Ch2_WeinfeldM

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toc
 * Constant Speed **

Class Notes (9/6): Constant Speed

 * distance - how far; not relative to starting place
 * position - where you're located, relative to a known location
 * displacement - requires direction, how far in that direction, relative to your starting point
 * speed and velcoity are often used as synonyms
 * speed - the rate of change of your position; how fast
 * velocity - total displacement
 * to measure speed, you need time and distance
 * reaction time - 10th of a second so it can be a flaw in measurements like a stop watch
 * hertz - measure of frequency
 * V = total distance/ total time
 * the smaller the distance, the more precise

Lab: A Crash Course in Velocity (Part 1)
September 8, 2011 Partner: Ben Weiss

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

**Hypotheses:** The CMV will move 30 centimeters/second. I tried to visualize how fast I could walk in a second and then assumed that the car would move slower than I walk. We can measure to the 10th of a centimeter the ruler is marked up to 10th of a centimeter. The Position Time graph will show where the CMV will be at a given time. I hypothesized this based off of the title.

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

**Data Table:**


 * Position- Time Table**

**Graph:**

**Analysis:** The graph is linear and the R2 value, the percentage of points described by the trend line, was 0.99144. This value shows that it was a good fit because the closer the value is to 1, the better fit it is. The equation of the line is y = 68.628x. 68.628 is the slope of the line, which is also equivalent to the average speed of the graph.

**Discussion questions:** **Conclusion**
 * 1) Why is the slope of the position-time graph equivalent to average velocity?
 * Slope is the change in y values over the change in x values. On our graph, the y value was change in position and the x value was change in time. Velocity is equal to change in distance over change in time so therefore our slope was equal to velocity.
 * 1) Why is it average velocity and not instantaneous velocity? What assumptions are we making?
 * Average velocity is over a course of time whereas instantaneous velocity is just a specific time so it would not be helpful in finding a constant speed. We are assuming that the average velocity will be the same as the instantaneous velocity as long as the points are incredibly close together.
 * 1) Why was it okay to set the y-intercept equal to zero?
 * It was okay to set the y-intercept equal to zero because at zero seconds, the position was zero. This means that the coordinate for this point would be (0,0) and thus the trend line will pass through the y intercept of 0.
 * 1) What is the meaning of the R2 value?
 * The R2 value is the percentage of points described by the trend line. We chose a linear graph and have a 0.99144 R2 value so it was a good fit because the closer the value is to 1, the better fit it is.
 * 1) 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?
 * It would lie under my trend line because it would have a lower slope and thus not be as steep.

Our CMV moved 68.628 centimeters/second. My hypothesis for the first question was inaccurate because I predicted that it would move 30 centimeters/second. My hypothesis for the second question was also proven wrong in the respect that we were able to measure the distance to the 100th of a centimeter rather than the 10th of one because we estimated the last decimal place. For the third question, I was correct in hypothesizing that the Position Time Graph would visually show us where our CMV was at a given time. There was a lot of room for error in this lab; for example, our results are not exact and can be off be one or two decimal places because of the meter stick we used and the way we read the measurements. To get a more accurate reading, we could have used a measuring device with more precise markings. We also could have used something that was flatter down because the metric sticks had some height, which may have thrown off our measurements. Another source of error is that we may have shifted the meter stick in the middle of measuring. To fix this problem, perhaps we could have taped it down the table or again, we could have used a measuring device that was flatter down.

September 8: Summarize Lesson 1 (a, b, c, & d)

 * ** What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. **
 * I read about distance and displacement, which we discussed in class. In class, I was able to understand the definitions of the two and could note the difference between them. The reading simply reemphasized that distance refers to how much ground an object has covered during its motion and displacement refers to how far out of place and object is or the object's overall change in position. The reading improved my understanding of the terms by explaining that distance is a scalar quantity and displacement is a vector quantity. The example about the physics teacher taking a walk mimics the example we saw in class. Velocity is a vector quantity that refers to the rate at which an object changes its position.
 * I also read about the distinction between speed and velocity, a concept which I felt I had understood from our lesson in class. Speed is a scalar quantity referring to how fast an object is moving. Like the distinction between distance and displacement, the difference is small so the two words are often used in place of one another. Therefore, it is important that I learn the distinction early on so that I can use the appropriate terms at the right time.
 * ** 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. **
 * In class, I was a bit overwhelmed by all of the new vocabulary and worried that I would mix things up. The reading helped clarify the definitions for me and allowed me to become more familiar with terms like mechanics, displacement, vectors, and scalers. I was shaky in class because I have used these words in the past, but never really thought about their actually meaning. For example, I always used speed and velocity interchangeable. The combination of our work in class and then reading about the words again has helped me feel more confident with my physics vocabulary.
 * ** What (specifically) did you read that you still don’t understand? Please word these in the form of a question. **
 * Can you have a negative displacement if you move backwards? Or is just the amount in any direction?
 * ** What (specifically) did you read that was not gone over during class today? **
 * The reading explained the difference between the terms mechanics and kinematics. In class, we didn't discuss the comparison. I now understand that mechanics is the study of the motion of objects while kinematics is the science of describing the motion of objects using words, diagrams, numbers, graphs, and equation. Basically, kinematics is the description of the mechanics of an object.

Class Notes - 9/9

 * kinematics is the study of motion
 * you can find the average speed from one trip or by averaging the speeds of two different trips
 * average speed v. instantaneous speed v. constant speed
 * average speed is average of how fast you've been going the whole time. average speed isn't always the same
 * instantaneous time will be different at different variables
 * constant speed means its not changing, its always the same value, therefore, the instantaneous speed is always the same.
 * we only have one equation for all of them, V = change in distance/ change in time
 * Types of Motion
 * at rest - not moving at all
 * constant speed - speed is not changing
 * increasing speed - acceleration (changing speed)
 * decreasing speed - acceleration (changing speed)
 * Motion Diagrams
 * at rest: velocity=0, a=0
 * constant speed: v --> v --> v --> (+) (velocity is the same b/c the arrows are the same length), a = 0
 * increasing speed: v -->, v >, v ->(+) (velocity is getting bigger), a --->(+) (pushing forward)
 * decreasing speed: v--->, v>, v-->(+) (velocity getting smaller), a <---(-) (pushing back)
 * you can also point the arrows in the other direction to show movement the other way or up or down to show the direction you are actually moving.
 * when drawing this on paper, put the acceleration and its arrow under or over the velocity arrows
 * [[image:35403689.jpg width="198" height="139"]]
 * Ticker Tape
 * at rest, there would just be one dot (......x......)
 * at constant speed, there are evenly spaced dots (x...x...x...x...x)
 * at increasing speed, there will be more space between dots (x..x...x....x......x)
 * at decreasing speed, there will be less space between dots (x....x...x..x.x)
 * Motion Diagrams v Ticker Tape Diagram
 * ticker tape diagrams can be used to get very precise measurements
 * flaw of ticker tape diagram is that you can't see direction so you can't tell where it starts and where it stops. you won't know if its increasing or decreasing
 * motion diagrams show direction, but don't show measurement
 * Signs are completely arbitrary and subjective
 * the assumption is that to the right and up is positive and to the left and down is negative
 * you can change it, but you have to then label it on the paper
 * sometimes things are moving up a hill, so we can tilt the axis

September 9: Summarize Lesson 2 (a, b, & c)

 * ** What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. **
 * The reading further enforced the importance of being able to visualize the results we are going to get in physics this year. I read about both the vector/motion diagrams and the ticker tape diagrams. I understood in class, and now understand even more, the difference between the two and the purpose they each have. Vector diagrams show direction, but do not show exact measurement. They can show if there is constant speed or acceleration, but can not get more specific than that. Ticker tape diagrams can be used to get very precise measurements, but do not show direction. They are both important, depending on what kind of information you are looking for.
 * ** 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. **
 * In the reading, I was able to see an example of ticker tape, when an object stops. In class, I understood what it would look like for an object to move at a constant velocity and when its accelerating, but not when it stopped. I now feel like I have a better understanding, which will help me interpret ticker tape in the future.
 * ** What (specifically) did you read that you still don’t understand? Please word these in the form of a question. **
 * I feel comfortable with all of the material.
 * ** What (specifically) did you read that was not gone over during class today? **
 * We went over both vector diagrams and ticker tape diagrams in class.

Class Activity: Graphing Acceleartion
Slope of x - t = speed Slope of v -t = acceleration v - t: area = displacement velocity time graph can tell you the most a - t: Change in velocity = area "under graph"
 * Data (graphs are inaccurate)**
 * At rest:**
 * Constant Slow Towards:**
 * Constant Slow Away:**
 * Constant Fast Towards:**
 * Constant Fast Away:**

Position time graphs shows where you are located. The y axis shows your position. On a constant speed position time graph has the same distance between each point. You are covering the same amount of distance in the same time intervals. The steeper the slope, the faster you are going. Slope is equal to the speed. Positive slope means you are moving away from starting point and negative slope, you are walking towards the origin.

Velocity time graphs show how fast you are going. The only place it can be at rest is a horizontal line at y = 0. A line that is higher up indicates faster speed than a line lower down. If there is a line below zero, you are moving backwards. Negative velocities don't mean that you are going slow so the bigger the absolute value, the faster you are.

Acceleration time graph should be horizontal at y = 0 because acceleration is 0.


 * Discussion questions **
 * 1) **How can you tell that there is no motion on a…**
 * 2) position vs. time graph
 * 3) The position remains constant for the entire graph, which is shown by a straight, horizontal line and there is equal space between each point.
 * 4) velocity vs. time graph
 * 5) The line should be straight and horizontal at y = 0 because velocity = 0. Because during our experiment, the radar picks up on other movements, the points go up and down, but they are all close to 0.
 * 6) acceleration vs. time graph
 * 7) The line should be straight and horizontal at y = 0 because there is no acceleration at rest. Because in our experiment, he radar picks up on other movement the points go up and down, but they are all close enough to 0.
 * 8) **How can you tell that your motion is steady on a…**
 * 9) position vs. time graph
 * 10) The amount of time between each point is equal. The line will be straight and there should be a consistent slope.
 * 11) velocity vs. time graph
 * 12) There is a straight, horizontal line, representing the velocity. This is because the velocity is not fluctuating.
 * 13) acceleration vs. time graph
 * 14) There is a straight horizontal line at y = 0, representing the acceleration. This is because there is no acceleration.


 * 1) **How can you tell that your motion is fast vs. slow on a…**
 * 2) position vs. time graph
 * 3) The steeper the slope is, the faster the motion.
 * 4) velocity vs. time graph
 * 5) The higher the absolute value for velocity is, the faster the motion is.
 * 6) acceleration vs. time graph
 * 7) You cannot tell on the acceleration vs. time graph because, assuming that the motion is at constant speed, there is no acceleration.


 * 1) **How can you tell that you changed direction on a…**
 * 2) position vs. time graph
 * 3) The sign of the slopes becomes opposite.
 * 4) velocity vs. time graph
 * 5) If the velocity is on both sides of the x axis.
 * 6) acceleration vs. time graph
 * 7) You cannot tell direction from the acceleration vs. time graph because acceleration does not depend on direction.


 * 1) **What are the advantages of representing motion using a…**
 * 2) position vs. time graph
 * 3) Position vs. time graph can show you how far you've moved in a certain amount of time. Position vs. time graphs can also show direction.
 * 4) velocity vs. time graph
 * 5) Velocity vs. time graph is useful because it can show you if you are moving at constant speed or if you are accelerating. It can also show direction because velocity=displacement/time. IT can help you calculate the rate because it is so exact.
 * 6) acceleration vs. time graph
 * 7) Acceleration vs. time graph is important for representing motion because it shows you how much or little you are accelerating. It shows if you are speeding up or slowing down. Acceleration can also help with calculations because it is so detailed.


 * 1) **What are the disadvantages of representing motion using a…**
 * 2) position vs. time graph
 * 3) Position vs. time graph cannot give you the exact velocity or acceleration; they can only show you an estimate about if there is a change in speed. It is not very useful for calculations.
 * 4) velocity vs. time graph
 * 5) The velocity vs. time graph doesn't show starting point or distance so calculations must be done to figure out other information. In our tests, it will difficult to use this graph because we cannot walk at a perfectly steady pace. Therefore, the results were not completely accurate.
 * 6) acceleration vs. time graph
 * 7) The acceleration vs. time graph doesn't show starting point or distance so calculations must be done to figure out other information. In our tests, it was difficult to use this graph because we cannot walk at a perfectly steady pace. Therefore, the results were not completely accurate.


 * 1) **Define the following:**
 * 2) No motion
 * 3) Absence of any acceleration or distance; at rest
 * 4) Constant speed
 * 5) Speed is not changing at all

Class Notes (9/13)
__Kinematics__ Acceleration - rate that velocity is changing - (a) m/s^2 A = the change in velocity/ change in time (V f - V i ) / t or V f =V i + at V = d/t __only__ for a average or constant speeds; cannot be used when there __is__ acceleration Another way to find average speed is by adding velocity 1 and velocity 2 and dividing it by 2; only for average speed





September 13: Summarize Lesson 1e
RULE OF THUMB **If an object is slowing down, then its acceleration is in the opposite direction of its motion.**

**1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.** **2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped clarify? Describe the misconception you were having as well as your new understanding.** **3. What (specifically) did you read that you still don't understand? Please word these in the form of a question.** **4. What (specifically) did you read that was not gone over during class today?**
 * I understood the concept of constant acceleration. If an object continues to change its velocity by the same amount between the same amounts of time, then it is constantly accelerating.
 * I also understood the difference between velocity and acceleration. Acceleration is the rate by which velocity is changing. Therefore, if the velocity is constant, there is no acceleration. If there is constant acceleration, the velocity is increasing.
 * I was a little bit shaky about the concept of acceleration. I understood that it meant changing speeds, but I think I had a hard time grasping that direction was important too. Reading it being explained helped me grasp the concept that acceleration is the rate at which an object changes its __velocity__. It depends on whether the object is speeding up or slowing down and whether the object is moving in a positive or negative direction.
 * I understand everything in the reading.
 * The reading discussed the motion of a free-falling object. A free falling object would have constant acceleration because as it falls down, it will keep going quicker.

Lab Acceleration Graphs (9/14)
Lab Partner: Ben Weiss


 * Objectives & Hypotheses:**
 * What does a position-time graph for increasing speeds look like?
 * The slope between points gets steeper as the speed increases
 * What information can be found from the graph?
 * How much speed is increasing between any time increments. The change in slope tells you the change in speed over an interval.

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

media type="file" key="My First Project - Mobile.m4v" width="300" height="300"
 * Procedure:**

<-- data table

<-- graph

a) Interpret the equation of the line (slope, y-intercept) and the R2 value. 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.) c) Find the average speed for the entire trip.
 * Analysis:**
 * The equation of the the line for the cart going down the incline is y = 0.073x 2 + 1.0832x. 0.072 is the A value which is equivalent to half the acceleration. Therefore, if you double this value, which is equal to 0.144, you have the acceleration of the cart. 1.0832 is the B value, which is equivalent to the initial velocity. The closer the B value is to 0 because it should be starting at 0 velocity. My B value was close, but not exactly there, which probably means I made an error when starting to record my data. The R 2 value is .99858 which means that the polynomial trend line was a great fit. The R 2 value was a good evaluation of our data, and it made us feel like we did a good job.
 * The equation of the line for the cart going up the incline was y = -0.207x 2 + 5.884x. The negative A value indicates that the cart was moving in the opposite direction. This makes sense because this cart was going up the incline instead of going down it. The A value is half the acceleration so the acceleration is negative, indicating opposite direction. The B value was farther away from 0 this time because we started measuring further down on the ticker tape to get the most accurate measurements. It wouldn't have been as exact earlier down because that was measuring the effort of our push. The R 2 value is 0.99999, which tells me that the polynomial trend line was a great fit!
 * [[image:Photo_on_2011-09-14_at_20.43.jpg]]
 * [[image:Screen_shot_2011-09-14_at_5.05.01_PM.png]]
 * [[image:Screen_shot_2011-09-14_at_5.07.24_PM.png]]


 * Discussion Questions:**

1. What would your graph look like if the incline had been steeper? 2. What would your graph look like if the cart had been decreasing up the incline? 3. Compare the instantaneous speed at the halfway point with the average speed of the entire trip. 4. Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense? 5. Draw a v-t graph of the motion of the cart. Be as quantitative as possible.
 * The slope would have gotten steeper on the graph for the cart going down the hill because with a steeper incline, the cart would have went faster.
 * The slope would have gotten less steep on the graph for the cart going up the hill because with a steeper incline, the cart would have had a harder time getting up the ramp, so it would be moving up a very slow pace.
 * [[image:Screen_shot_2011-09-14_at_5.16.40_PM.png]]
 * The cart accelerates at first because we pushed it up the hill. But as it starts to go up more of the incline on its own, the speed decreases so the slope starts becoming less steep.
 * The instantaneous speed at the halfway point was 20 cm/s while the average speed was 18.31 cm/s for the cart going downhill. It makes sense that these numbers are close together. In order for the average speed to be 18.31 cm/s, at some instants the speed would have had to have been faster and at others slower. This instant is an example of a moment when the speed was above average.
 * Because our graph is not linear, there is not a consistent slope, or velocity. Finding instantaneous speed means that we want to know what the speed is at one given moment. Therefore, if we use a tangent line, it will touch that one point we want to find the speed of. You can use the tangent line to then find the speed because the tangent line will be straight and you can use any two points to find the slope.
 * [[image:Screen_shot_2011-09-14_at_6.46.52_PM.png]]
 * The graph above is velocity time graph for the cart moving down hill.
 * [[image:Screen_shot_2011-09-14_at_6.44.03_PM.png]]
 * The graph above is the velocity time graph for the cart moving up hill.

The results of this lab supported my hypotheses. I discovered that, like I predicted, the slope between points gets steeper as the speed increases. The position time graph was a polynomial function and resembled a curve. The equation of the line was y = -0.2075x 2 + 5.884x and the R 2 value was 0.9999 which indicates that the trend line was a great fit. The graph looks like the way it does because as speed increases, more distance is covered in less time, thus giving you a steeper slope. I was also correct in hypothesizing that a position time graph can show how much speed is increasing between any time increments. The change in slope tells you the change in speed over an interval. A possible source of error is that we lined up the ramp at a different part of the textbook then everybody else did. This was make our incline different and therefore make the cart move either slower or faster. Another source of error could have came from our measuring device. We had to estimate the measurements because the ruler only measures up to millimeters. We were also measuring from an angle which could have thrown things off. To minimize these issues, we could have measured how much of the ramp was on top of the book. That way, everybody in the class could be sure there books were in almost the exact same location. To eliminate the measuring problems, we could have used a more precise instrument and one that is flat.
 * Conclusion:**

September 15: Summarize Lessons 3 and 4
Lesson 3: Lesson 4
 * **What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.**
 * The slope of a position time graph reveals important information about the velocity of an object. If an object is moving at constant speed, then the slope will be constant and if the object is changing speeds then the slope will change. If an object is moving in the rightward (+) direction, then the slope of the line will be positive. The larger the velocity, the larger the slope. The slope of a position time graph is equivalent to the velocity of the object.
 * In order to find the slope, you must use this formula: [[image:U1L3c2.gif]]
 * **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 felt pretty confident with all of this information before the 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?**
 * We have discussed everything in class besides for how to calculate slope, but I already knew how to do that.


 * **What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.**
 * Velocity time graphs show whether an object is moving with constant velocity or accelerating. At constant velocity, acceleration equals 0, and the graph has a straight, positive line. When the object is accelerating, the graph is positive with a changing slope.
 * The slope of a velocity time graph represents the acceleration of the object. Therefore, a positive slope represents positive acceleration. A straight line has a slope of 0, which represents no acceleration.
 * **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 an object is moving away from the origin, it is considered positive velocity. On the other hand, when an object is moving towards the origin, the velocity is negative. The reading helped clarify this for me, because in class I was a little bit unsure about how to identify negative v positive velocity on the graph.
 * **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?**
 * The reading went in depth about how to find the area on a velocity time graph. The area that you calculate, whether it be of a rectangle or a triangle, represents the displacement. Velocity is equal to the total displacement/time so that is why this graph can show you the displacement.

Lab: A Crash Course in Velocity (Part II)
September 21 Partners: Ben Weiss, George Souflis, and Kaila Solomon

Objectives: Both algebraically and graphically, solve crashing and catching up problems. Then set up each situation and run trials to confirm your calculations.

Procedure (Part A)

media type="file" key="New Project 3 - Mobile.m4v" width="300" height="300"

Procedure (Part B)

media type="file" key="New Project 4 - Mobile 1.m4v" width="300" height="300"

Observations:

Part A: Part B:
 * Trial || CMV1 distance (cm) || CMV2 distance (cm) ||
 * 1 || 413.05 || 186.95 ||
 * 2 || 420.24 .................. || 179.76 ||
 * 3 || 415.12 || 184.88 ||
 * 4 || 414.63 || 185.27 ||
 * 5 || 417.54 || 182.46 ||


 * Trial || CMV1 distance (cm) || CMV2 distance (cm) ||
 * 1 || 173.25 ..................... || 82.25 ....................... ||
 * 2 || 178.53 || 76.97 ||
 * 3 || 174.68 || 80.82 ||
 * 4 || 170.99 || 84.51 ||
 * 5 || 176.81 || 78.69 ||

Calculations (Part A) (the unclear number at the bottom reads 182.025 cm)

For part A, we first set up an equation to find out after how many seconds the cars would collide. The equation we used is the first equation shown in the picture above. Once we knew how many seconds it would take, 6.08 seconds, we multiplied each velocity by 6.08 seconds to find out the distance the car would move in this given amount of time. The calculations are showed next on the paper. Underneath the math is a statement which reads, CMV1 and CMV2 will hit in 6.08s. CMV1 will move 417.26 cm (to the right) and CMV2 will move 182.025 cm (to the left).

Calculations (Part B)



For part B, we were trying to calculate where CMV1 would catch up to CMV2 if CMV1 started 100 cm behind CMV2. In order to do this, we used the equation above, which essentially set the two velocities times the amount of time equal, subtracting 100 from CMV1's side because CMV1 had to travel an additional 100 cm.

Analysis: The results we collected during our tests were very close to the results we got with our calculations. In order to verify how accurate our predictions really were, we calculated both percent error and percent difference. For the percent error calculations, we chose to use an average of all five trials for the experimental data because all of our numbers were within a 7 number range. The equation for percent error is: and the equation for percent difference is:. Percent error compares the results you get in the lab to the results you should have got according to your calculations. Percent difference compares your average experimental results to each individual result. Below is a picture of all of my work and the results I got for both percent error and percent difference. In both part A and B, I had very small percentages for both percent error and percent difference which leads me to believe that I was successful.

Part A:

Part B:

Discussion Questions
 * Where would the cars meet if their speeds were exactly equal?
 * If their speeds were exactly equal, they would meet 3 meters (or 300 centimeters) away from where they started. This is exactly in the middle and because they would be traveling at the same velocity, they would cover the same amount of distance in the same amount of time.
 * The car in the back would never catch up to the first one because they are traveling at the same speed and therefore will continue to be the same distance apart.
 * 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.
 * [[image:Photo_on_2011-09-21_at_20.29.jpg]]
 * The cars met after 6.08 seconds at 417.26 cm.
 * >> [[image:Photo_on_2011-09-21_at_20.46.jpg]]
 * The cars met at 2.59 seconds at 177.75 cm. CMV1 traveled 177.75 cm in 2.59 seconds and CMV2 travelled 77.75 cm in the 2.59 seconds, but started at 100 cm.
 * 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?
 * [[image:Photo_on_2011-09-21_at_20.51.jpg]]
 * A velocity time graph does not show position so you can't see when they are going to be in the same place at the same time.
 * A velocity time graph does not show position so you can't see when they are going to be in the same place at the same time.

Conclusion

This lab has taught me how to calculate when two objects will collide and when a faster object will catch up to a slower one. It also showed me how I can then test my theoretical results and then how to calculate the percent error and percent difference. In part A, we had predicted that CMV1 would move 417.26 cm. After running 5 trials, our average experimental result was 416.116. The percent error was very small, only 0.27%! The percent difference for each trial was small as well: they ranged from .24% to .99%. For part B, we predicted that CMV1 would catch up to CMV2 after it had moved 177.75 cm. The average of my experimental results was 174.85 cm. These were very close as well and the percent error was 1.63% and the percent differences for the 5 trials were between 0.92% and 2.21%. There were many sources of error. The most significant one was that our cart broke in the middle and we had to borrow another group's. Therefore, the velocity of the cart may have been different than what we had recorded. To eliminate this problem, we could have redone part I of the lab to find the velocity of this new cart. Another source of error could have been our technique to measure. It may not have been precise because were basing it off of what we saw. To be more accurate we could have been closer to the course in order to see the measurements better. We also could have used a more precise measuring tool.

Egg Drop Lab
Picture of Final:

Brief Discussion of Results: Our final egg drop was unsuccessful. Our design was made out of straws. The egg rested in a box of straws, between 4 straws placed in X with newspaper in between the parts of the x. This box had a straw roof on top. Underneath this box, we had another row of straws with scrunched up balls of paper separating the two. We designed it this way so that when it dropped, the paper balls would crush rather than the egg on top. However, when we dropped our project, this was not the case. The structure stayed in tact, but the egg inside completely broke.

Calculation for Acceleration:

Brief Analysis About Why Egg Broke: Our project broke because even though we tried to give it comfort underneath, the impact of the crash still broke the egg. Even though the egg didn't come into direct contact with the ground, it still did hit hard from the drop inside the box. In order to alleviate some of the pressure of the fall, we should have stuck with our original design of a cone. This way, there would be more space in between the egg and the fall and the impact would not have had such an impact. Our acceleration was 6.44 m/s/s, which is very fast. Acceleration does not go above 9.8 m/s/s. If we had been able to slow down the rate of acceleration with a parachute that would have also helped.

What We Would Do Differently: If we were to make a new prototype, we could probably combine all three of our prototypes to make a successful egg drop. Our first two prototypes were cones. They didn't work however because they were very short and just made out of paper. If we were to combine the cone idea with our straw structure, we could make a longer cone that had some extra comfort of straws. We would also make it taller than our other cones so that there would be a better chance of success. We also would add a parachute to slow down our acceleration rate.

Lesson 5: Summarizing Technique 1
A free falling object is an object that is falling under the sole influence of gravity. 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 A [|ticker tape trace] or dot diagram of free falling motion would depict an acceleration. 9.8 m/s/s is denoted as the symbol ** g **. Acceleration is the rate at which an object changes its velocity. Another way to represent this acceleration of 9.8 m/s/s is to add numbers to our dot diagram. One means of describing the motion of objects is through the use of graphs - [|position versus time] and [|velocity vs. time] graphs. In this part of Lesson 5, the motion of a free-falling motion will be represented using these two basic types of graphs. The line on the graph curves. A curved line on a position versus time graph signifies an accelerated motion. The position-time graph reveals that the object starts with a small velocity (slow) and finishes with a large velocity (fast). The negative slope of the line indicates a negative (i.e., downward) velocity. The line on the velocity vs. time graph is a straight, diagonal line. The velocity-time graph reveals that the object starts with a zero velocity and finishes with a large, negative velocity. An object that is moving in the negative direction and speeding up is said to have a negative acceleration. Free-falling objects are in a state of [|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 ** v **** f ** g * t where g is the acceleration of gravity. The value for g on Earth is 9.8 m/s/s. The distance that a free-falling object has fallen from a position of rest is also dependent upon the time of fall. This distance can be computed by use of a formula; the distance fallen after a time of t seconds is given by the formula d0.5 * g * t 2 where g is the acceleration of gravity. The answer to the question (doesn't a more massive object accelerate at a greater rate than a less massive object?) is absolutely not! That is, absolutely not if we are considering the specific type of falling motion known as free-fall. 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 actual explanation of why all objects accelerate at the same rate involves the concepts of force and mass. 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. Thus, 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 Fall Lab 10/5
Objective: What is the acceleration of a falling body?

Hypothesis: The velocity will increase as the weight is dropping. Therefore, the velocity time graph will be a linear graph. The slope should be 981cm/m 2 because the slope of a velocity time graph = acceleration and the rate of acceleration of a free falling object is 981cm/m 2. The position v time graph will have a nice curve starting at zero and then decreasing because the free falling object is dropping, or decreasing its height/position.

Data Table: Sample calculation for how I found the instantaneous velocity:

Sample calculation for how I found the mid-time: Class Data and Average for Slope: Analysis:

Discussion and interpretation of the equation of the graphs:
 * The equation of the velocity v time graph is y=754.43x-7.9786. The equation y=mx+b is derived from the equation, v f = v i +at. The y intercept is not equal to 0 because the timer may have mad a dot right before or after we dropped it. Therefore, by the time we got the first dot, the tape was already in motion. Our y intercept is -7.9786 cm/s 2, which is equivalent to .007 m/s 2 , which is very close to 0. The slope of the graph should have been 9.81 m/s 2 or 981 cm/s 2 because the slope of a velocity time graph is equivalent to the acceleration and the acceleration of a free falling object is 981 cm/s 2 . Our slope, however, was 754.43. The slope is not as steep as it should be, but it is still reasonable considering the many different sources of error we encountered during the lab (discussed in conclusion). The graph is linear because the slope is constant. This makes sense because the slope of the v-t graph, or the acceleration, should be constant for a free falling object. The R 2 value is 0.99277 which shows that the linear graph was a great fit and our results were very good. My hypothesis was therefore partially correct.
 * The equation of the position v time graph was y = 381.24x 2 - 13.85x +1.3508. The graph is a nice curve with an R 2 of 0.99971 which shows that the polynomial trend line was a good fit. This graph curves up, rather than down like I had hypothesized, because we did not mark the different positions as negative, or decreasing. Instead, we kept it all positive, which is okay, because direction is all relative. Technically, we dropped the weight from 0 and the values should have been negative, but it is okay that we did it this way. The equation for this position time graph, with variables, is y=Ax 2 + Bx, which is derived from y = 1/2at 2 +v i t. A = half the acceleration. Our value for A was 381.24. According to this, our acceleration, or slope, should have been 762.48cm/s2, instead of the 754.43 cm/s 2 that I got from the v-t graph. This is because on the y intercept = 0 on the position time graph. This is just another analysis and the fact that both accelerations are very close shows that I did my calculations correctly. B = initial velocity. Our B value is not equal to zero. This is probably because the weight was in motion before we actually released it and before the timer started making dots.

Percent Error and Percent Difference:

Percent Error: Percent Difference:

Discussion Questions Conclusion My lab results were very successful and supported my hypothesis. The v-t graph of a free falling object is linear because of constant acceleration and the position time graph is a curve because of increasing velocity. The equation of the line of my velocity time graph was y =754.43x - 7.9786 and I used a linear trend line. The R2 value was 0.99277, which shows that my results were very, very close to being exactly linear, therefore proving the hypothesis true. The equation of the position time graph was y=381.24x2 - 73.85x + 1.3508, and the trend line was polynomial with an R2 value of 0.9997, which again indicates that the hypothesis was correct. Due to many potential sources of error, my values were not exactly what they should have been and I did have a significant percent error and percent difference. A major problem was friction. The friction of the ticker tape in the spark timer definitely slowed down the rate at which the weight was accelerating and therefore my value was lower than 981 cm/s 2. Another very likely cause of error was when we were measuring, either the ticker tape or the measuring tape could have moved. Even the smallest movement would make a difference in our data. In order to avoid some of these errors, we could have tried using a different device to avoid friction. Of the tools we have used so far, we could try using the motion detectors. This might cause its own problems, but is an option. In order to account for the measuring problems, we could have used an instrument with less height and more exact measures. It is inevitable that there will be some error when performing an experiment, but our results lead me to believe that we did do a pretty good job.
 * Does the shape of your v-t graph agree with the expected graph? Why or why not?
 * Yes, the v-t graph's shape does agree with what I had expected. Although the line is not perfect, the shape of the graph is still linear, which is what I had anticipated. The slope of a velocity-time graph is equal to the acceleration and because the acceleration of a free falling object is constant, it makes sense that the slope of the v-t graph would be constant.
 * Does the shape of your x-t graph agree with the expected graph? Why or why not?
 * Yes and no, the shape of the x-t graph does agree with the expected graph. The graph is a curve starting at zero and increasing. I expected the graph to be the same shape, but in the opposite direction. Neither is incorrect, it just depends on how you interpret the data. The shape is what I had expected because the object moves quicker as it is getting farther away from the initial location. Therefore, the slope of the line, which is equal to the velocity, gets steeper as time goes on.
 * Did the object accelerate uniformly? How do you know?
 * No, the object did not accelerate uniformly. I know this because the graph is not 100% linear. Therefore, the slope between points, which is equal to the acceleration, is not the same throughout the entire time. If the acceleration had been uniform the whole time, the slope would have been constant, the trend line would have touched all points, and the r
 * value would have equaled 1.
 * What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?
 * It is hard to say what could make the acceleration due to gravity higher than 981 cm/s 2 . Other than errors made with the procedure, the acceleration could have been greater if there was some force of air pressure pushing down on the object.
 * The acceleration could have been lower than it should have been for a few reasons. In this lab, the most likely answer is friction. The ticker tape had to slide through the spark timer, and in many cases, also over the top. This definitely slowed down the rate of acceleration.

Class Notes (10/3) Free Fall
*ACCELERATION AT MAX HEIGHT DOES __NOT__ EQUAL 0!!!!
 * free fall - gravity is only force acting on it; no other forces; ignores air resistance completely!
 * acceleration of free falling object = -9.5 m/s/s
 * always write down the information you have
 * initial velocity
 * final velocity
 * time
 * acceleration
 * displacement
 * you always know that the acceleration is -9.8 m/s/s and that the initial velocity when something is dropped = 0
 * when you throw something up and then it falls down, it is symmetrical around max height
 * for free fall, you can never use v=d/t because it is not in constant motion
 * make sure to:
 * draw correctly
 * list what you have and what you want
 * use equation that makes sense