in the quadratic equation they come from. As for my hypotheses go, it is safe to say I was accurate. Considering I've been learning this in Calculus the past few weeks, I knew that a position-time graph with increasing speed had both "+ acceleration" and "+ velocity" creating this quadratic that curves up. I was also right in saying that the position-time graph position vs. time, velocity vs. time, and acceleration vs. time. Because the graph is a polynomial and not linear, we can find the derivative of position to find velocity and the derivative of that to find acceleration. If there was no curve to the graph, we wouldn't be able to see acceleration. However, they told us that the graph had increasing speed leading me to this conclusion I said above. There are numerous sources of error that may have contributed to inaccuracies. For starters, some carts may be heavier/lighter than others leading to a change in acceleration. To minimize this issue, we can all weigh our carts and take the weight into account. Another error may be due to track regarding the EXACT position on the physics book and friction. Some carts may ride smoother along certain tracks than others.
 * (Discuss, in essay format:** **What results did you get? Was your hypothesis accurate? (Be specific, using data from the lab to support claims.)** **What sources of error may have contributed to inaccuracies?** **What could you do to minimize these issues if you had to redo this lab?)**

**__Lab: A Crash Course in Velocity (Part II)__** **9/21/11**  **Hypothesis:** **Catch-Up Hypothesis (a)**
 * with Andrew Chung, Amanda Fava, and John Chiavelli**

I conclude that at the point where the slow CMV will travel 81.70 cm and the fast CMV will travel 181.70 cm, is where the fast CMV will pass the slow CMV. **Crash Hypothesis (b)**

I conclude that the slow CMV will travel 186.10 cm and the fast CMV will travel 413.90 cm at which point they will crash. From the start, it will take both CMVs 6.40 seconds to reach that point. **Observations:**  ﻿**Crash Observations**  media type="file" key="good vid crash (iPhone & iPod).m4v" width="300" height="300" **Catch-Up Observations** media type="file" key="good catch up (iPhone & iPod).m4v" width="300" height="300" **Data Table:** **Percent Difference** **:** ((Average Experiment - Individual Experiment)/Average Experiment) X 100 = Percent Difference (A) is percent difference for Trial 1 of Crash tests. The rest are in the above chart. (B) is percent difference for Trial 1 of Catch-Up tests. The rest are in the above chart. ** Percent Error: (**(theoretical - experimental)/theoretical) X 100 = Percent Error (A) is percent error of Crash tests. (B) is percent error of Catch-Up tests. For the crash tests, we got a percent error of 2.26% meaning that the results were very good in comparison to our orignal hypothesis of 186.10 cm. For the catch-up test, we got a perfect 0% error meaning that our test results were exactly what we predicted, 181.70 cm.

**D iscussion questions ** If the CMVs have similar speeds, that means they are traveling the same distance over the same amount of time. So, if they are seperated 600 cms apart, both CMVs will travel to middle until they meet each other. So the cars would meet at 300 cm (the middle) in this case. Then, in the part for both CMVs going the same way separated by 100 cm, they would never meet.
 * **Where would the cars meet if their speeds were exactly equal?**

After 2.81 seconds, both CMVs will meet at 181.72 cm from the Blue CMV and 81.72 cm from the yellow CMV. After 6.4 seconds, the blue CMV will have traveled 413.90 cm and the Yellow CMV will have traveled 186.10 cm to make them crash. From this graph, you cant find the point when they are at the same place at the same time. Velocity time graph tells nothing about position and that is what we need to find the answer. Over the course of the lab, my group and I were fortunate to find both accurate and precise results in comparison to our pre-lab algebra. For the "crash" test, we calculated that it would take 6.4 seconds for the slow CMV to travel 186.10 cm to crash with the fast CMV. After doing the test 5 times, the smallest distance the slow CMV had to travel was 177.67 cm and the fastest was 184.15 cm. Therfore, not only were our tests precise by being a close distance to each other, but they also were very accurate with the calculations. And just like the "crash" test, we got great results for the "catch-up" test. For this, the algebra said that the fast CMV would travel 186.10 cm. After 5 tests, the results showed a low of 180.17 cm & high of 185.37 cm. Once again, the results showed that our tests were both accurate and precise in comparison to the pre-lab algebra. As for the source of error goes, there could be some error in regards to the function of the CMVs. For example our blue (fast) CMV had a tendency to curve to the left making it very difficult for us to gather our data leaving us to make educated guesses. So instead of doing that, we lined both CMVs against the wall which kept the blue CMV on straight path. Of course, this may slow it down a little bit and affect the results; however, it is better than our options before. Another source of error may be the surface that the CMVs were on. When we originally tested them for velocity, we were on one hallway floor. But while doing Part 2, our CMVs were on a different hallway floor while still using the data from the first time. These floors may differ in smoothness/friction and cause a slight change in results in the data. To eliminate this problem, we could have gone to the same spot both times, even though there still may be a difference in friction due to dust and other particles. And a third source of error is due to the batteries. When we first did the tests, the CMVs had a certain battery giving these cars a certain speed. However, when we did the second tests, it was over a week later and the batteries had lost power causing the CMVs to lose power. In response, we could have used pefectly new batteries both times or complete the tests one of another (which wasn't an option).
 * **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.**
 * **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?**
 * Conclusion :**

**__Egg Drop Project__**
**Our Project** The base of our project was made of a sheet of paper rolled up into a cone. Around the outside of the cone, about halfway up, we put four folded-up sheets of papers. They were attached by scotch tape. As for the inside of the cone, on the bottom there were many crumpled pieces of paper. We made it so the egg fit perfectly snug right were the paper cushions were. And the we put more crumpled paper on top of the egg. Then, we attached a sheet of aluminum foil to the cone with sewing thread to make a parachute.

**Calculations:**

**Results/Analysis:** Our project turned out successful, as the egg stayed in tact after the drop. When we dropped our project, the parachute played a major role in keeping our egg unbroken. The parachute took up enough surface space in comparison to the cone that it was able to catch air and really slow down the fall. It kept our project at a slow 2.93 m/s^2, which allowed for a gentle fall. And then, there was the cone shape that first hit the ground. The bottom of the cone crumpled, which was what we wanted. And the egg was high enough up the cone that it was able to slide down into the paper cushion, which it was expected to do, with enough space not to crack.

**Conclusion:** Due to the fact that our project was a success, I would not change anything about it. If we were allowed to make it larger than one sheet of paper, I would have increased the parachute size greatly. We did this for the 2nd prototype (forgot about the guidelines) and the fall was so much smoother and slower because the parachute could catch more air. But other than that, I wouldn't change our project.

**__Quantitative Graph Interpretation (Examples)__** **Example 1:**
 * Example 2:**

__**Homework: 1-D Kinematics - Lesson 5: (Freefall and the Acceleration of Gravity) - 10/3**__

Free falling object is object falling under sole influence of gravity. Free-falling objects don't encounter air resistance. All free-falling objects accelerate downwards at rate of 9.8 m/s^2. Acceleration is the same for all free-falling objects regardless of how long they have been falling, or whether they were initially dropped from rest or thrown up into the air.

Ticker tape trace or dot diagram of its motion would depict acceleration.



Acceleration of Gravity = g = 9.8 m/s/s, downward

**__EQUATIONS__**
 * vf = g * t**
 * d = 0.5 * g * t2**

Position versus time graph for free-falling object.



Velocity versus time graph for free-falling object.



__**Notes: Freefall**__


 * Freefall** - only gravity is acting on object, no other forces (IGNORE AIR RESISTANCE)

Velocity is symmetrical around maximum height.


 * Practice Problem:**

= ﻿﻿__Lab: Freefall__ = 10/4/11. with John Chiavelli

**Objectives:** ﻿What is the acceleration of a falling body?

**Hypothesis: The** acceleration of a falling body of mass will be 9.8 m/s^2.

**Procedure:**
 * 1) ﻿Gather up the supplies, which included a ticker tape timer, timer tape, masking tape, mass a clamp, and a ruler (don't know the name of it).
 * 2) Find a long strand of the timer tape that had no wrinklies or dent and cut it off from the rest of the roll.
 * 3) Attach one end of the timer tape to the mass (ours was 100 g) with masking tape.
 * 4) Set the ticker tape timer to 0.1 seconds.
 * 5) Over the ledge, run part of the tape through the timer so that you have the weight hanging off.
 * 6) Start the timer and let the weight drop, carefully allowing for the timer tape to go through smoothly.
 * 7) Receive the tape after the drop.
 * 8) Measure the distances between each dots to show what gravity does to the velocity and accerlation of an object in freefall.

**Data:** Time and Distance of Weight from Initial Point of Drop to the Ground:

The Mass Object from Initial Point of Drop to Ground: **Graphs/Analysis:** Looking at a velocity-time graph of an object in freefall can tall you a lot. Because the data points should form a linear trendline, the equation of this line is going to be y=mx+b. Because "m = slope" and "acceleration of a v-t graph = slope", then "m = acceleration". In this graph, x = time, y = instantaneous speed, and initial velocity equals "b". So y=mx+b --> vf = at +vi. From this graph we can tell that the acceleration of the mass is 891.38 cm/s^2 and the intitial velocity is -26.952 cm or -0.027 m. The equation of a position-time graph for freefall is supposed to be y=Ax^2+Bx. For this graph, y = distance and x = time. As proven in a previous lab "A = 1/2(acceleration)" and as shown above in the v-t graph, "b = initial velocity". So y=Ax^2+Bx --> y=.5at^2+(vi)t. Using these two equations, we can tell that acceleration equals 890.36 cm/s^2 while the initial velocity is -27.085 cm or -0.027 m. **Class Data of Acceleration (cm):**

**Calculations:** **Percent Difference:**


 * Perecent Error:**


 * Sample Calculation of Finding Instantaneous Velocity:**

**Discussion Questions:**
 * **Does the shape of your v-t graph agree with the expected graph? Why or why not?**
 * Yes. Being that the acceleration of gravity is 9.8 m/s^2 and acceleration is the slope of velocity, I expected the slope of this velocity-time graph to be around 9.8 (m/s) or 981 (cm/s). "M" equals the slope of a straight line; therefore, according to my graph, I had a slope of 891.38, which is fairly close to what I expected. Although people usually look at the acceleration of gravity as -9.8 m/s^2, I chose it to be in the opposite direction by making the maximum height where I dropped the mass to be at 0 cm and the floor to be a higher number of centimeters. Therefore, down was positive.


 * **Does the shape of your x-t graph agree with the expected graph? Why or why not?**
 * Yes. I expected the position-time graph to be in an polynomial form. The slope of a postion-time graph is gradually velocity. So, because velocity gradually increases as the mass approaches the floor, it would make sense that so does the slope of the x-t graph. This lead me to believe that as time goes on, the position will increase at a faster rate.

 (Perecent difference graph is above in the calculations section)
 * **How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.)**
 * Our results were pretty close to the class as we got a percent difference of 6.19%. Our experiment yielded an acceleration of 891.38 cm/s^2 while the average experiment yielded an acceleration of 849.417 cm/s^2.


 * **Did the object accelerate uniformly? How do you know?**
 * You know the object accelerated uniformly by looking at the velocity-time graph. Because acceleration is the slope of the a v-t graph, you want the point on the v-t graph to form a consistent linear trendline. If this is the case, then the a-t graph will show the acceleration to be horizontal meaning the object accelerate uniformly. By looking at our v-t graph, taking measuring errors into account, we have a pretty uniform, linear slope showing that our objects accelerates uniformly.


 * **What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?**
 * When an object is made as dense as it could be, this causes its acceleration to incease; visa-vera. A higher altitude would cause acceleration due to gravity to be lower than usual; visa-versa.

**Conclusion:** As we should have due the natural idea of gravity, our results turned out to be fairly accurate. From the velocity-time graph we obtained an acceleration of 891.38 cm/s^2 and from the position-time graph we obtained an acceleration of 890.36 cm/s^2. The slight variability in this is due to different calculations, nothing more. Compared to my hypothesis which said that the acceleration of an object in freefall is 891 cm/s^2, we were very accurate. Now the reason the acceleration is positive and not negative is all based on how I made my direction. Instead of making the initial drop 0 and everything below it to the floor negative, I chose to make it positive. As long as I stick to the same direction there is no difference. As mentioned earlier, the percent difference from the class average of 839.417 cm^s/2 was 6.19 % while the percent error from the acceleration due to gravity (g) of 98.1 cm/s^2 was 9.14%. This means that our data was both precise and accurate. However, it is not perfect and therefore it means there is some sort of error. One possibility is in the measurements. Due to the fact that we laid both the ticker tape and measuring device on the ground and only taped them both with masking tape at a few stops meant that there was a possibility that one of these two pieces would move. Unfortunately, this would cause inaccuracy in our measurements. The only way to fix this would be to spend a good amount of time really taping them both up to ensure no movement. Another source of error could be from the friction in the ticker tape timer. By going through the machine, friction is created between the timer and ticker tape causing the tape and weight to be slowed down a little bit in its freefall. If there was a timer that caused no friction by letting the paper through loosely while stilling plotting the dots would be a way to solve this source of error.