Ch2_SteeleM


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** Maddi's Physics Wiki - Ch2 ** =__ Lab: A Crash Course in Velocity __= Madison Steele and Ben Sherman - 9/6/2011

**Objective(s):**
 * 1) How Fast does a CMV go?
 * 2) What does a position-time graph tell us?
 * 3) How precisely are you required to measure?

**Materials**:
 * Constant Motion Vehicle
 * Tape measure /or metersticks
 * Spark timer
 * Spark tape

**Hypothesis:**
 * 1) Blue = 1/2 meters/second, yellow = 1/4 meters/second
 * 2) Position-time us the rate of the CMV
 * 3) All numbers that you know for sure __plus__ a guess number (2 decimals; last decimal = guess)

**Procedure:**
 * 1) Tape spark-tape to CMV
 * 2) Put the piece of spark-tape through the spark timer
 * 3) Adjust the spark timer to 10 hz, so that it sparks every 1/10 of a second
 * 4) Turn on the CMV, letting it pull the spark-tape all the way through the spark timer
 * 5) Measure the distance from the time=0 mark to each of 10 subsequent marks
 * 6) Graph data points on excel (see below)

**Data: x-t for CMV**

Our data is precise because the R 2 value is very close to 1. The R 2 value is how close my data points are to the line of best fit. The slope of a line on a x-t graph is the velocity. The line is straight and slope is constant, meaning that t he CMV has a constant velocity.
 * Analysis:**

y axis = change in position x variable = time Change in distance over time represents change in slope

**Discussion questions:**
 * 1) Why is the slope of the position-time graph equivalent to average velocity?
 * We found the slope of the line of best fit, which fits the trend of the CMV’s average velocity. The line of best fit/slope takes into account the all of the points on the chart, which were our car’s time and position. Slope is defined by change in position divided by change in time. This value is in cm/s, which is a measure of average velocity. This value, in meters per second, is 0.47999.
 * 1) Why is it average velocity and not instantaneous velocity? What assumptions are we making?
 * It is average velocity because we want to find the speed of the CMV. To find the speed of the CMV, it is necessary to use a line of best fit (which takes each data point into account). Instantaneous velocity only shows the velocity of an object at one point. By calculating the average velocity instead of the instantaneous velocity, we are assuming that the speed of the CMV is in fact not constant (some fluctuation).
 * 1) Why was it okay to set the y-intercept equal to zero?
 * The y-intercept was set to zero because the car started at zero milliseconds, and zero feet traveled. Regardless, the line of best fit/slope/speed does not change.
 * 1) What is the meaning of the R 2 value?
 * The R 2 value shows how close the plotted points are to the line of best fit. The closer your R 2 value is to 1, the more precise your plotted points are (relative to the line of best fit).
 * 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?
 * I would expect the graph of the slower CMV to be underneath my CMV graph. Slope is defined by change in position divided by change in time. Since the slower CMV changes position less frequently than my CMV does over the same amount of time, the slower CMV will have a smaller slope. Therefore, the graph of the slower CMV would fall below the graph of my CMV.

**Conclusion:**

Our line of best fit was 47.999, which means that we traveled at 47.999 centimeters per second. My (Ben's) hypothesis was that the blue CMV would travel at 2 ft/s, or .6096 m/s, and Maddi predicted that the blue CMV traveled at .50 m/s. Our results show that Maddi’s hypothesis was more accurate, and mine was much faster than our actual results. Other blue CMVs in the class traveled at .40196 m/s and .67186 m/s. My hypothesis was closer to the .67186 m/s CMV, and around the median for the three blue CMVs. There are several sources of error that could come into play during this lab. When the car travels, instead of going in a straight line, it can veer either left or right, which will cause the CMV to slow down and therefore lower the average velocity. Another factor is that the batteries in the CMVs may have been lower or higher than in other cars, which would cause the CMV to have a slower or faster velocity. Lastly, the floor that the CMV was tested on may have not been level, which would result in the CMV having to travel at an angle, which would impact its velocity. If the experiment was conducted again, these issues could be minimized by putting fresh batteries in each CMV, putting each CMV on the same, level surface, and by making sure that each CMV travels in a straight line, rather than turning.

**Homework** - 9/8/2011
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * I already understood that a scalar quantity, such as speed, is ignorant of direction and that a vector quantity, such as velocity, is "direction aware."
 * I also already understood the difference between distance and displacement. Distance (a scalar quantity) is how for an object has traveled in total. Displacement (a vector quantity) is an object's overall change in position.
 * 1) What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.
 * I was not confused about anything in class.
 * 1) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * If an object is moving in the negative direction, and is speeding up, why does it have negative acceleration?
 * 1) What (specifically) did you read that was not gone over during class today?
 * I read about acceleration, which was not gone over during class today. The reading explained constant acceleration, how to calculate an object's average calculation, and how to determine the direction of an accelerating vector.

**Notes** - 9/9/2011

Average speed - total distance + total time (sped-o-meter), instantaneous speed might be all over the place Constant speed - going the same speed the whole time, instantaneous speed is the same as the constant speed Instantaneous speed - speed at any given point V=change in distance/ change in time (m/s)

a=0 || a=0 || a ---> || <--- a || Acceleration
 * Types of Motion || Motion Diagram ||
 * At rest || v=0
 * Constant || --v--> --v--> --v-->
 * Increasing || -v-> --v--> ---v--->
 * Decreasing || ---v---> --v--> -v->
 * Causes velocity to get bigger
 * Increasing speed - points in same direction as velocity
 * Decreasing speed - points in opposite direction as velocity

Signs are arbitrary

Ticker tape diagram
 * Increasing speed - dots get further apart
 * Decreasing speed - dots get closer together
 * Constant speed - dots stay same distance apart

**Homework** - 9/9/2011
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * I understood how to interprate the dots on a ticker tape diagram. A large distance between dots means that the object was moving fast during that time interval. A small distance between dots means the object was moving slow during that time interval. A changing distance between dots represents changing velocity (acceleration), and a constant distance between dots represents constant velocity (no acceleration).
 * I also understood that when making a vector diagram, the magnitude of a vector quantity is represented by the size of the vector arrow. Arrows that are the same size represent constant velocity. Arrows getting progressively bigger or smaller in size represent acceleration.
 * 1) 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 little bit confused about the direction of acceleration. I didn't understand how an object could have negative acceleration (in certain situations), even if it was speeding up. The animation in the reading helped me understand that the direction of acceleration (negative or positive) is not only based on whether an object is speeding up or slowing down, but also on the direction of velocity.
 * 1) 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.
 * 1) What (specifically) did you read that was not gone over during class today?
 * I did not read anything that wasn't gone over in class.

=**__ Activity: Graphical Representations of Equilibrium __**= Madison Steele and Ben Sherman - 9/12/2011


 * 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?


 * Materials allowed**: Motion detector and USB link


 * Instructions:**
 * 1) Watch the demonstration for setting up and using Data Studio and the motion detector.
 * 2) Be sure to take screenshots of your graphs in order to document your data.


 * Data:**

The graph above shows a person walking (at a slow, constant speed) away from the motion detector, and then towards the motion detector. Run #1 = person walking away from motion detector. Run #2 = person walking towards motion detector.

The graph above shows a person walking away from the motion detector. Run #1 = person walking away quickly. Run #2 = person walking away slowly.

The graph above shows a person at rest. The person is standing less than 1.0 m away from the motion detector.


 * Discussion questions: **
 * 1) How can you tell that there is no motion on a…
 * __position vs. time graph__ - points will stay on zero m, so the line will have a slope of zero (horizontal and straight)
 * __velocity vs. time graph__ - points will stay at zero m/s, so the line will have a slope of zero (horizontal and straight)
 * __acceleration vs. time graph__ - the points will stay at zero m/s/s. The line will have a slope of zero (horizontal and straight)
 * 1) How can you tell that your motion is steady on a…
 * __position vs. time graph__ - the line will have a constant slope (either increasing or decreasing)
 * __velocity vs. time graph__ - positive value, no slope
 * __acceleration vs. time graph__ - slightly positive value with no slope
 * 1) How can you tell that your motion is fast vs. slow on a…
 * __position vs. time graph__ -<span style="background-color: transparent; color: #000000; text-decoration: none; vertical-align: baseline;"> the slope will be steeper on the faster graph, while the slower graph will rise more slowly and more steadily
 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">__velocity vs. time graph__ -<span style="background-color: transparent; color: #000000; text-decoration: none; vertical-align: baseline;"> if the motion is fast, the slope will be steeper. if the motion is slow, the slope will be less steep/more horizontal.
 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;"><span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">__acceleration vs. time graph__ - can't detect change if going quickly or slowly, but at a constant speed.
 * 1) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">How can you tell that you changed direction on a…
 * <span style="color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 12px;">__position vs. time graph__ - If you are heading away from the sensor, the line will have a positive slope. If you are heading towards the sensor, the line will have a negative slope.
 * <span style="color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 12px;">__velocity vs. time graph__ - Walking away from the motion detector, the velocity is above the x-axis. Walking toward the motion detector, the velocity is below the x-axis. The two lines are reflections of each other.
 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">__acceleration vs. time graph__ - You can’t detect change in direction with this graph. This is because acceleration=change in velocity/change in time. Acceleration does not take direction into account.
 * 1) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">What are the advantages of representing motion using a…
 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">__position vs. time graph__ - This graph shows an object’s position relative to the motion detector at a certain time. Because the graph shows whether an object is moving towards or away from the motion sensor (direction), it is possible to calculate the object’s velocity (velocity=change in position/change in time).
 * <span style="color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 12px;">__velocity vs. time graph__ - This graph shows an object’s velocity per unit of time. Therefore, it is easy to distinguish between an object moving quickly and an object moving slowly.
 * <span style="color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 12px;">__acceleration vs. time graph__ - This graph shows us an object’s acceleration (change in velocity/change in time). It also shows us if the change in velocity is positive or negative.
 * 1) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">What are the disadvantages of representing motion using a…
 * <span style="color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 12px;">__position vs. time graph__ - It doesn’t accurately show the acceleration (acceleration=change in velocity/change in time) of the walking person.
 * <span style="color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 12px;">__velocity vs. time graph__ - This graph doesn’t show the position of an object relative to the motion sensor.
 * <span style="color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 12px;">__acceleration vs. time graph__ - While this graph shows acceleration (+/- changes in velocity), it does not show the direction in which an object is moving.
 * 1) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">Define the following:
 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">__No motion__ - An object being stationary (no change in position) in relation to another object; velocity = 0, acceleration = 0
 * <span style="color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 12px;"> __Constant speed__ - Consistent change in position per unit of time; no acceleration.

**Notes** - 9/13/2011

**Notes** - 9/14/2011

=__**Lab: Acceleration Graphs**__= Madison Steele and Ben Sherman 9/14/11


 * Objectives:**
 * What does a position-time graph for increasing speeds look like?
 * Hypothesis – The points gradually increase and the slope is positive.
 * What information can be found from the graph?
 * Hypothesis – The graph shows an object’s position relative to its starting point per unit of time.

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


 * Important Notes:**
 * Be sure to write your hypothesis BEFORE doing the experiment.
 * All data should be recorded in an organized spreadsheet in Excel.
 * You need to document your procedure. Decide with your lab partner how you intend to do this.

media type="file" key="VIDEO0004 copy.mov" width="300" height="300"
 * Procedure:**

- y-axis = change in position x-axis time = time Equation on chart: Δ d = A(t2) + B(t) Theory equation: Δ d = 1/2(a)(t2) + Vi(t)
 * Data: x-t cart going down ramp**


 * Analysis:**
 * 1) Interpret the equation of the line (slope, y-intercept) and the R2 value.
 * The equation is y = 10.892 + 6.2919x
 * The slope represents the average velocity of the cart. Average speed = Δ distance/ Δ time
 * The R2 value represents how close the points are to the line of best fit. R2 = 0.99994, meaning that our data is precise (precise ≥ 0.95).
 * 1) 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.)
 * Slope at halfway point = 15.5 cm/s
 * Slope at end = 29 cm/s
 * 1) Find the average speed for the entire trip.
 * 17.1 cm/s


 * Discussion Questions:**
 * 1) What would your graph look like if the incline had been steeper?
 * The slope would be steeper.
 * What would your graph look like if the cart had been decreasing up the incline?
 * 1) Compare the instantaneous speed at the halfway point with the average speed of the entire trip.
 * The instantaneous speed at the halfway point (15.5 cm/s) is similar to the average speed of the entire trip (17.1 cm/s).
 * 1) Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense?
 * Instantaneous speed is the speed of an object at one specific point in time. A tangent line is one that touches a curve at one point only. We drew a line tangent to one specific point on the acceleration curve. Because the slope of the tangent line is constant, we knew that the specific point on our curve must share the same slope.
 * 1) Draw a v-t graph of the motion of the cart. Be as quantitative as possible.


 * Conclusion:**

Our hypothesis proved to be correct. In our results, the points gradually increased, which meant that the slope was positive and steadily increased (avg. speed = 17.1 cm/s), which is exactly what we said in our hypothesis. The real world version of this was that as the cart moved sped up going down the ramp, the ticker tape spaces became closer together, indicating higher speed and velocity. The second part of our hypothesis, in which we said that the purpose of a p-t graph was to show an object’s position relative to its starting point, is also true. For example, we were able to tell that after 0.5 seconds, the cart was 5.85 cm away from the starting point. I think that for the second part of our hypothesis, it would have been sensible to also say that a p-t graph can show the speed of an object by finding the instantaneous slopes. Here are some sources of error that may have contributed to inaccuracies or inconsistencies: 1. The rulers we used (the colored, slanted ones) did not allow for completely accurate measurements of the spaces between ticker-tape dots. 2. The data will be different for each group, depending on how many textbooks were used under the ramp and how close the textbooks were to the start of the ramp. 3. The data will be different for each group, depending on how many points were used on the graph and how much elapsed time was included on the graph (ex: our group only plotted 10 points over 1.0 second, whereas other groups may have decided to use 15 points over 1.5 seconds). In the future, it would be a good idea to use the flat, white rulers to allow for more accurate measurements. Also, each group should agree to use the same number of textbooks under the ramp, put the textbooks in the same, specific place under the ramp (ex: 1 inch from the beginning), and plot the same number of points over the same increment of time.

**Notes** - 9/15/2011



**Homework** - 9/15/2011 __Lesson 3__
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * I already understood how to calculate slope. In class we have been discussing that slope=change in distance/time.
 * I also understood that when an object is moving at constant speed, the slope of this object on a position time graph is constant (straight line).
 * 1) 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.
 * While I understood everything discussed in class, I didn't really absorb and feel //comfortable// with the material. After reading, I now feel more comfortable with the information and more confident about applying it to practice problems. I now have a better understanding of the importance of slope and am familiar with each position-time graph discussed in class.
 * 1) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * I understand everything.
 * 1) What (specifically) did you read that was not gone over during class today?
 * I didn't read anything that was not gone over during class today.

__Lesson 4__
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * I understood that the slope line on a velocity-time graph represents acceleration. If the line on the v-t graph is horizontal, it means that there is no acceleration.
 * I also understood that If the line is above the x-axis, the velocity is positive. If the line is below the x-axis, the velocity is negative.
 * [[image:U1L4a7.gif]]
 * If the line is getting further away from the x-axis (@ v=0), then the object is speeding up. If the line is getting closer to the x-axis (@ v=0), the object is slowing down.
 * [[image:U1L4a6.gif]]
 * 1) 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 understood everything in class but the reading helped //solidify// my understanding of it.
 * 1) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * I understand everything that I read.
 * 1) What (specifically) did you read that was not gone over during class today?
 * We didn't go over how to determine the area on a v-t graph.

=**__ Lab: A Crash Course in Velocity (Part II) __**= Madison Steele, Ben Sherman, Remzi Tonuzi, Julie

**Objectives**:

Both algebraically and graphically, solve the following 2 problems. Then set up each situation and run trials to confirm your calculations.
 * 1) Find another group with a different CMV speed. Find the position where both CMV’s will meet if they start //at least// 600 cm apart, move towards each other, and start simultaneously.
 * [[image:Screen_shot_2011-09-23_at_1.42.33_PM.png]]
 * [[image:a.JPG width="560" height="417"]]
 * 1) Find the position where the faster CMV will catch up with the slower CMV if they start //at least//1 m apart, move in the same direction, and start simultaneously.
 * [[image:Screen_shot_2011-09-23_at_1.41.16_PM.png]]
 * [[image:photo.JPG width="560" height="417"]]

**Available Materials**: Constant Motion Vehicle, Tape measure and/or metersticks, Masking tape (about 30 cm/group), Stop watch, spark timer and spark tape


 * Procedure: **
 * Crashing:
 * media type="file" key="VIDEO0004 (1).mp4" width="300" height="300"
 * Catching up:
 * media type="file" key="Movie on 2011-09-21 at 11.46.mov" width="300" height="300"

**Discussion questions:**
 * 1) Where would the cars meet if their speeds were exactly equal?
 * __For problem crashing situation__: At exactly 300 cm (middle). If the CMVs had equal speeds, they would be covering the same amount of distance over the same amount of time. Therefore, if they started at opposite ends, and the distance between them was 600 cm, they would meet at 300 cm.
 * __For problem catching up situation:__ They would never meet because they start 1 m apart and have the same speeds. There would be no way for one to catch up to the other.
 * 1) Sketch position-time graphs to represent the catching up and crashing situations. Show the point where they are at the same place at the same time.
 * Position-time graph crashing
 * [[image:photo_pic.jpg width="511" height="381"]]
 * Position-time graph catching up
 * [[image:photo_(4).JPG width="560" height="417"]]
 * 1) Sketch velocity-time graphs to represent the catching up situation. Is there any way to find the points when they are at the same place at the same time?
 * Velocity-time graph catching up
 * [[image:photo_(3).JPG width="560" height="417"]]
 * Is there any way to find the points when they are at the same place at the same time?
 * No. It is not possible to do this by using a velocity-time graph. This is because velocity-time graphs do not show the position of a moving object. The slope of a line on a velocity-time graph represents acceleration, not position.

Data Tables:
 * [[image:Data_chart_cmv_lab_picture_1.png]]
 * [[image:Data_chart_cmv_lab_picture_2.png]]
 * Sample calculations:
 * [[image:61f1291e-5b83-40e8-9ae5-a2e672975008.gif]]
 * [[image:%dif.png width="569" height="77"]]


 * Conclusion:**

For the crashing situation our hypothesis proved correct. Using algebra, we hypothesized that the CMVs would crash at 239.80 cm. Our experimental results were 195.56 cm (Trial 1) and 220.50 cm (Trial 2). To ensure that our data is accurate and precise, we calculated the percent difference and percent error. Percent difference compares each individual trial to the average of all trials. Percent error shows how close our experimental values are to our theoretical values. The percent difference for both Trial 1 (195.56 cm) and Trial 2 (220.50 cm) was 5.99%, which is relatively low and very consistent. The percent error for Trial 1 (195.56 cm) was 18.44 %, which isn’t that low. However the percent error for Trial 2 (220.50 cm) was 8.05%, which is very low. If I am remembering correctly, we messed up Trial 1 because the ramp we used was crooked. This may account for the difference in percent error between Trial 1 and Trial 2.

For the catching up situation our hypothesis proved incorrect. Using algebra, we hypothesized that the blue CMV would catch up with the fast yellow CMV (100 cm away from the blue CMV) at 299.19 cm. Our experimental results were 165.5 cm (Trial 1), 175.25 cm (Trial 2), and 167.25 cm (Trial 3). We did further analysis situation by calculating the percent difference and percent error. The percent error for Trial 1 (165.50 cm) 44.68%, the percent error for Trial 2 (175.25 cm) was 41.43%, and the percent error for Trial 3 (167.25 cm) was 44.02%. The percent error for all three trials was high, yet extremely consistent. This led me to think that something was wrong with one of my CMVs, rather than my calculations (which I checked meticulously!). The percent difference for Trial 1 (165.50 cm) was 2.26%, the percent difference for Trial 2 (175.25 cm) was 3.50%, and the percent difference for Trial 3 (167.25 cm) was 1.23%. The percent difference for all three trials is low and extremely consistent, which confirmed my suspicion of something being wrong with the CMVs. Since the percent difference is so low and consistent, the experiment must have been conducted correctly.

Upon watching my procedure video a few times, I noticed that the yellow CMV was very slow. I knew that each metal track I used in my experiment (to keep the CMVs from veering off) was 120 cm long. While watching the video, I counted 4 seconds from the time the yellow CMV started moving to the time the blue CMV caught up to it. In that time, the yellow CMV only traveled about 70 cm. I know this because the yellow CMV started 100 cm away from the blue CMV, and ended at about 170 cm (170-100=70). I then calculated the speed of the yellow CMV (v = 170 cm – 100 cm / 4 seconds) and got 14.5 cm/s. This did not correlate with the speed of the yellow CMV we used for the other trial (v = 31.95 cm/s). This makes total sense because I had to do the catching up trial a day later. I found my yellow CMV in Mr. Vanucci’s room, which could explain why it ran slower a day later (other students used it, therefore the batteries lost charge).

Sources of error include the CMV batteries and friction created by the track we used. Because I had to do the catching up trial a few days later, my CMV was used by other classes and the batteries were run down. To eliminate this problem in the future, I would make sure to use fresh batteries before each of the two trials. Also, I used a track to keep my CMVs from veering off. Sometimes, the tracks didn't match up perfectly, which created friction. Also, the CMVs had the naturally tendency to veer and the track prevented them from doing so. This resistance between the CMVs trying to veer and the track also caused friction, which would have slowed the CMVs down. To avoid this problem, I would make sure to use CMVs that didn't have the tendency to veer off.

=__Project: Egg Drop__= Madison Steele & Andrea Aronsky

**Procedure**:
 * 1) Design and build a package that has the lightest possible weight, and yet durable enough to protect an egg when dropped from a height of about 14 m. The contraption should result in the egg decelerating slowly upon impact.
 * 2) Test the 1st prototype. Make changes to design.
 * 3) Test the 2nd prototype. Make changes to design.
 * 4) Drop egg on due date from Mr. Saxon’s room. Record the time it takes for the contraption to hit the ground after release.
 * 5) Record and analyze results.


 * Description:**
 * [[image:egg.jpg]]
 * The cone is made out of construction paper. There are straws going the through the bottom of the cone which served as a base for the egg. The straws going through the top of the cone served to protect the egg from shifting or falling out. There is tape around the straws to prevent the straws from shifting when the contraption hit the ground. The parachute is made of two pieces of printer paper connected by sewing string. The parachute was connected to the contraption by two pieces of string threaded through holes.
 * **Final weight:** 28 grams


 * Drop and Time:**
 * We dropped the contraption from a height of about 14 m.
 * It took 2.03 seconds for the contraption to hit the ground.
 * Our egg survived the fall intact!


 * Analysis:**
 * Calculation of acceleration
 * [[image:fk.JPG width="560" height="417"]]
 * Comparison of acceleration to g = 9.8 m/s 2
 * Near Earth’s surface, an object in free fall in a vacuum will accelerate at about 9.8 m/s 2 (due to gravity), independent of its mass. However, my contraption had an acceleration of 4.13 m/s 2 . This is because there is air on earth, and therefore air resistance. Air resistance refers to the forces that oppose the relative motion of an object through a fluid, liquid, or gas. Because my contraption had a parachute, there was more air resistance which caused it to accelerate at a slower rate during free fall.
 * What would you do differently?
 * I would use printer paper instead of construction to make the contraption lighter. However, I would not change the structure of my design because the egg remained intact!


 * Discussion: **
 * Our first prototype was two cones (one inside the other) with newspaper separating them. It had a parachute made of one sheet of printer paper. This prototype saved the egg from cracking when it hit the ground, however the egg rolled out and cracked after initial impact. Our second prototype was the same design, except that it had tape over the top to prevent the egg from rolling out. However, our egg broke when we dropped this prototype. Our last prototype was a totally different design than the previous two. We wanted something lighter and more compact. We decided not to use any newspaper for cushioning, and instead used straws to prevent the egg from making contact with the ground.

=__Quantitative Graph Interpretation__=

**Homework** - 10/4/2011 Summary Method #1
 * 1) **//T//**//rivia//: Draw a line through or delete anything that seems trivial or frivolous (adjectives, similar examples, transition words)
 * 2) **//R//**//edundancies//: Draw a line through or delete any repetitive information or examples.
 * 3) **G**//eneralize//: Replace lists of specific items with general terms and phrases. (Ex: if text lists mirrors, lenses, prisms and thin films, you can substitute "optical materials" for all of those.)
 * 4) **//T//**//opic// **//S//**//entence:// Write a good topic sentence for the material (the subject and the author’s claim about it)

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">__Lesson 5: Free Fall and the Acceleration of Gravity__ <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">a) Introduction to Free Fall <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">b) The Acceleration of Gravity <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">c) Representing Free Fall by Graphs <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">d) How Fast? and How Far? > falling object after a time of <span class="boldred" style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">t <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">seconds is vf = g * t.
 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">**The author explains what it means for an object to be in free fall.** 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. Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s, a ticker tape diagram of its motion would depict an acceleration. Free-fall acceleration can be demonstrated by a strobe light. Instead of seeing a stream of water free-falling from the medicine dropper, several consecutive drops with increasing separation distance are seen.
 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">**The author introduces the symbol “g” for the acceleration of gravity.** The numerical value for the acceleration of a free-falling is known as the <span class="boldred" style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">acceleration of gravity <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;"> - the acceleration for any object moving under the sole influence of gravity. The symbol to denote the acceleration of gravity (9.8 m/s/s downward) is g.
 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">**The author explains the position-time and velocity-time graphs for an object in freefall.** A position versus time graph for a free-falling object is shown below:
 * [[image:ama.png]]
 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">Since a free-falling object is undergoing an acceleration (g = 9.8 m/s/s), it would be its position-time graph is curved. The object starts with a small velocity (slow) and finishes with a large velocity (fast). The negative slope of the line indicates a negative velocity. A velocity versus time graph for a free-falling object is shown below:Since a free-falling object is undergoing an acceleration (g = 9,8 m/s/s, downward its velocity-time graph is diagonal. The object starts with a zero velocity and finishes with a large, negative velocity; the object is moving in the negative direction and speeding up. An object that is moving in the negative direction and speeding up has a negative acceleration.
 * [[image:asdf.png]]
 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">Since a free-falling object is undergoing an acceleration (g = 9.8 m/s/s) downward its velocity-time graph is diagonal. The object starts with a zero velocity and finishes with a large, negative velocity; the object is moving in the negative direction and speeding up. An object that is moving in the negative direction and speeding up has a negative acceleration.
 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">**The author claims that that time plays a role in the velocity of an object in freefall.** The velocity of a free-falling object is changing by 9.8 m/s every second. 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
 * [[image:aaaaa.png width="182" height="280"]]

**Notes** - 9/9/2011
 * Free Fall**



=__<span style="font-family: Arial,Helvetica,sans-serif; font-size: 18px;">Lab: Free Fall __= <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">By Madison Steele and Ben Sherman


 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">Objectives: **
 * 1) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">What is the acceleration of a falling body?
 * 2) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">Predict the velocity-time graph of acceleration due to gravity.
 * 3) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">How will you find "g" from this graph?


 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">Hypothesis: **
 * 1) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">-9.8 m/s2
 * 2) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">Velocity-time graph:
 * [[image:U1L5c2.gif]]
 * 1) I will find g by the slope of the line on the v-t graph (acceleration).

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">**Data & Analysis**:
 * Data table
 * [[image:Free_Fall_Data_Screenshot.png]]
 * Position-time graph
 * [[image:Free_Fall_x-t_graph.png]]
 * Velocity-time graph
 * [[image:Free_Fall_v-t_graph.png]]
 * Analysis
 * Equation of line on x-t graph: y = 360.15 x 2 + 64.924 x or [y = A (t 2 ) + B (t)]
 * Equation of line on v-t graph: y = 708.28x + 68.25
 * Theoretical equation: Δ d = 1/2(a) (t 2 ) + Vi (t)
 * B = Vi
 * A = 1/2(a) --> 360.15 cm = 1/2(708.28 cm [slope of v-t graph]) --> 3.6015 m = 1/2(7.0828 m) --> 3.6015 m = 3.5414 m)
 * However, 3.6015 m does not equal 1/2(7.0828 m)
 * The slope of the v-t graph isn't exactly the same as doubling the coefficient. This is because Excel preforms different statistical analyses for polynomial and linear graphs.
 * Percent error and percent difference
 * [[image:stuff123.jpg]]
 * % error = 27.76 %
 * % difference = 12.15 %
 * Sample calculation of velocity and mid-time
 * [[image:abcdefg.jpg]]
 * Note for velocity calculation: 8.81 //cm// and 0.1 //seconds//
 * Note for mid-time calculation: 0.1 //seconds// and 0 //seconds//


 * Discussion questions: **
 * 1) Does the shape of your v-t graph agree with the expected graph? Why or why not?
 * No. Because we measured everything as a positive value. For example, we considered the distances between the dots on the ticker tape diagram to be positive distances.
 * 1) Does the shape of your x-t graph agree with the expected graph? Why or why not?
 * Yes. The graph is a j-curve with a positive slope, which is what I predicted. As time increases on the x-axis, the distance from the starting point increases on the y-axis. It is a j-curve because the weight is in free fall, which means that it accelerates as it falls.
 * 1) How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.)
 * The percent difference for my experimental trial was 12.2%. This signifies that our acceleration of gravity (7.08 m/s2) was 12.2% lower than the class average (8.06 m/s2). Our result was moderately precise, as our value for acceleration of gravity was close to the class average. However, our result wasn’t accurate because it deviated greatly from the actual value of acceleration of gravity (9.08 m/s2).
 * 1) Did the object accelerate uniformly? How do you know?
 * Yes the object accelerated uniformly. I know this because the line on the v-t graph has a constant slope, which means that acceleration is uniform. The velocity changes at a constant rate.
 * 1) What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?
 * There is no way to make the acceleration due to gravity higher except for error while doing calculations. This is because acceleration due to gravity can never exceed 9.8m/s2 on earth. However, the acceleration due to gravity would be lower if there was friction on the ticker tape while it was being dropped.

The execution of this experiment yielded results that disproved my hypothesis. I hypothesized that the acceleration due to gravity of my free falling weight would be a uniform 9.8 m/s2. However, because of friction it was still uniform but less than 9.8 m/s2. Our object fell at an acceleration of 7.08 m/s2, which is 27.76% less then 9.8 m/s2 (due to friction). The percent error was 12.15%, meaning that our results were precise (close to class average). This means all groups experienced friction, which slowed down the falling weight. In the future, I would make sure no parts of my body were touching the ticker tape, in order to reduce friction. Also, the initial velocity didn't equal exactly zero. This is because the first ticker tape dot wasn't exactly at zero. In order to get the initial velocity to equal exactly zero, I would factor the time and distance elapsed into my calculations.
 * Conclusion:**