Ch2_BehrensB

toc __ Chapter 2 - Motion __

= ** Class Notes - September 9, 2011; Constant Speed ** =

// average speed- // the average of all speed readings // constant speed- // when an object stays at a consistent rate // instantaneous speed- // the speed reading of one instance of time.

__ Types of Motion __ * four types of motion * 1. at rest 2. constant speed 3. increasing speed 4. decreasing speed

Ways to Represent the Types of Motion 1. Motion Diagram - at rest; v = 0, a = 0; - constant speed; speed is neither increasing nor decreasing --->--->---> [arrows all the same length]; a = 0 - increasing speed; --->->--->-> [arrows increasing in length]; a = ---> - decreasing speed; ->--->->---> [arrows gradually decreasing in length]; a = <--- * throwing an object upwards = decreasing speed * throwing an object downwards = increasing speed

// Signs are arbitrary [random] // - up is positive, to the right is positive. - down is negative, to the left is negative.

2. Ticker Tape Diagram / Spark Diagrams (reference to lab 9 / 6) - used for taking measurements

=__ ** 1-D Kinematics - Lesson 1; a, b, c, & d ** __= Summary Questions What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. From the class discussions, I understood the concepts of displacement v. distance as well as scalar v. vector. Displacement is defined as the distance between the beginning and ending points, while distance itself measures the entire length traveled. Displacement, which is more specific to change in position, is labeled as being a vector quantity; and distance, a scalar quantity. Vectors are quantities concerned with both distance and direction; scalars pertain to quantities in terms of distance only. These two terms are used to describe not only displacement and distance, but velocity and speed, too. 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. Tonight’s reading clarified the difference between velocity and speed. The original issue I’d had with this concept was mentally separating them into two different types of motion. From the lesson, I’ve learned that velocity, in relation to displacement, is concerned with change in position, and the rate at which an object does so. Therefore, if the position never changes, the velocity will remain at zero. Speed, however, is oriented around distance—a scalar quantity—and the rate at which an object moves. I’ve also used both average speed and average velocity formulas correctly in practice problems, as a result of the clarified information. What (specifically) did you read that you still don’t understand? Please word these in the form of a question. Because the reading was so helpful and served as a clear distinction between all concepts we’ve covered so far, I feel that I understand the reading. What (specifically) did you read that was not gone over during class today? The reading covered the topics of average speed v. instantaneous speed, which had not yet been discussed in class, however was clearly explained on the website. Also, the introduction with definitions to mechanics and kinematics, had not been mentioned in class yet, but were simple enough to understand after reading.

=__ ** 1-D Kinematics - Lesson 1; e ** __= Summary Questions What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. From our class discussion today, I understood the meaning of acceleration, and how it differs from velocity; acceleration is defined as the rate at which an object’s speed changes, whereas velocity is the rate at which an object’s position changes. I also understood how to rearrange the formulas in order to find a specific factor. Because we had been introduced to the formula for acceleration, it helped to make the reading more productive and successful. 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. After the class discussion, I had been a little unsure of difference between negative acceleration and positive acceleration, as well as how to determine an object’s direction of acceleration. I had remembered that direction of acceleration was the opposite of the object’s velocity, however the reading clarified the details of this concept. The opposing direction of acceleration only applies to an object that is decelerating. When the object is slowing down, the acceleration is the opposite direction of the velocity. For example, if a cart is slowing down with a negative velocity, the acceleration is positive. What (specifically) did you read that you still don’t understand? Please word these in the form of a question. Regarding free-falling objects, is the proportional relationship between the time interval and the distance traveled the same? What (specifically) did you read that was not gone over during class today? I read about constant acceleration, a form of acceleration in which the object’s velocity is proportional to the time interval it travels. If an object has a constant velocity, then it is not accelerating, however if the velocity is changing at an even rate, then it is exhibiting constant acceleration.

=**__ Speed of a CMV Lab; Part I __**=

Lab Partners: Joe Miller & Andrea Aronsky Date: September 6, 2011

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

Hypothesis: - The Yellow CMV is moving 0.5 m/s and the blue CMV is moving 1 m/s. - The position of time graph will show us how quickly our CMV will travel a certain distance. - We should measure to the closes millimeter or second decimal place.

Procedure Outline: - Pick up all of our materials - Measure out approximately an arms length of spark tape - Place the spark tape partially through the spark timer and tape it to the CMV - Start the spark timer and the CMV - Allow the CMV to run and retrieve the spark tape - Begin to record and measure the length between the dots on the spark tape - Each dots represents .1 second so go for 10, or 1 second - Make sure to start a little bit into the tape to help account for the time needed for the CMV to reach its constant speed

Excel Data Tables:
 * Yellow Constant Motion Vehicle ||  ||
 * Time (s) || Position (cm) ||
 * 0.0 ||  0.0  ||
 * 0.1 ||  1.38  ||
 * 0.2 ||  2.83  ||
 * 0.3 ||  4.15  ||
 * 0.4 ||  5.62  ||
 * 0.5 ||  7.06  ||
 * 0.6 ||  8.51  ||
 * 0.7 ||  10.08  ||
 * 0.8 ||  11.59  ||
 * 0.9 ||  13.82  ||
 * 1.0 ||  14.61  ||


 * Blue Constant Motion Vehicle ||  ||
 * Time (s) || Position (cm) ||
 * 0.0 ||  0.0  ||
 * 0.1 ||  5.35  ||
 * 0.2 ||  11.35  ||
 * 0.3 ||  18.25  ||
 * 0.4 ||  24.95  ||
 * 0.5 ||  32.21  ||
 * 0.6 ||  39.43  ||
 * 0.7 ||  46.73  ||
 * 0.8 ||  54.12  ||
 * 0.9 ||  61.34  ||
 * 1.0 ||  69.43  ||

Excel Graphs:

Discussion questions 1.Why is the slope of the position-time graph equivalent to average velocity? // - The slope of the line of best fit takes into account all parts and components of the graph; the slope is the average of the distance and time. This means that it is also velocity. // Why is it average velocity and not instantaneous velocity? What assumptions are we making? - //Instantaneous velocity is the measure of the speed of a CMV at one instance, the distance between two points, during the specified period of the given time, whereas average velocity is the speed of the CMV over the entire period of time. We use average velocity as opposed to instantaneous velocity because of the fact that the vehicle needs to be in constant motion as it is on the line of best fit, in order to find the velocity at which it moves.// Why was it okay to set the y-intercept equal to zero? // - It is ok to set the y intercept for this type of graph to zero because the constant motion vehicles are starting at certain position, the constant motion vehicle has at this point not moved it was immobile, and than we are to begin our measuring from the first location. If there has been no time and the vehicle has not moved and it is assumed that the constant motion vehicle’s position (cm) should start at zero. // 1.What is the meaning of the R2 value? // - The R2 value is a value that ranges from 0-1. The value itself represents the line of best fit or the perfect fit between the data and the line drawn through them. It represents how close to 100% accuracy the equation of the line of fit is. // 1.If you were to add the graph of another CMV that moved more slowly on the same axes as your current graph, how would you expect it to lie relative to yours? // - Comparing the resulting graphs of the two CMVs shows that the blue CMV, which was ultimately faster, had a steeper slope than the yellow CMV. The y of the yellow CMV had a lesser value, causing the slope to lie below the slope of the blue CMV. //

Conclusion We hypothesized that the blue constant motion vehicle would go 1 meter per second but the results showed that the car actually went 67.186 centimeters per second, which is much slower. For the yellow car, we thought the speed would be half a meter per second, which was also faster than the actual 14.6 centimeters per second. There are quite a few sources of error that may have contributed to inaccuracies such as the different battery lives; some batteries in a constant motion vehicle could have been used more often, therefore the car would run slower. Depending on the person reading the measurements, people may have had different viewpoints of the ruler marks, thus different measurements were calculated between groups. Another error could have resulted by someone beginning the measurements on the spark tape at different locations, as there wasn’t a constant speed yet when the car first started moving. Ways to minimize these issues could be done by buying new batteries and replacing the old ones from all the constant motion vehicles, using a ruler that would be easier to read like measuring tape or a foot ruler because they lie flat against the spark tape, and using the same leveled floor for each group.

=__CMV Lab; Part II__ =

September 20, 2011

Lab Partners: Joe Miller, Andrea Aronsky

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


 * 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.
 * 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.

Calculations:



- Position of the yellow (slower) CMV was 104.32 centimeters, while the position of the blue (faster) CMV was -495.93 cm.

- The blue CMV (faster) will catch up to the yellow CMV after 126.36 cm.

Data:



Video:

Crashing

media type="file" key="Crashing Experiment 1.mov" width="300" height="300"

Catching Up

media type="file" key="Catching Up Experiment 1.mov" width="300" height="300"


 * Discussion questions **


 * 1) Where would the cars meet if their speeds were exactly equal?

If the speeds of the cars were exactly equal, they would meet at the halfway point, three hundred centimeters. It is implied that if the cars travel at equal speeds, they also travel the same distance in the same amount of time (however not necessarily in the same direction). Because their initial positions are six hundred centimeters apart and their directions are opposite, the two cars would collide at three hundred centimeters.


 * 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?



Velocity-time graphs cannot show when the CMVs are in the same place at the same time, because though they appear to both begin at zero, in reality their initial positions are separated by one meter.

Percent Error:



Percent Difference:



Conclusion:

Judging by the percent error of this lab experiment, the results were not as accurate, in a general sense, as they could have been. The percent errors for the collision of the CMVs experiment were greater than those of the catching up experiment, in comparison to the calculations previous to the lab. The crashing percent errors for each trial ranged from 16.47%, being the least, to 22.38%, being the greatest. The percent errors for catching up ranged from .079% to 3.83%. Percent difference values for each situation were lesser in value than the percent errors. The range for the crashing percent difference was from 0.6% to 2.87%, while the range for the catching up percent difference was from 0.11% to 4.15%. After completing the equations, we expected to observe the CMVs colliding at the positions of 104.32 cm and -495.73 cm, yellow being the former and blue, the latter; the blue CMV would catch up to the yellow CMV after 1.82 seconds at 126.36 cm. Running the experiment proved difficult to actually match the mathematical solutions because of the sources of error that surfaced throughout this lab. For example, the blue car was defected, as it veered off the course to which it was originally set. As a result, it took many trials to obtain sufficient measurements. Changing different aspects, such as replacing the batteries in each CMV and choosing a leveled surface instead of the hallway floor, could have perfected this lab. We could have also found a better way to film the collision and catching up positions, which would provide better data. If we could redo the lab to expel these errors and acquire better results, we would insert new batteries into each car. Also we would execute this lab on top of a leveled lab table, which would eliminate the extra trials that were needed to get the measurements and ensure that there would be no obstructions in the path of the CMVs, such as bumps or dirt and rocks from shoe bottoms. We would also change this lab by filming the crash and catching up scenarios at different angles in order to capture a clear reading of the positions.

­­­ =** At Rest & Constant Speed Graph Notes **=

= ­­­ **__ 1-D Kinematics - Lesson 2; a, b, & c __**= Summary Questions What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. Lesson 2 served as a review of the information that had been discussed in class on Friday. I understood both methods of representing types of motion, called Ticker Tape and Vector diagrams. Ticker Tape diagrams are used for taking measurements of a moving object’s velocity, as used in Tuesday’s lab. Vector diagrams, the second method reviewed in the reading and also known as Motion diagrams from Friday’s notes, give a visual representation of an object’s velocity in the form of arrows. The arrows depict the acceleration, deceleration, and constant speed of the moving object. This type of diagram can serve as a visual for all types of vectors. 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. Neither of the methods describe in Lesson 2 were confusing or unclear. The reading helped to fortify the understanding I had already gained from the class notes. What (specifically) did you read that you still don’t understand? Please word these in the form of a question. I feel that I understand both methods well. The notes taken in class, providing background information on the reading, as well as what I read on Physics Classroom gave me a sufficient understanding of the diagrams. What (specifically) did you read that was not gone over during class today? Everything that I had read was discussed in class on Friday.

­ =** Class Notes - September 14, 2011; "The Big Five" **= ­­­

=**__ Graphical Representations of Equilibrium Activity __**= // "At Rest" Graph: // // "Forward Then Back" Graph: // // "Fast Then Slow" Graph: // Discussion Questions 1.How can you tell that there is no motion on a… position vs. time graph velocity vs. time graph acceleration vs. time graph // When there is no motion, this is represented on all three graphs with a horizontal line from the start of each run through the end of it. //

2.How can you tell that your motion is steady on a… position vs. time graph // You can tell that the motion is steady because each line is relatively the same in regards to slope. The pattern for walking away, which shows that the line gradually increases, is the opposite with the same rate for walking back towards the motion censor. // velocity vs. time graph // Steady motion is represented on a velocity v. time graph because the line for each run is almost horizontal, and a pattern can be seen as well. There are not any points that are too severely out of place in comparison to the entire line. They are both fairly horizontal. // ◦ acceleration vs. time graph // As with the previously described graph, the lines on this graph follow a horizontal pattern too. Because the scale is much larger, it is easier to read the lines of each run in regards to the patterns that they follow, which shows that the motion is steady. //

3. How can you tell that your motion is fast vs. slow on a… ◦ position vs. time graph // Upon reading the data collected on the graphs, the pattern is evident in that the first run inclines at a steeper slope, while the second line lies below the first, representing the difference in motion of fast v. slow. // ◦ velocity vs. time graph // The lines on the velocity v. time graph have a similar pattern, however the line of the second run lies below the first run, showing that the motion of the latter has a faster velocity. // ◦ acceleration vs. time graph // The lines on this graph overlap several times, therefore revealing no specific pattern. However, both runs accelerate at proportional rates, causing the lines to be similar in slope. //

4. How can you tell that you changed direction on a… ◦ position vs. time graph // When walking away from the motion detector, the slope of the line is positive. The distance between the person and the motion detector is increasing. While walking toward, the slope is negative. The distance between the person walking and the motion detector is decreasing. // ◦ velocity vs. time graph // The line will reflect over the x-axis when the direction is changed. It was positive while walking away from the motion detector, but negative walking to the detector. // ◦ acceleration vs. time graph // You can't tell the change of direction in this graph because acceleration does not keep track of direction; it keeps track of displacement. //

5.What are the advantages of representing motion using a… ◦ position vs. time graph // You can see the slope, or the change in position over the change in time. It clear to see how far the person is from the motion detector relative to how long they were walking. The slope of the line is the average speed. // ◦ velocity vs. time graph // This graph is useful when seeing the velocity difference between a person moving fast and slow. The faster the person is walking, the greater the velocity is. Therefore, it shows the relationship between speed and motion. It also shows you the velocity of an object at a certain interval of time as you are able to see the velocity when a person is walking towards and away from the sensor. // ◦ acceleration vs. time graph // You can determine if something is speeding up or speeding down, thus it is a representation of acceleration that is both increasing and decreasing. It also shows if velocity is negative or positive. //

6. What are the disadvantages of representing motion using a… ◦ position vs. time graph // I don’t think there is a disadvantage unless the walking path of the person is disrupted, but this graph it doesn’t give you acceleration. // ◦ velocity vs. time graph // This graph was not truly accurate because the sensor was sensitive to other movements which interfered with how the graph depicted the lines and created inaccuracies. // ◦ acceleration vs. time graph // It doesn’t show the exact velocity, or the change in position of the object over time, thus direction is not shown. //

7. Define the following: ◦ No motion ◦ Constant speed
 * 1) // At rest, the object is not moving //
 * 2) // 0 velocity and 0 acceleration //
 * 1) // Constant change in position per time interval //
 * 2) // acceleration is 0 //

// ­­­ // ** Class Notes - September 15, 2011; Increasing & Decreasing Speed Graphs **

=**__ 1-D Kinematics - Lesson 3; a, b, & c __**= Summary Questions What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. After the class discussion, I understood the difference between increasing speed and decreasing speed represented on position-time graphs and velocity-time graphs. Comparing the graphs along side each other distinctively showed the difference in the shapes of the slopes. I also understand what a slope looks like when velocity is increasing and decreasing. 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. Everything I’ve read in this lesson I understood from class, as it only discusses position-time graphs, which I originally felt confident about after the class discussion. What (specifically) did you read that you still don’t understand? Please word these in the form of a question. I feel that I understand what I’ve read in lesson three. What (specifically) did you read that was not gone over during class today? Everything in the reading was at least briefly discussed and class and reinforced with specificity.

=**__ 1-D Kinematics - Lesson 4; a, b, c, d & e __**= Summary Questions What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. From our class discussion, I understood that the slope of a velocity-time graph is equal to the acceleration of the moving object. Positive slopes indicate positive acceleration; negative slopes indicate negative acceleration; horizontal slopes indicate that the object is not accelerating. 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. After the class discussion, the relationship between slope and acceleration was unclear; however, the reading provided illustrations and explanations that clarified the confusion I’d had. What (specifically) did you read that you still don’t understand? Please word these in the form of a question. I currently do not have any questions about the information I read on the website. What (specifically) did you read that was not gone over during class today? The reading included a section that described how displacement is found through velocity-time plots. This concept was not confusing, as it reviews what we’ve learned in geometry regarding area of figures.

=** Acceleration Graphs Lab **=

Lab Partner: Andrea Aronsky

Objectives: 2.What does a position-time graph for increasing speeds look like? 3.What information can be found from the graph?

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

Hypothesis: // A position-time graph for increasing speed curves upwards in a "U" shape. The graph gives information about the object's velocity. //

Procedure:

media type="file" key="Cart Acceleration Lab.mov" width="300" height="300" Data / Excel Graph: time (s) position (cm)

Analysis: a) Interpret the equation of the line (slope, y-intercept) and the R2 value. It is a polynomial equation with a R2 value of 0.99891, which is very close to 100%.

b) Find the instantaneous speed at halfway point and at the end. (You may find this easier to do on a printed copy of the graph. Just remember to take a snapshot of it and upload to wiki when you are done.) Halfway: 10 cm/s // End: 20.75 cm/s //

c) Find the average speed for the entire trip. // Speed: distance/ time = 10.85/ 1 = 10.85 cm/s // This position time graph contains both the linear and polynomial trend line along our points. The orange line represents the line tangent to the halfway point while the green line represents the line tangent to the last point.
 * polynomial equation: y= 8.2662x2 + 2.4565x; R2: 0.99907
 * linear equation: y= 8.9613x; R2: 0.91725

Discussion Questions: 1.What would your graph look like if the incline had been steeper? 2.What would your graph look like if the cart had been decreasing up the incline?

3.Compare the instantaneous speed at the halfway point with the average speed of the entire trip. They are almost the same value being that the instantaneous speed is 10 cm/s while the average speed is 10.85.

4.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 the object is at during that specific time. The tangent line only intersects with one point out of the ten on the graph; therefore, when taking the slope of that line you are figuring out the velocity, and the velocity will be the same throughout the straight line because it remains constant.

5.Draw a v-t graph of the motion of the cart. Be as quantitative as possible.

Conclusion: The results of this lab proved our hypothesis to be accurate. We originally thought that the graph would have the shape of a “J” and that we would be able to gather information about the cart’s velocity from the results, which was true. The slope, or y-value, of the linear trendline was 8.9613x, while the R2 value was 0.91725. The slope of the polynomial trendline was 8.266x2 – 2.4665x, and the R2 value was 0.99907. Many different errors could have had possible effects on the results of this lab; as in the previous lab, inaccuracies can be accounted to misreading of the ruler during the process of taking measurements; and unleveled work station, textbook, and ramp; and mistiming of starting the spark timer. The lab procedures and results can be perfected by including a second person in the measurement readings, in order to get a more exact number. A flatter workstation would have also corrected another source of human error. Lastly, taking multiple runs during the lab would have minimized the errors that stem from mistiming of the spark timer.

=**__ Egg Drop Project __**=

Description: The final device that we used on the day of the egg drop in class was a combination of both prototypes that we had previously tested. The design was a hollow cube shape made solely of straws and held together by tape and sewing thread. The parachute, which aided in slowing the device's acceleration, was a sheet of paper attached by straws. The egg capsule was made of and secured by straws as well.

Mass of Project: 84.44g

Results: The final project was successful. The egg remained intact throughout the drop, and the time was 2.18 seconds.

Analysis: Calculation of "a"

In comparison to 9.8 m/s/s, the acceleration of the final device is a lesser value, which can be attributed to many factors. 9.8 m/s/s represents the average gravitational force exerted on objects that are dropped, or free falling, towards the Earth's surface. However, because the acceleration value calculated for the final device was affected by the use of the parachute, which partially counteracted the increasing velocity of the object, the number was lower than 9.8. Also, the size of the device could have affected the acceleration too.

Reflecting on the project, there isn't anything that necessarily needed to be changed. If there were one aspect that could be altered, it would be the weight of the device. Instead of using four straws on all sides, three would have been sufficient, and the weight would have ultimately been less.

=** Quantitative Graph Interpretation **=

= **__ 1-D Kinematics - Lesson 5; a, b, c, d & e __** =

Topic Sentence: An object experiencing free fall moves solely under the force of gravity.


 * Free-falling objects do not encounter air resistance.
 * All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s

A ticker tape or dot diagram would depict acceleration. If an object travels downward and speeds up, then its acceleration is downward.


 * Acceleration of gravity ** - acceleration for any object moving under the sole influence of gravity (the symbol ** g ** ).

A position versus time graph for a free-falling object is shown below.

A velocity versus time graph for a free-falling object is shown below.

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 **

The distance that a free-falling object has fallen from a position of rest is also dependent upon the time of fall. This distance can be computed by use of a formula; the distance fallen after a time of ** t ** seconds is given by the formula. ** d = 0.5 * g * t2 **

The answer to the question (doesn't a more massive object accelerate at a greater rate than a less massive object?) is absolutely not, if we are considering free-fall. More massive objects only fall faster if there is air resistance present. All objects free fall at the same rate of acceleration, regardless of their mass.

=** Class Notes - October 3, 2011; Freefall **=

= __Freefall Lab__ = October 4, 2011

Lab Partner: Andrea Aronsky

Objectives: What is acceleration due to gravity? What does a velocity-time graph of the falling object look like? How are you going to get acceleration due to gravity from a velocity-time graph?

Hypothesis & Rationale:
 * Acceleration due to gravity is the rate at which an object accelerates, or increases its velocity, under the force of the Earth's gravity.
 * A velocity-time graph of the falling object would have a negative slope towards the origin, because the object would be accelerating towards the ground.
 * You can get the acceleration due to gravity from a velocity-time graph by taking the slope of the trendline of the data, because on a v-t graph the slope is equivalent to the acceleration of the object.

Data:

Class Data:

Sample Calculations:

Graphs: Position-Time Graph:

Velocity-Time Graph:

Analysis: In this equation, the coefficient "A" is equivalent to half the acceleration of the object. For example, to find the acceleration, set 0.5a equal to the value of A and solve. Our value for A was 445.97, and when the equation was solved, a equaled 891.94 which was relatively close to 894.54 cm/s/s. The coefficient "B" represents the initial velocity of the object. For this lab, B did not equal zero because, as discussed in class, by the time the spark timer had marked the first dot on the tape, the weight had already been falling. Therefore, the initial velocity could not be zero. For example, our value for B was 83.709 cm. The graph shapes were affected by the slope of the data; in the velocity-time graph, the slope equaled the acceleration of the object. However, our graph had a positive slope and was moving away from the origin, when in reality it should have been negative and moving towards the origin, since the object was being dropped towards the ground. Because acceleration can be either positive or negative, the direction isn't of as much importance as the actual slope, which does remain the same in both situations. In the position-time graph, the slope is equivalent to the velocity, which is increasing in consistent intervals. Therefore, the shape of the graph is a "J" curve. The value for "b" is equivalent to the y-intercept, which, for this lab, cannot be set at zero. The reason for this, as previously discussed, is that object had begun falling before the spark timer had marked the first dot. The initial velocity is not zero at the first dot, since the weight was moving already; it is the instantaneous velocity of the position at which the weight was when the first dot was made, and as a result the y-intercept is not starting at zero. For our graph, the y-intercept is 82.984 cm/s. The R2 value for our position-time graph was .99999, and for the velocity-time graph was .99892. These values show the correctness and accuracy of the trendlines, and in this case the lines were very close to being completely correct.

Discussion Questions: The shape of the velocity-time graph was relatively similar to that of our hypothesis. The only difference between the graphs is that the final velocity-time graph has a positive slope, depicting an object moving in the opposite direction. Therefore, the final graph has a positive slope, in contrast with the original expectation, which was hypothesized to have a negative slope. However, the shape of the trendlines is the same.
 * 1) Does the shape of your v-t graph agree with the expected graph? Why or why not?

The shape of the position-time graph does match the original expectation. Because acceleration due to gravity is increasing velocity by 9.8 m/s2, the prediction was that the shape of the graph would be a “J” curve, which expresses exponential growth. After graphing the data, the resulting shape agreed with our original thoughts.
 * 1) Does the shape of your x-t graph agree with the expected graph? Why or why not?

In comparison to the class average, which was 805.9 cm/s2, our percent difference was 10.99%. The results showed that the falling object’s acceleration was 894.54 cm/s2, which is lower than the value of “g” at 981 cm/s2. The average is lower than the acceleration of gravity measurement because everyone’s individual experiments were lower as well as a result of human errors discussed in class. Further calculations show that the smallest percent difference was 0.16%, whereas the largest was 12.59%. In reflection, our percent difference fell on the higher end of individual values.
 * 1) How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.)

Generally, the object did fall in a consistent interval in regards to acceleration. By calculating the instantaneous speed at each position, you see that it increases by a constant decimal in the range of four or five each time. Closer examination shows that the increasing interval is not exact at each instantaneous speed, which might have also been affected by human errors, however in a general sense the object did accelerate uniformly.
 * 1) Did the object accelerate uniformly? How do you know?

Because free fall is motion based only on the force of gravity, it is nearly impossible to execute this lab under that condition. In order to ensure that the tape was straightened out during the object’s fall, it was necessary to have one person hold it while the other used the spark timer and dropped the weight. As a result, friction between the tape and the spark timer or the person holding the tape had slowed the object’s fall, causing the results to be slightly inaccurate. Also, the final value for “g” could have been ultimately lower due to mistaken measurements, which would have affected the graph as well.
 * 1) What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?

Conclusion: Our hypothesis was correct because we had said that acceleration due to gravity was the rate of increasing velocity of the object under the sole influence of gravity, which we had discussed in class on previous days. We also thought that the velocity-time graph for the falling object would have a negative slope, moving towards the origin, which was also true. However, when we graphed the data collected from the lab, the slope was positive and moving away from the origin. This is because the acceleration of 9.8 m/s/s was used positively rather than negatively, causing the graph to be the opposite of what we thought. The value for g, though, can be used in both directions, so it does not ultimately change the slope of the graph. And lastly, we hypothesized that you can get acceleration due to gravity from the slope of the trendline, because on a velocity-time graph, the slope equals the acceleration. For example, the equation of the line on the v-t graph was y = 894.54x + 82.984, where m is the slope of the line, and in this case, the acceleration of the object that was dropped. Our experimental errors are shown in the percent difference and percent error. The percent difference for this lab was 10.99%, and the percent error was 8.81%. The percent error is not as low was we would have wanted it to be due to experimental errors made during the lab procedures. The value for g is 981 cm/s/s, whereas our value was 894.54 cm/s/s. The errors could have occurred when we dropped the weight, with the presence of friction, as previously discussed, between the tape and the spark timer. Therefore, all of our procedures, including measurements and graphing the data afterwards, were not as accurate as they could have been. If our measurements were off, then the graphs would not display the proper acceleration value of 981 cm/s/s. As a result, it is extremely difficult to complete the free fall lab with the weight moving under the sole influence of gravity. To address the errors made during the lab, we could edit our procedures in order to acquire better data. Instead of actually holding the tape the entire time the weight is being dropped, the person holding it will let go as the object begins moving. This will allow the results to depict the object moving almost completely with the force of gravity, providing a more accurate value for acceleration due to gravity. However, to attain the most accurate results possible, we would use a motion sensor to show the weight's acceleration and motion in a graph on Data Studio. By replacing the spark timer with the motion sensor, the friction between the ticker tape and the timer would no longer be an issue in inaccurate results. First we would set the motion sensor at the bottom of the school lobby, the site of the drop, and the weight would be dropped from the floor above. Once the process of dropping the weight is refined, the measurements will inevitably be better as well. Overall our lab and results would benefit if we made these changes.