Ch2_WeissB

=__Speed, Velocity, Acceleration__=

**//Notes//**
toc

//Constant Speed Notes//
__Speed and Velocity Formulas__ Speed- total distance over total time Velocity- total displacement over total time

__Types of Motion__ 1. At rest- where no movement is being made

2. Constant Speed- where movement is being made at a constant rate

3. Increasing Speed

4. Decreasing Speed

Acceleration is any change in speed (i.e. increasing and decreasing speed)

**//Motion and Ticker Tape Diagram Notes//**
__Motion Diagram__ At Rest
 * V=0
 * A=0

Constant Speed
 * Appears as: à à à because velocity is the same (therefore same arrow length)
 * Have a=0 underneath

Increasing Speed
 * Start w/ shorter arrows that gradually get longer
 * Have an arrow pointing in same direction as first arrows w/ an a underneath

Decreasing Speed
 * Start with longer arrows that gradually get smaller
 * Have an arrow pointing in opposite direction as first arrows w/ an a underneath

When an object in a different direction than left or right (for example, up or down), turn the motion diagram towards that direction



__Ticker Tape Depictions__ At Rest
 * One dot

Constant Speed
 * Evenly spaced dots

Increasing Speed
 * More and more space between each dot

Decreasing Speed
 * Less and less space between each dot



//**Graph Shapes Notes**//
__Positive and Negative Signs__ Signs are arbitrary

__Position Time Graph__ At Rest- will appear as a straight horizontal line Constant speed- will appear as linear, slope (speed) will always be the same

//Instantaneous Speed will be equal to Average Speed at the midpoint of the graph//

__Velocity Time Graph__ At Rest- straight horizontal line (on x-axis) Constant Speed- straight horizontal line above or below x-axis (above means moving away from origin, below means moving towards origin)

__Acceleration Time Graph__ Will appear as the x-axis at rest and at constant speed

**//Increasing and Decreasing Speed Notes//**
Increasing Speed: . slope of position-time graph will get steeper over time . velocity-time graph will appear as a linear graph, whose slope is equal to the acceleration; starts off low, ends up high . acceleration-time graph appears as a straight horizontal line with no slope Decreasing Speed: . slope of position-time graph will lessen over time . velocity-time graph will appear as a linear graph, whose slope is equal to the acceleration; starts off high, ends up low . acceleration-time graph appears as a straight horizontal line with no slope
 * positive- movement away from origin
 * negative- movement towards origin
 * positive- movement towards origin
 * negative- movement away from origin

//Lab: A Crash Course in Velocity (Part 1)//
Date: 9/7/11 Partner: Maddie Weinfeld


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

2. If you measure distances, distances can be measured precisely up to tenths of centimeters. 3. If you use a position-time graph, it will tell you where the CMV is at a given time.
 * Hypotheses ** : 1. If the CMV moves, it will move at a rate of 2 centimeters per minute.

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

The graph appears to be a linear graph, whose slope is the velocity of the CMV. The R2 value is the percent accuracy of the results that were gotten from the experiment compared to the ideal results. Because the percentage is about 99.1%, the graph is extremely accurate.
 * Analysis**


 * Discussion questions **
 * 1) Why is the slope of the position-time graph equivalent to average velocity? The slope of the position-time graph is equivalent to average velocity because the slope is the change in position over the change in time, which is equivalent to average velocity.
 * 2) Why is it average velocity and not instantaneous velocity? What assumptions are we making? Average velocity is the velocity over time, while instantaneous velocity is the velocity at a certain point in time. The slope is for the entire graph as a whole, as opposed to a single point; therefore, it determines average velocity. We are assuming that the average and instantaneous velocity are the same.
 * 3) Why was it okay to set the y-intercept equal to zero? Because at the point at which 0 seconds have passed, the CMV has traveled a distance of 0 cm. As a result, the y-intercept should be set at zero, as that is equivalent to (0,0).
 * 4) What is the meaning of the R2 value? The R2 value is equivalent to the percentage of accuracy of the results of the experiment compared to the ideal results.
 * 5) If you were to add the graph of another CMV that moved more slowly on the same axes as your current graph, how would you expect it to lie relative to yours? I would expect the graph of a slower CMV to be below the graph of my CMV.


 * Conclusion **

I found that my CMV moves at a velocity of 62.628 cm per second. Therefore, my first hypothesis that the CMV would travel a distance of 2.00 cm per minute is inaccurate because the distance the CMV would actually travel a distance of 4117.68 cm per minute. During the experiment, we calculated distance with a ruler measuring centimeters with intervals of .10 centimeters. As a result, my second hypothesis is accurate because .10 centimeters is the equivalent of a tenth of a centimeter. During the experiment, not only could I determine the distance the CMV would travel at a given time, but I could also determine the velocity traveled. Because of this, my third hypothesis is partially accurate because I was correct in my prediction that I could determine traveling distance, but could have included other possible determinations. Sources of error could have resulted from the ruler that was used. The ruler did not allow us to measure hundredths of centimeters (causing us to estimate) and caused perspective issues (the ruler was thicker than the tape), which made length difficult to determine. These issues could have been avoided by using a different measuring tool, one that measures in hundredths and/or was as flat as the tape.

//Summarization (Version 2a) of 1-D Kinematics Lesson 1a-d//
Date: 9/8/11


 * 1) **What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.** By the time I had read Lesson 1, I had already understood the difference between distance and displacement. Distance is the total amount of motion made by an object, while displacement is the total change in position compared to the object's original position; these definitions were confirmed in the reading. I also had already understood the definition of speed, being the rate at which an object moves across a certain distance over a certain amount of time.
 * 2)  **What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding**. The reading helped me understand what the exact definition of the word // kinematics // was. The word had been mentioned in class, along with the use of graphs, diagrams, etc. to chart out motion, but I did not connect the two in class. Now, I realize that kinematics // is // the science of using graphs, diagrams, etc. to chart out motion. Also, I was a little uncertain of the equation for average speed. During class, I was unsure of whether or not to take the total distance and time into account. Now, I do understand that the average speed is total distance/total time.
 * 3) **What (specifically) did you read that you still don’t understand? Please word these in the form of a question.** When is instantaneous velocity a different value than average velocity?
 * 4) **What (specifically) did you read that was not gone over during class today?** I read about the definitions of scalars and vectors, which were not gone over in class. According to the reading, scalars are quantities that are described magnitudes or numerical value alone, while vectors are quantities that are also described by direction.

// Summarization (Version 2a) of 1-D Kinematics Lesson 2 //
Date: 9/10/11


 * 1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.** By the time of my reading, I already understood how to read a ticker tape. Specifically, I knew that the distance between dots demonstrated the change (or lack of change) in a moving object's velocity. Furthermore, I knew how to set up a ticker tape diagram as a result of the first lab we performed in class. As explained in the reading, you are supposed to attach the tape to the moving object and then insert it into a machine that produces the ticks as movement is made.


 * 2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.** While studying ticker tape diagrams, I was slightly confused about the a potential result of the diagram. I was not sure how the ticks would appear if there was no movement being made by the object attached to the ticker tape. However, based on the visual examples given, I understand now that either the ticks would appear to be on top of each other or extremely close to each (if barely any movement was being made).


 * 3. What (specifically) did you read that you still don't understand? Please word these in the form of a question.** Do the lengths of the arrows in a vector diagram matter? For example, would a certain change in velocity need to be represented by a particular length of arrow?


 * 4. What (specifically) did you read that was not gone over during class today?** There was nothing in the reading that we did not go over in class.

//Activity: Graphical Representations of Equilibrium//
Date: 9/12/11
 * 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

At Rest: Constant Slow Towards: Constant Slow Away: Constant Slow Towards:
 * Data:**

Constant Fast Away:


 * Discussion questions: **
 * 1) How can you tell that there is no motion on a…
 * 2) position vs. time graph- the graph appears as a straight, horizontal line that is identical to the x-axis or extremely close to it
 * 3) velocity vs. time graph- velocity will be at 0
 * 4) acceleration vs. time graph- acceleration will be at 0


 * 1) How can you tell that your motion is steady on a…
 * 2) position vs. time graph- the graph appears to be linear; the slope (speed) will always be equal regardless of what points you are calculating it with
 * 3) velocity vs. time graph- the graph will appear as a horizontal straight line that is above or below the x-axis (positive velocity means you are moving away from the origin, negative means you are moving towards the origin)
 * 4) acceleration vs. time graph- acceleration will be at 0.


 * 1) How can you tell that your motion is fast vs. slow on a…
 * 2) position vs. time graph- when motion is fast, intervals between points will be longer, when motion is slow intervals between points will be short
 * 3) velocity vs. time graph- when motion is fast, you will have a horizontal straight line that is at a high y-value, when motion is slow, you will have a horizontal straight line that is at a low y-value
 * 4) acceleration vs. time graph- when motion is fast, acceleration will be greater than 0; when motion is slow, acceleration will be less than 0.


 * 1) How can you tell that you changed direction on a…
 * 2) position vs. time graph- Position will increase or decrease
 * 3) velocity vs. time graph- Velocity will increase or decrease
 * 4) acceleration vs. time graph- You cannot tell


 * 1) What are the advantages of representing motion using a…
 * 2) position vs. time graph- you can determine the total distance you are aware from your starting point at a certain period in time
 * 3) velocity vs. time graph- you can determine your speed, you can determine direction, you can determine displacement
 * 4) acceleration vs. time graph- you can determine by how much you are increasing or decreasing your rate of motion


 * 1) What are the disadvantages of representing motion using a…
 * 2) position vs. time graph- does not immediately tell you the difference between points and acceleration
 * 3) velocity vs. time graph- you cannot determine your position
 * 4) acceleration vs. time graph- you cannot determine your original speed


 * 1) Define the following:
 * 2) No Motion- when no acceleration, displacement, or change in distance takes place; at rest
 * 3) Constant Speed- when acceleration, displacement, or change in distance occurs at a constant rate

**//Summarization (Version 2a) of 1-D Kinematics Lesson 1e//**
Date: 9/13/11 1. **What (specifically) did you read that you already understood from the class discussion? Describe at least 2 items fully.** By the time of the reading, I had already understood the definition of acceleration. Acceleration is the rate at which an object's velocity changes. Also, I had already understood the difference between constant acceleration and non-constant acceleration. In essence, constant acceleration takes place when an object's velocity increases by the same amount over equal intervals of time, while non-constant acceleration takes place when an object's velocity increases by different amounts over equal intervals of time.

2. **What (specifically)** **did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.** Today in class, I was slightly confused about what the equation for acceleration meant. However, now I realize that acceleration is basically the change in velocity over time.

3. **What (specifically) did you read that you still don't understand? Please word these in the form of a question.** What exactly is the relationship between the distance travelled during a time interval and calculations of total time after this interval?

4. **What (specifically) did you read that was not gone over during class today?** In class today, we did not go over the acceleration of a free-falling object. According to the reading, free-falling objects usually accelerate as they get closer to the ground.

**//Lab: Acceleration Graphs//**
Parter: Maddy Weinfeld Date: 9/14/11


 * Objectives and Hypotheses:**
 * What does a position-time graph for increasing speeds look like? It will look linear with an increasing slope.
 * What information can be found from the graph? You can find the position of a moving object at a given time, how much the speed of the object changes at a certain point (due to changes in slope), velocity

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

Procedure: First, the track is set up with one end firmly on the edge of a textbook, so that the track has an incline. Next, spark tape is run through the spark timer in the direction of the arrow; at the same time, make sure the spark timer is set so that it will record the distance of an object every second. After that, use scotch tape to connect the spark tape with the dynamics cart, so that the tape will run through the spark timer in the direction of the arrow as the cart moves. Then, put the cart at the top of the track, turn the spark timer on, and gently release the cart. After this is done, measure the distance (in cm) between the dots made on the spark tape. Finally, use time (as your x-axis) and the distance (as your y-axis) to make a graph.

Data Table:
 * Data and Graphs:**

Graph:

a. The equation of the line is y= 0.073 x 2 + 1.0832x, fitting the equation y=Ax^2+Bx. This equation is equivalent delta d=1/2*a*time^2+initial velocity*time. Therefore, A is equal to 1/2 of the acceleration, while B is equal to the initial velocity. Because of this, the acceleration of the cart is .146 cm/s^2. The value of the y-intercept of this graph is (0,0). This makes sense because when 0 seconds have passed, the cart has not moved at all or has moved 0 cm. At the same time, the R2 value of the graph is .99858, meaning that there is about 99.9% accuracy between the results gotten from the lab and the ideal results; therefore, they are almost completely identical.
 * Analysis:**

b. Instantaneous speed at the halfway point is about 1.67 cm/s; Instantaneous Speed at the endpoint is about 3.38 cm/s.

c. Average Speed= (18.31-0)/(10-0)=1.83 cm/s

1. The steepness of the ramp would affect the incline at which the cart accelerates on, which changes the acceleration. As a result, if the ramp were to become steeper, then the acceleration would increase, meaning a greater length between distances at certain times.
 * Discussion Questions:**

2. The graph would take on the shape of a curve that begins higher up on the y-axis (with a greater distance), then ends much lower on the y-axis (once the distance decreases). See the graph of the cart moving uphill for an example.

3. The average speed of the entire trip is about .16 cm/s more than the instantaneous speed at the halfway point.

4. The instantaneous speed of the midpoint is the slope of the tangent line because the midpoint is a point on the tangent line. Furthermore, when you find the slope of this line using the midpoint of the graph of the cart's movement and another point on the tangent line, you would be finding the change in distance between these two points over the change in time. Velocity is the change in distance over the change in time. Therefore, you are finding the velocity of the midpoint when you are calculating the slope of the entire tangent line.

5.

During this experiment, one hypothesis was disproven, while the other one was proven. My hypothesis that the graph of a position-time graph for an accelerating chart would appear to be linear was incorrect. Instead, according to the final graph, it will be a curved graph, due to the fact that the slope increases over time. However, my hypothesis that I could determine speed at a certain time, how much the speed changed over a certain span of time, and velocity was correct. I could do all these things by looking at individual points on the graph, finding slope between points, and finding the displacement over time. In this experiment, there could have been various sources of error. First, it was difficult to measure in hundredths of centimeters because the ruler we were given measured only up to tenths. This could have been corrected by using a measuring tool that could measure in hundredths of centimeters. Another source of error could have been the slant of the track. During this experiment, the class did not establish a uniform tilt for the ramp, possibly leading to different results for different groups, and perhaps increasing or decreasing the speed of the cart by too much. This could have been corrected by establishing a set slant for the track and then using that as our incline.
 * Conclusion:**

**//Summarization (Version 2a) of 1-D Kinematics Lesson 3//**
Date: 9/15/11

1. **What (specifically) did you read that you already understood from the class discussion? Describe at least 2 items fully.** By the time of the reading, I already understood that a position-time graph for an object moving rightward with constant velocity appeared linear with a positive slope. This is the case because the slope remains consistent throughout the entire graph. I also understood that a position-time graph for an object moving rightward with changing velocity would appear to be a curved line. This is the case because the slope changes from each point.

2. **What (specifically)** **did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.** Initially, I was confused by how velocity affects acceleration. However, based on the reading's description of this topic and the use of graphs as demonstrations, I realize now that

3. **What (specifically) did you read that you still don't understand? Please word these in the form of a question.** When an object is acceleration, how exactly does velocity change from one point to the next?

4. **What (specifically) did you read that was not gone over during class today?** There was nothing in this reading that we did not go over in class today.

**//Summarization (Version 2a) of 1-D Kinematics Lesson 4//**
Date: 9/15/11

1. **What (specifically) did you read that you already understood from the class discussion? Describe at least 2 items fully.** By the time of the reading, I had already understood that a velocity-time graph showing constant, rightward velocity would show a straight horizontal line. This is the case because the acceleration (or the slope) is 0; when the slope is 0, a line is horizontal. I also understood that a velocity-time graph showing changing, rightward velocity would appear as a linear line, since the slope is greater than 0, but maintains consistency.

2. **What (specifically)** **did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.** During class, I was unsure why a velocity-time graph with constant velocity would appear to be a straight, horizontal line, even though I knew it was true. The reading's description and diagrams helped me understand the reason behind this. A velocity-time graph does not show total displacement at a certain point. Instead, it shows how much displacement is made from one point in time to the other. Thus, if an object were to maintain a velocity of 10 m/s throughout its movement, it would make sense for the entire graph to be at the y-axis value of 10.

3. **What (specifically) did you read that you still don't understand? Please word these in the form of a question.** Would decreasing velocity begin to increase again after it passes the negative version of its magnitude?

4. **What (specifically) did you read that was not gone over during class today?** We did not go over the fact that finding the area of a v-t graph would lead to the totla displacement.

**//Lab: A Crash Course in Velocity (Part 2) REDO//**
Date: 9/21/11 Partners: Maddy Weinfeld, Kaila Solomon, George Souflis

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

**Purpose:** Algebraically and graphically, find the positions of two carts moving towards each other at the point in time when they crash into each other and the positions of the two carts when the faster cart passes the slower cart when both move in the same direction. Then, run trials to confirm these calculations.

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 * Procedure:**

**Diagrams:** Crash: Passing: //Note: CMV 1 is the faster blue cart, while CMV 2 is the slower yellow cart.//


 * Equations:** Before beginning any experimentation, our group created equations to represent the point in time at which the blue and yellow carts would crash into each other when traveling towards each other, as well as the point in time at which the blue cart would pass the yellow cart when both were moving in the same direction. In the case of the crash scenario, the equation would be 98.653t=600. As explained in the instructions for this lab, there is a distance of 600 cm separating the starting points for the blue and yellow carts. At the same time, a previous experiment (see Ben and Maddy's and George and Kaila's Part 1 Lab) demonstrated that the speed of the blue cart is 68.628 cm/s, while the speed of the yellow cart is 30.025 cm/s. The above equation shows how much time in seconds it will take for both carts to displace an added total of 600 cm. Meanwhile, we also came up with the equation: 68.628t-100=30.025t to represent the passing scenario. This equation indicates the time it takes for both carts to be at the same position with the "-100" reflecting the -100 cm displacement of the blue cart.


 * Calculations and Data:**

After creating the equations for the two scenarios, we found that the two carts would crash into each other in an ideal time of 6.08 seconds. This means that the blue cart would have a total distance of 417 cm to the right, while the yellow cart would have a total distance of 183 cm. We also found that the blue cart would pass the yellow cart in an ideal time of 2.59 seconds. This means that the blue cart would have a total distance of 177.75 cm, while the yellow cart would also have a total distance of 77.75 cm.

To test the accuracy of these calculations, we ran five trials of both situations and calculated the displacement of both carts at the time when they crashed and when the blue cart passed the yellow cart.

__Crashing__ Below are the displacements of the blue and yellow carts at the time they crashed in each of the five trials:

To test the accuracy of these findings, we calculated percent error for the average of the distances of CMV 1 (the blue cart), as well as the percent difference for each individual distance. We chose to do this for the blue cart only because the displacement of the blue cart directly affects the distance of the yellow cart. Therefore, calculating percent error and difference for the yellow cart would lead to the same (or very similar) results. We also chose to do percent error for the average (and not each individual displacement) because all the distances are extremely close to one another. Therefore, there is no outlier value that would significantly tamper with the average and make the results less valid. Our findings of percent error and difference are below: Based on our calculations, our findings are extremely close to what we found as the ideal displacement of the blue cart, with little percent error or difference. As a result, our calculated ideal displacement (and therefore ideal time) are accurate.

__Passing__ Below are the displacements of the blue and yellow carts at the time the blue cart passed the yellow cart in each of the 5 trials.

To test the accuracy of these findings, we calculated percent accuracy and difference in the same manner as the crash situation for the same reasons.

Based on our calculations, our findings are extremely close to what we found as the ideal displacement of the blue cart, with little percent error or difference. As a result, our calculated ideal displacement (and therefore ideal time) are accurate.

1. Where would the carts meet if their speeds were exactly equal? If the carts were expected to crash when they met, they would meet after each moving the same distance. They would meet 3.00 m or 300 cm away from where they each started. This is because is 3.00 m is the midpoint and since each car would have the same velocity, they would each travel the same distance in the same time. If the situation was similar to the passing scenario, the cars would never actually meet. This would be the case because the cart that starts off 100 cm behind the other would always be 100 cm behind the other cart. As a result, the cart that starts off 100 cm behind the other would never gain on the other cart.
 * Discussion Questions:**

2. Crashing:

Catching Up:

3. One is not able to determine the time at which the two carts reach the same point at the same time because velocity-time graphs do not have position on it.

During this lab, I learned how to calculate the time/displacements at which two objects moving at different speeds will crash and the time/displacements at which a faster moving object will pass a slower moving object. I also learned how to use percent error and difference in order to test how accurate lab results are. In this lab, I used both of these skills to find these times, then test how accurate my calculations were. I was able to successfully able to do this; I found an ideal distance of 417.26 cm and an average of calculated displacements from 416.116 cm for the blue cart in the crashing situation**.** I also found an ideal displacement of 177.75 cm and an average of calculated displacements from 174.85 cm. After the tests were done, I found that there was an average percent error of .27% and a range of percent difference from .24-.99% in the crash situation and an average percent error of 1.63% and a range of percent difference from .097 to 2.21% in the passing situation. In short, we had extremely accurate results! However, there were various sources of potential error in this experiment. A major one was the fact that our original blue cart broke mid-testing and we had to use a different one. As a result, we may have been using a cart with a slightly different acceleration than the acceleration calculated in the Part I Lab. The potential difference in velocity and acceleration between the two could have made results less accurate. Another potential source could have been the measuring tool we used. We were not able to calculate hundredths of centimeters on this tool, causing us to estimate and leading to less accurate results If I were to redo this lab, I would redo Part I with the new blue cart, so that I could incorporate the actual velocity of that cart and use a measuring tool that could measure hundredths of centimeters.
 * Conclusion:**

//Egg Drop Conclusion//
Date: 9/30/11 Partner: George Souflis Purpose: To build a device that is supposed to prevent an egg from breaking during free-fall and calculate the acceleration of that device.

Picture: The time the device hit the ground was 1.638 seconds and the total mass of the device (including the egg) was 158.02 g.

Calculation of Acceleration of Device As shown in the above picture, using the equation delta d= (initial velocity * time) + (.5 * acceleration * t), we calculated the acceleration of the device as being 6.34 m/s2

Discussion of Results: The egg drop device shown in the picture ended up not giving the egg proper support. This box, constructed from pieces of wood (with straws and rubber bands around the inner walls to provide padding) was supposed to be able to cushion the impact and protect the egg from breaking. However, this did not work as planned and the egg ended up completely shattering.

Conclusion/Sources of Error: While watching the device make its descent, we noticed that the device began to tumble and spin in the air as it went further down. This potentially was caused by the two thin planks at the bottom of the box, which were intended to keep the box stable once it hit the ground, but may have ended up throwing the box off balance. Furthermore, the pitfalls of using a box shaped device were discussed in class. We were told in class that a box could not properly slow the momentum of the egg and that a cone-shaped device would be better at doing that. This is because as a cone device travels downward, the egg also travels deeper into the cone, gaining even more support from the device. If given the opportunity to do this experiment differently, my partner and I would have used a cone device as opposed to a box, in order to provide the egg with beneficial support.

**//Quantitative Graph Interpretation//**
Graph A (please ignore work shown in picture): Work for Graph A:

Graph E and Work: Graph F and Work: Graph G and Work:



//Summarization (Version 1) of 1-D Kinematics Lesson 5//
A free falling object is an object that is falling under the sole influence of gravity. Free-falling objects do not encounter air resistance and accelerate at a rate of -9.8 m/s2 (denoted with symbol ** g ** ). Because a free-falling object travels downwards and speeds up, its acceleration is downward; therefore it is negative. A free-falling object’s x-t graph looks like a curve that starts high and finishes with a large downward velocity; its v-t graph looks like a diagonal line that starts at the origin and has a negative slope. The total velocity of a free-falling object is equal to ** g * t **, while the total distance is equal to ** .5 * g * t^2 **. All objects free fall at the same rate of acceleration, regardless of their mass.

//Freefall Lab//
Date: 10/5/11 Partner: Maddie Weinfeld


 * Purpose**: To calculate the acceleration of a free-falling object (in this case a 100 g weight).

Ticker Tape Timer, Timer tape, Masking tape, Mass, clamp, meterstick.
 * Materials**:


 * Hypotheses**: The acceleration of the weight will be -9.81 m/s/s (or -981 cm/s/s). Also, the v-t graph will appear as a linear graph from which the value g (-981 cm/s^2) can be determined through its slope (ideally 981).

Position and Time of Free-Falling Weight: Instantaneous Velocity and Mid-Time of Free-Falling Weight at Given Times and Positions: Average of Calculated Accelerations in Class:
 * Data**:

X-T Graph:
 * Graphs:**

V-T Graph:

Example of Finding Instantaneous Velocity: In this example, the instantaneous velocity from 0 to .1s is being calculated. By finding the change in distance at this point divided by the change in time, the instantaneous velocity was found to be 49.5 cm/s^2
 * Calculations:**

Percent Error: According to this calculation of percent error between the ideal acceleration of 981 cm/s^2 and the calculated acceleration of 754.43 (the slope of the trendline of the V-T graph; see ANALYSIS OF GRAPHS for more info) there is 23.1% error.

Percent Difference: According to the calculation of percent difference between the average experimental acceleration (839.417 cm/s^2) and my individual experimental acceleration (754.43), there was 10.1% total percent difference.

The equations of the X-T and V-T graphs fit the formulas y=Ax^2+Bx and y=mx+b, respectively.
 * Analysis of Graphs:**

Shapes: The shapes of the X-T and V-T graphs fit the expectations of objects under acceleration. For example, the X-T graph should appear as a curved line with gradually increasing slope because it is subject to acceleration; therefore, it does not have a constant velocity and should appear as a linear graph. Similar logic applies to the shape of the V-T. This graph should appear as a linear graph, which it does, because velocity is increasing at a constant rate. Therefore, it will have a slope that is greater than 0, which leads to a slanted linear graph.

m: The meaning of m in the V-T is the acceleration of the free-falling weight. This is the case because the slope (like acceleration) is found as the change in velocity over the change in time. Therefore, the slope can tell you (like acceleration does) how much the velocity is increasing by at any given moment. As a result, we can tell that the calculated acceleration of the weight is 754.43 cm/s^2 or 7.5443 m/s^2 downward direction.

b: The meaning of b in the V-T graph is the velocity of the free-falling weight at 0 seconds. According to this graph, the b value is -7.9786. Therefore, the velocity at 0 seconds is -7.9786 cm/s^2 (or -.007 m/s^2). However, this is not the ideal value of b. The value b should instead be 0 because when 0 seconds have passed, the weight should have a velocity of 0 m/s^2. However, -.007 m/s^2 is extremely close to 0 m/s^2, so the value of b from the graph is still reliable.

A: The meaning of A in the X-T graph is half of the total acceleration. This is because the equation y=Ax^2+Bx is equivalent to an equation that finds the change in distance when given initial velocity and acceleration (**delta d= (vi)t+ .5at^2**). In this case, t=x.Therefore, according to the equation of the trendline, the acceleration is 746.58 cm/s^2 or 7.4658 m/s^2 in the downward direction. While this isn't exactly to the calculated acceleration from the V-T graph (this is because the y-interecept is (0,0) on this graph), they are both close enough to each other that both can be considered accurate.

B: The meaning of B in the X-T graph is the initial velocity of the free-falling weight (see A for reasoning). According to the equation of the trendline, the initial velocity is -6.6985 cm/s^2 (or about -.066985 m/s^2).

R2: The R2 values of the graphs compare how the calculated measurements fit the ideal trendlines of each graph. With .99962 on the X-T graph and .99277 on the V-T graph, all of our results are extremely accurate compared to the trendlines.

1. Does the shape of your v-t graph agree with your expected graph? Why or why not? Yes, the shape of my v-t agrees with my expected graph. As I mentioned in my hypothesis, I believed that the graph would end up as a linear line with a consistent slope (i.e. a consistent acceleration). This was the case in the actual graph, since it was a linear graph with a slope of 754.43 (according to the trendline).
 * Discussion Questions:**

2. Does the shape of your x-t graph agree with your expected graph? Why or why not? Yes, the shape of my x-t graph agrees with my expected graph. I had expected the x-t graph to take the shape of a curved graph that has a constantly increasing slope (due to the object's acceleration). This was how the x-t graph ended up looking because of the object's acceleration of about 746.58 cm/s^2 (according to the trendline).

3. How do your results compare to the rest of the class? (Use percent difference to discuss quantitatively). Compared to the class average of 839.417 cm/s^2, my calculated velocity of 754.43 cm/s^2 has a percent difference of 10.1%. See **Calculations**.

4. Did the object accelerate uniformly? How do you know? In actuality, the object did not accelerate uniformly because it did not exactly fit the trendline. In other words, the slope of the graph was not exactly that of the trendline throughout. If it had been, the R2 value would have been 1.

5. What factors would cause acceleration due to gravity to be higher than it should be? Lower than it should be? Acceleration due to gravity could have been made higher than it should be with some form of pressure or force pushing down on the object as it made its descent. Acceleration due to gravity could have been lower through friction caused from the spark tape moving through the spark timer (sometimes over the top of the timer). The slowing of the movement of the spark tape through the spark timer could have slowed down the acceleration of the weight.


 * Conclusion**: Several of my hypotheses were proven by this experiment, while another was not. Just as I stated in my hypothesis, the overall shape of the V-T graph is linear, with a slope that can be used to determine the acceleration of the moving object (in this case the weight). However, the results within the experiment did not fit the hypothesis that the acceleration of the weight would be -981 cm/s^2. While I know this to be true from previous class work I had done in class, according to the V-T graph, the acceleration is instead 754.43 cm/s^2 in the downward direction (or -754.43 cm/s^2). This error in the calculation of **g** is reflected in the calculations for percent error and difference (23.1% and 10.1% respectively). This error can be attributed to several sources. One potential source of error could have been the possible lapse between dropping the weight and measuring its displacement on the sparktape. While the sparktimer was supposed to begin when the velocity of the weight was at 0, this may not have been perfectly synchronized, thus the sparktimer may have begun to calculate displacement as the weight was in the middle of movement. As a result of this error, displacements that are attributed to certain times may not have been found at those times. Another potential source of error could have been from the measurement of the distance between points on the sparktape. While applying a measuring tool to measure these distances, it is possible that the measuring tool and/or the spark tape shifted. This could make measurements less accurate, leading to less accurate overall results. If this experiment were to be repeated, I would have better coordinated the use of the spark timer with the dropping of the weight and taped down the measuring tool and/or the spark tape to gain better results.