Ch2_ShermanB

= = toc

= 9.06 Classnotes =



Ben Sherman Honors Physics - Chapter 2

= Lab: A Crash Course in Velocity (Part 1) = Honors Physics ** Hypothesis- **1.) The CMV (blue) appears to move at 2 fps, and the yellow CMV appears to move at 1 fps. -A position time graph tells us how fast something moves. This is because it tells us the difference in position (distance) over time. -We need to measure to the tenth m/s

** Data **



** Objectives- **1.) How fast does a CMV go? 2.) What does a position-time graph tell us? 3.) How precisely are you required to measure? Our line of best fit was 47.999, which means that we traveled at 47.999 centimeters per second, or .47999 meters per second. My hypothesis was that the the blue CMV traveled at 2 ft/s, or .6096 m/s, and Maddi predicted that the blue CMV traveled at .5 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, which will lower the average velocity. Another factor is that the batteries in the CMVs may have been lower than in other cars, which would cause the CMV to have a slower 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 the CMV, putting the CMV on a level surface, and by making sure that the CMV travels in a straight line, rather than turning.
 * Discussion Questions **
 * 1) Why is the slope of the position-time graph equivalent to average velocity?
 * 2) 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 .47999.
 * 3) Why is it average velocity and not instantaneous velocity? What assumptions are we making?
 * 4) 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).
 * 5) Why was it okay to set the y-intercept equal to zero?
 * 6) 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.
 * 7) What is the meaning of the R 2  value?
 * 8) 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).
 * 9) <span style="background-color: transparent; color: #000000; font-family: Times New Roman; font-size: 16px; text-decoration: none; vertical-align: baseline;">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?
 * 10) <span style="background-color: transparent; color: #000000; font-family: Times New Roman; font-size: 16px; text-decoration: none; vertical-align: baseline;">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 **

= CLASS NOTES - 9-7-11 =



1-D Kinematic, HW 9-8-11, Lesson 1abc
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) From class, I fully understood vector and scalar units. A scalar unit describes something that has a number value, such 1 mile per hour, or 1 cookie, while a vector is something that has a number value as well as a direction. For example, a vector would be 10 miles per hour west. I also understood the distance between distance and displacement. Distance is a scalar unit describing how far something has traveled in total (ex: 10 miles). Displacement is a vector that describes how far something is from the origin, and can be canceled out of the object travels in the opposite direction that it originated from. For example, you could walk 10 miles north and then walk 10 miles south, with a displacement of zero miles.
 * 3) 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.
 * 4) I understood all the topics in the reading that were discussed in class.
 * 5) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 6) I understood all the topics that were discussed in the reading.
 * 7) What (specifically) did you read that was not gone over during class today?
 * 8) In the reading, the topics that were introduced were the definition of mechanics and kinematics.

= __Types of motion (9.09.11)__ =

-Average Speed – Total distance divided by time -Constant Speed – going at a same speed for an extended period of time -Instantaneous Speed – speed at a certain point, rather than over an extended period of time -all of the speed calculations can be found with the equation v = ∆D/∆T -four types -at rest -constant -increasing -decreasing -How to display the types of motion -motion diagram -v = 0 -a = 0 -constant speed v v v -> -> -> a = 0 -arrows are same length because velocity isn’t changing, acceleration is 0 -increase in speed v v v -> --> > a -> -arrows getting longer because speed is increasing -acceleration is causing the arrows to get larger, points in the same direction as velocity -decrease in speed v v v ---> --> -> <- a -acceleration for decreasing speed always points in opposite direction from velocity -increasing speed downward is the increase in speed diagram rotated -decreasing speed updard is the same as decreasing speed diagram rotated with the velocity arrows going upward

-Signs are arbitrary (random) -look at a coordinate plane for positive or negative velocity/speed/acceleration -when velocity and acceleration have the same sign, the object is increasing in speed

-Ticker tape diagram -dots have a constant difference between them when speed is constant -dots have increasing distances between them when increasing -dots have decreasing distances between them when decreasing

Motion and constant acceleration applet []

= CLASS NOTES 9-12-11 =



= **__ 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

Hypothesis -The blue CMV appears to move at 2fps, and the yellow CMV appears to move at 1fps. A position-time graph tells us how fast something moves. This is because it tells the difference in position (distance) over time. We need to measure to the 10th of a ms.

**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. This graph shows the person returning. Run 1 is slower, run 2 is faster.

** Discussion questions: **
 * 1) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">How can you tell that there is no motion on a…
 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">__position vs. time graph__ - <span style="background-color: transparent; color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 12px; text-decoration: none; vertical-align: baseline;"> points will stay on zero m, so the line will have a slope of zero (horizontal and straight)
 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">__velocity vs. time graph__ - <span style="background-color: transparent; color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 12px; text-decoration: none; vertical-align: baseline;"> points will stay at zero m/s, so the line will have a slope of zero (horizontal and straight)
 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">__acceleration vs. time graph__ - <span style="background-color: transparent; color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 12px; text-decoration: none; vertical-align: baseline;"> the points will stay at zero m/s/s. The line will have a slope of zero (horizontal and straight)
 * 1) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">How can you tell that your motion is steady on a…
 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">__position vs. time graph__ - <span style="background-color: transparent; color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 12px; text-decoration: none; vertical-align: baseline;"> the line will have a constant slope (either increasing or decreasing)
 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">__velocity vs. time graph__ <span style="background-color: transparent; color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 12px; text-decoration: none; vertical-align: baseline;"> - positive value, no slope
 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">__acceleration vs. time graph__ - <span style="background-color: transparent; color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 12px; text-decoration: none; vertical-align: baseline;"> slightly positive value with no slope
 * 1) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">How can you tell that your motion is fast vs. slow on a…
 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">__position vs. time graph__ - <span style="background-color: transparent; color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 12px; 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; font-family: Arial,Helvetica,sans-serif; font-size: 12px; 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;">__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__ - <span style="color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 12px;">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__ <span style="color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 12px;"> - 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__ - <span style="color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 12px;"> 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.

= CLASS NOTES 9-13-11 =



= CLASS NOTES 9-14-11 =



= __**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.mp4" width="300" height="300"
 * Procedure:**

Data:


 * 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

> 2.)What would your graph look like if the cart had been decreasing up the incline? = =
 * Discussion Questions:**
 * 1) What would your graph look like if the incline had been steeper?
 * The slope would be steeper
 * If the cart had been increasing up the incline, the line would have started out towards the bottom, because the ticker tape box would have been at the bottom, so the position would start at zero. Then, it would have a sharp increase (steep slope) until the car started to slow down, where the line would become more gradual and the slope would start to decrease.
 * 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

__Kinematics, 9.14.11__
 * V = ∆D/∆T
 * Constant and average speed only
 * V = Vi+VF/2
 * Average speed only
 * ∆D/∆T = Vi+VF/2
 * ∆D = ½(Vi+VF) ∆T
 * used for acceleration, when initial and final speeds are different from one another
 * Acceleration -> rate that your speed changes
 * a= VF-Vi/∆T
 * VF= Vi + at <- same as above equation, rearranged∆D, Vi, Vf, t
 * Vi, Vf, t, a
 * Use each equation depending on what info you have
 * Equations can be subbed in if you need
 * ∆d = Vit + 1/2at2
 * Vf2= Vi2 + 2a∆d
 * Vi and Vo are the same

__Kinematics HW, 9.14, Lesson 1e__
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) From class, I fully understood that acceleration is the rate at which an object’s velocity changes, instead of how fast the object is moving. That means acceleration is the change in an object’s velocity, over time, and is generally measured in m/s/s. I also very well understood that an object with a constant velocity has no acceleration. This makes sense, because if the velocity is constant, there is no change, so there can’t be acceleration.
 * 3) 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.
 * 4) I fully understood everything discussed in class that was discussed in the reading.
 * 5) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 6) I fully understood everything from the reading.
 * 7) What (specifically) did you read that was not gone over during class today?
 * 8) In class, we didn’t go over the motion of a free falling object. I’m not sure that if either today, or in previous days, if we went over the fact that if an object is slowing down, it has negative acceleration.

= CLASS NOTES 9-15-11 =



__P-T Graphs, 1-D Kinematics, Lesson 3, 9.15.11__
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) From class, I fully understood the shape of p-t graphs and how they change based on the situation. For example, a p-t graph that of an object traveling quickly in the positive direction with constant velocity is a straight line with a constant slope that travels at about a 45 degree angle starting at point 0,0. A graph that has a an object that is increasing speed will start at 0,0 and have a gradually increasing slope/line. I also fully understood how the slope of a p-t graph is the velocity of the object. This means that any change in slope is a change in velocity of the object. For example, the slope is increasing, then so is the velocity.
 * 3) 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.
 * 4) I fully understood everything from class, so the reading didn't clarify.
 * 5) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 6) I fully understood the reading.
 * 7) What (specifically) did you read that was not gone over during class today?
 * 8) In class, we discussed everything that was covered in the reading.

__P-T Graphs, 1-d Kinematics, Lesson 4, 9.15.11__
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) I fully understood the link between v-t graphs and acceleration. Similar to a p-t graph and velocity, the slope of a v-t graph is an object's acceleration. If the v-t graph has positive slope, the object has positive acceleration. If it has a negative slope, it has negative acceleration, and the same holds true for constant velocity/slope and when at a standstill. I also fully understood how to take the slope of the line of a v-t graph. This is the same as for any straight line, so I had experience with finding the slope of a line from previous times (as well as in class).
 * 3) 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.
 * 4) I understood everything from class that was covered in the reading.
 * 5) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 6) I understood everything discussed in the reading.
 * 7) What (specifically) did you read that was not gone over during class today?
 * 8) In the reading, the only topic that wasn't discussed in class is finding the displacement of an object on a v-t graph. To do this, you find the area of the trapezoid, rectangle, or triangle that is made by the line

= LAB: CRASH COURSE IN VELOCITY =


 * Objectives ** :

Both algebraically and graphically, solve the following 2 problems. Then set up each situation and run trials to confirm your calculations.

a. 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.

b. 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.


 * Available Materials ** :

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

Catching up - media type="file" key="VIDEO0001.3gp" width="300" height="300" Crashing- media type="file" key="VIDEO0004.mp4" width="300" height="300"
 * Procedure **


 * Discussion questions **

DATA: Blue CMV: Yellow CMV:
 * 1) Where would the cars meet if their speeds were exactly equal?
 * 2) If the cars had equal speeds, they would collide at half their starting distance.
 * 3) If the cars had equal speeds and were started 1m apart, the CMV2 would not overtake with CMV1. If they are traveling at the same speed, it is impossible for one to overtake the other. The only location where the cars would meet is at the finish, 1m apart.



>
 * 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.
 * 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?

PERCENT ERROR AND PERCENT DIFFERENCE

CONCLUSION After using algebra to solve both parts, our hypothesis was found to be not exact, but somewhat correct In the colliding scenario, we found that the crash would occur at 238.58 cm. For out 3 experimental runs, our results were A.) 195.16 cm, B.) 220.5 cm, and C.)200 cm. For these three runs, our percent error was A.) 18.03%, B.) 7.58%, and C.) 16.3%. These three results ranged from fairly inaccurate (Run A and B), to an more acceptable percentage (Run B). All of these results weren't extremely accurate, and could have been better. For our percent difference, the average was 205.35cm, and the results were more precise. For run A, the percent difference was 4.77%, for B.) 7.38%, and C.) 2.61%. Ultimately, our percent difference seemed to be more precise than our percent error was accurate for the colliding question (part A).

For part B, the catching up question, we found that the blue CMV should have to travel 294.23 cm to catch up to the yellow CMV. Our results led us to conclude that part B's results weren't very accurate. For the three runs that we ran on part B, we had the results A.) 220 cm, B. ) 240 cm, and C.) 239 cm. For these experiments, our percent error was fairly significant. For A.) it was 25.22%, for B.) 18.43%, and for C.) 18.77%. These numbers were too high, which means that there was a variable that was changed in our trials. For the percent difference of the catching up situation, we had an average of 233. Our results were, for A.) 5.57%, B.) 3%, and C.) 2.58%. Our results for part B were very precise, but not very accurate.

In our experiment, there were several factors that could have led to our fairly inaccurate results. The most preeminent is that for the blue CMV, Maddi and I had labeled our CMV to be able to find it later. After searching through the CMV's, we were unable to find a yellow CMV labeled with Julie and Remzi's names. We used the yellow CMV that we figured was most likely theirs, but it may have not been, which meant that their data wouldn't fit that particular CMV, which would lead to incorrect results. Another issue is that found many groups to have issues with the CMV's veering, which would change the velocity of the CMV to be inaccurate. Our group decided to use track for the cars to drive on so that they wouldn't change direction and would thus "stay on track." In some failed trials, the car would fall off the track. In the rest, it would stay on. I think an issue could have arose with the track. Since the CMV's tried to veer when not on the track, they had to be trying to change direction on the track as well. This could lead to the CMV creating friction on the track, which would slow it down. On the catching up run, one CMV was on the track, but one was not. The car that wasn't on a track would ride up on the side of the track, which could cause it to also have friction created, which would slow down one CMV. That could explain why question B was so inaccurate. Lastly, the CMV's were used by different groups in different periods, which could have led to a significant decrease in battery, which would cause the CMV's to be much slower.

To make the results of this lab more accurate and reliable, I think there is an important change that should be made. The first is that there should be an apparatus to keep the CMVs straight when they tare traveling in both the overtaking and passing runs. This device would need to be similar to what my group used to force the CMVs to travel straight, but it should have sides that hold the CMV on the ramp. Also, the platform on which the CMV travels should be level with one another, which was an issue that we faced. When the CMV would pass from one platform to another, the CMV would slow down because of differences in height, which would change our results. Without the ramps, the CMVs would diverge from going strait, which would change their velocity as well as prevent the CMVs from crashing into one another or catching up. If they were on platforms, we would be better able to point out the crashing/passing locations.

= Egg Drop Project/Results =

Procedure - The procedure of the egg drop was relatively simple. An egg was placed in or device, and was dropped out a third story window (8.5m).

Weigh in - On our final device, the total weight of the device and egg was 70.45g, with the egg weighing in at 55.25g. Our device weighed 15.2g.

Results:

A picture of our final design:

-On our final drop, the was considered smashed. After removing the covering foil, I saw that the egg was leaking slightly, with a large crack. It did take a large amount of time for our egg to fall (1.58 and 1.75 seconds, averaged to 1.67 seconds). Our acceleration came out to be 6.09 m/s^2.

Calculations and Analysis: ΔD=(Vo)(t)+1/2(a)(t^2) 8.5m=(0)(1.67s)+1/2(a)(1.67s^2) 8.5=1/2(a)(2.79) 17=2.79a 17/2.79=a a=6.09m/s^2 This acceleration is an acceptable value, as it is below 9.8m/s^2, which is acceleration due to gravity.

Since our egg was broken, there were obviously flaws in our final design. In our first prototype, we used a double circular parachute with aluminum balls inside the carrier, yet it failed because the aluminum wasn't evenly covering the egg. On our second prototype, we used a hexagonal parachute with thread to connect the parachute and the carrier, and also used an aluminum cocoon to surround the egg. This device failed because of the weak thread, which broke as soon as we deployed the device. In our final model, we tried to take the best of both worlds - the hexagonal parachute to hold in more air and to slow the device down more, tape to attach the carrier to the parachute, aluminum that was neither a cocoon-like shell or torn pieces, but rather an egg enclosed in aluminum, but not too tight. This design was flawed, leading to the breaking of the egg. I believe that the rest of the device was strong, but the aluminum covering of the egg was the main point of failure. If we had redesigned egg enclosure I think our device would have been perfect, mainly because it was extremely light. I think that a tight aluminum shell for the egg surrounded by something like what we had used on the final model.

= Class 10-03 =


 * Freefall - an object moving under the influence of gravity only.
 * v=d/t <- DON'T USE, no acceleration.
 * falling objects have the same time for falling and increasing acceleration

HW, 10/3/11 - 1D Lesson 5

5.1 A free falling object is an object that is falling under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. Two important motion characteristics that are true of free-falling objects: Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s, a [|ticker tape trace] or dot diagram of its motion would depict an acceleration. The dot diagram at the right depicts the acceleration of a free-falling object. Object at regular time intervals - say, every 0.1 second - is shown. ball is speeding up as it falls downward. If an object travels downward and speeds up, then its acceleration is downward. Free-fall acceleration strobe light demonstration. The room is darkened and a jug full of water is connected by a tube to a medicine dropper. The dropper drips water and the strobe illuminates the falling droplets at a regular rate - say once every 0.2 seconds. Instead of seeing a stream of water free-falling from the medicine dropper, several consecutive drops with increasing separation distance are seen. The pattern of drops resembles the dot diagram shown in the graphic at the right. __//**-Free falling objects are falling solely under gravity and accelerate at 9.8 m/s/s.**//__
 * Free-falling objects do not encounter air resistance.
 * All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s (often approximated as 10 m/s/s for //back-of-the-envelope// calculations)

5.2 It is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. The symbol g. The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s/s. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude. Occasionally use the approximated value of 10 m/s/s g = 9.8 m/s/s, downward( ~ 10 m/s/s, downward) The value of the acceleration of gravity (**g**) is different in different gravitational environments. Even on the surface of the Earth, there are local variations in the value of the acceleration of gravity (**g**). These variations are due to latitude, altitude and the local geological structure of the region. To accelerate at 9.8 m/s/s means to change the velocity by 9.8 m/s each second.If the velocity and time for a free-falling object being dropped from a position of rest were tabulated, then one would note the following pattern. Object’s velocity is changing by 9.8 m/s each consecutive second. That is, the free-falling object has an acceleration of approximately 9.8 m/s/s __//**The acceleration of a free falling object is a constant 9.8m/s/s, but has slight variations depending on where in the world an experiment is done.**//__
 * ** Time (s) ** || ** Velocity (m/s) ** ||
 * 0 || 0 ||
 * 1 || - 9.8 ||
 * 2 || - 19.6 ||
 * 3 || - 29.4 ||
 * 4 || - 39.2 ||
 * 5 || - 49.0 ||

5.3 A position 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), it would be expected that its position-time graph would be curved.Object starts with a small velocity (slow) and finishes with a large velocity (fast). Slope of any position vs. time graph is the velocity of the object ( [|as learned in Lesson 3] ), the small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity. Finally, the negative slope of the line indicates a negative (i.e., downward) velocity. A velocity versus time graph for a free-falling object is shown below.

Observe that the line on the graph is a straight, diagonal line. Since a free-falling object is undergoing an acceleration (g = 9,8 m/s/s, downward), velocity-time graph would be diagonal. Object starts with a zero velocity (as read from the graph) and finishes with a large, negative velocity; that is, the object is moving in the negative direction and speeding up. An object that is moving in the negative direction and speeding up is said to have a negative acceleration .Slope of any velocity versus time graph is the acceleration of the object the constant, negative slope indicates a constant, negative acceleration. An object moving with a constant acceleration of 9.8 m/s/s in the downward direction. __//**A position-time graph representing free fall has a curve -it starts slow, then finishes with a high velocity. A velocity-time graph of a free falling object starts at zero then has a diagonal line in the negative direction with a slope of -9.8; it starts slow then ends fast.**//__

5.4 If dropped from a position of rest, the object will be traveling 9.8 m/s (approximately 10 m/s) at the end of the first second, 19.6 m/s (approximately 20 m/s) at the end of the second second, 29.4 m/s (approximately 30 m/s) at the end of the third second, etc. Thus, the velocity of a free-falling object that has been dropped from a position of rest is dependent upon the time that it has fallen. The formula for determining the velocity of a falling object after a time of t seconds is vf = g * t where g is the acceleration of gravity. The value for g on Earth is 9.8 m/s/s. The above equation can be used to calculate the velocity of the object after any given amount of time when dropped from rest. Example Calculations:At t = 6 svf = (9.8 m/s2) * (6 s) = 58.8 m/sAt t = 8 svf = (9.8 m/s2) * (8 s) = 78.4 m/sd = 0.5 * g * t2 g is the acceleration of gravity (9.8 m/s/s on Earth). Example Calculations:At t = 1 ad = (0.5) * (9.8 m/s2) * (1 s)2 = 4.9mAt t = 2 sd = (0.5) * (9.8 m/s2) * (2 s)2 = 19.6mAt t = 5 ad = (0.5) * (9.8 m/s2) * (5 s)2 = 123 m(rounded from 122.5 m)

__//**Free falling objects travel at constant acceleration, so their final velocity can be calculated by multiplying time by 9.8m/s/s. The distance of travel of a free falling object is based on the time of fall, and can be calculated with the equation d = (.5)*(g)*(t2)**//__

5.5 Value (known as the acceleration of gravity) 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. 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-fall is the motion of objects that move under the sole influence of gravity; free-falling objects do not encounter air resistance. More massive objects will only fall faster if there is an appreciable amount of air resistance present. The actual explanation of why all objects accelerate at the same rate involves the concepts of force and mass. 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. __//**All free falling objects will accelerate at the same rate; an object with a large mass and surface area can be faster or slower than an object with a smaller mass and smaller surface area, but their acceleration is the same. This is because of the ratio of force to mass, and that more mass will offset the amount of force on an object.**//__

= FREE FALL LAB = **Objectives:** What is acceleration due to gravity? What will an acceleration v-t graph look like? How will you get acceleration due to gravity off a v-t graph.

**Hypothesis:** Acceleration due to gravity is 9.8m/s^2, an acceleration v-t graph will start at zero and have a slope of 9.8 in the positive direction, and to find the acceleration due to gravity on a v-t graph, you find the slope of the line.

**Data:** **Analysis:** V-T Graph: On our v-t graph, we had the equation y=708.29x+68.25, with an R^2 value of .9958. The “708.29x” value represents the slope of our free falling object, which was 7.08. Had our experiment encountered no errors and was free of any added factors, such as friction caused by the spark an ideal slope would be 9.8. This is because the slope of a v-t graph is an object’s acceleration, and a slope of 9.8 means that the acceleration of the free falling object was 9.8m/s^2, or acceleration due to gravity. In our experiment, this was not the case, as we had a percent error of 27.8%, which is excessive and means that our experiment wasn’t accurate. Such a large percent error means that at some point during our experiment, there was an added factor that caused our results to be well under what they should be. In comparison to the rest of the class, our percent difference was 12.2%, which means that the average results of our peers weren’t extremely accurate either, but our results were somewhat precise. The average experimental value was 805.9 cm/s@\^2, which was much lower than the theoretical value of 980 cm/s^2. Our y-intercept is 68.25, and was not set to zero. This is because at our first point, the object wasn't stopped, and was moving at 68.26 cm/s. Our R^2 value was .9958, which shows that our points were very close to fitting the trend line and increased gradually, at a steady rate. This high value shows that we had a very good fit on our trend line, which shows that the difference in our data points wasn’t flawed by very much.

X-T Graph: The position-time graph that was created from our data had an equation of y=360.15x^2+64.924x. This equation can be compared to the equation d=Vi+1/2at^2, where Vi is initial velocity, a is acceleration, and t is time. In the equation of out x-t graph, 360.15x^2 refers to the A value, which is equivalent to 1/2at2 in the second equation. 1/2at^2 means half the acceleration over a certain time period. Ultimately, that value should be half of the slope of the v-t graph, whose slope is the acceleration. Our value was 360.15 cm/s^2, which when doubled is 720.32 cm/s^2, which is very close to the slope of our v-t graph (the object’s acceleration). The ideal value would be 490m/s^2, which is half of acceleration due to gravity. Since our percent error was 27.8%, we weren’t accurate, and thus not close to the 490m/s2 value. Since the class’s average experimental value was 805.9cm/s^2, the average value for the x-t graph should be very close to 402.3cm/s^2, which were fairly close to, as our percent difference was 12.2%. The second part of our equation, 64.924x is the same as the Vi value in the second equation. Vi is the initial velocity of our object, which should be the similar as the y-intercept of our v-t graph. In our v-t graph, the y-intercept is 68.25, which means the initial velocity of our object was 68.25cm/s^2. The initial velocity in the x-t graph is also very similar to the initial velocity in the v-t graph. Lastly, our R^2 value of the x-t graph’s line of best fit is 0.99991, which is extremely close to 1, or 100% accurate. Such a value shows us that our position-time graph’s points fit the trend very well.

**Discussion Questions-** **Conclusion** The first part of my hypothesis proved to be inaccurate. Our object’s acceleration was 708.28cm/s^2, which was 27.80% lower than the theoretical value, which is 980 cm/s&2. All the groups were faced with significant friction, and thus decreased speeds and acceleration because of the ticker tape traveling through the spark timers. Although faced with this issue, the average class acceleration was 809.5cm/s^2, which was much higher than ours. Since that value is an average, other groups had been able to get much closer to the theoretical value. The second part of my hypothesis, which was that the v-t graph of a free falling object will start at zero and move in the positive direction with a slope of 980, also proved to be partially incorrect. Although this would be true in a perfect experiment conducted in a lab, it was not in my case. The slope of our trend line was 708.28, which meant that the line was going in the positive direction with an increasing, constant slope. The line did not start at zero, because we found that the initial velocity of the object was 68.25. Lastly, I hypothesized that to find the acceleration due to gravity on a v-t graph, you find the slope. Although we didn’t have the same slope as the theoretical experiment, we still were able to calculate the acceleration of our object by finding the slope of the v-t graph.
 * 1) Does the shape of your v-t graph agree with the expected graph? Why or why not?
 * 2) The general shape of our v-t graph was similar to the expected graph, and agreed with it. Both our v-t graph and the expected graph had a positive slope increasing in the positive direction, but our slope was smaller than the expected graph (708.28 instead of 980).
 * 3) Does the shape of your x-t graph agree with the expected graph? Why or why not?
 * 4) The shape of our x-t graph does agree with the expected graph. Both graphs start out with low Y values, and have a fairly sharp slope upward in the positive direction away from the origin, representing the quickly increasing acceleration of the object.
 * 5) How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.)
 * 6) Our percent difference was 12.2%, which means that our acceleration of gravity was 12.2% lower than the class average. This means that our results were fairly precise, but weren’t accurate.
 * 7) Did the object accelerate uniformly? How do you know?
 * 8) Yes. I know the object accelerated uniformly because from looking both graphs, the line steadily increased throughout the measured times and wasn’t constant. This means that the acceleration was constant, which the lines of best fit prove. Although our percent error was high, it didn’t have an impact on whether or not the object was uniformly acceleration.
 * 9) What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?
 * 10) Factors that would cause acceleration due to gravity to be lower than the theoretical value include holding the spark timer vertically instead of horizontally, which would cause the tape to drag and create friction, which would slow the object down. Another factor would be that the tape wasn’t fully straight and clear of knots/kinks. This would also cause the tape to drag when going through the spark timer, which would create friction and slow down the object. A factor that would cause the experimental value to be higher than the theoretical value is if the experimenter collected incorrect data. For example, writing a distance that was off be a certain amount, or missing a point completely would cause the acceleration to spike.

In our experiment, our acceleration was calculated to be 708.28cm/s^2, which was 27.80% lower than the theoretical value of acceleration due to gravity. The first reason why I think our results proved to be so inaccurate was because of the large amount of friction that was put on the ticker tape, and that ultimately slowed down our object and its acceleration. When we dropped the object, ticker tape was fed through the spark timer as the object fell. When the tape was going through the spark timer, although it was held vertically, the tape was entering at almost a 90-degree angle, which meant that it would have to bend to go through the timer, and caused significant drag. Also, the ticker take was found to not only bunch up, but also to coil and bend. As a result, when going through the spark timer, small or large inconsistencies in the ticker tape could have caused drastic amounts of friction to be created, which would lower our acceleration value.

I think the best way to get more accurate results in this experiment would be to change the way the data is recorded. I think the time measurement device should be changed from a spark timer, which allows for drag and friction to be created and prevent a theoretical result, to something that doesn’t create friction, like the motion sensors. They can record the object’s distance/position over time, but without having anything attached to the object. If the motion sensor isn’t a viable solution, then another change to help get stronger data would be to change the way the ticker tape is fed through the spark timer. Since it has to enter at an angle, the tape should be repositioned at a different angle to reduce the amount of friction.

= __ Quantitative Graph Interpretation __ =



= Free Fall Classnotes =