Ch2_VanMaldenJ

toc ** Chapter Two **

** CMV Lab **

 * Julie and Remzi **

** Objective- ** 1. How precisely can you measure with a meter stick?How fast does your CMV move?What information can you get from a position time graph? ** Hypothesis ** Based on all of the information we have acquired this far, we estimate that the yellow car will move at average velocity of 15 cm/s while the blue, which is seemingly faster than the yellow, will move at an average velocity of 20 cm/s.

** Procedure- ** ** Data and Graphs- ** CMV Position Time Graph Analysis- In this graph the line is constant. This is because the distance between all of the points is almost the same. Which makes sense because a CMV is a Constant Motor Vechile so it should be expected to keep a consistent cm/s. ** Conclusion ** In our CMV lab we tested for the average speed of the fast yellow car. We found that the CMV had an average speed of 31.95 cm/s, ruling our hypothesis to be inaccurate. We hypothesized that the CMV would have an average velocity of about 15 cm/s, but our result proved this hypothesis to be incorrect. Sources like the age of batteries, obstacles in the way of the experimental path, and unleveled testing surface may have contributed to inaccuracies in our data. These sources could have lead to a change in the result and possible error in the data. To minimize the chance of error we could ensure the batteries in the CMV are functioning well by changing them and replacing them with new ones, make sure the testing path is level and clear of any obstacles that could interfere.
 * Gather all required materials
 * String the spark tape through the spark timer.
 * Attach the end of spark tape to the CMV
 * Make sure the spark timer is on 10 Hz, and then turn on both the car and the spark timer.
 * When all of the spark tape has run through the timer, turn off both.
 * The tape will now have dots on it that represent the distance the car moved per every 1/10s.
 * Measure the distance between the dots and create a scatter graph interpreting your results.
 * Discussion questions **
 * 1) Why is the slope of the position-time graph equivalent to average velocity?
 * 2) The slope of the position-time graph is equivalent to the average velocity because the slope is equal to rise over run, which is equal to change in y over the change in x. ∆Y and ∆X are the change of the position (∆Y) over the change in time (∆X), which is the definition of average velocity.
 * 3) Why is it average velocity and not instantaneous velocity? What assumptions are we making?
 * 4) Because we are assuming that velocity remains constant or relatively close to constant throughout the experiment, there would be no unexpected change in speed.
 * 5) Why was it okay to set the y-intercept equal to zero?
 * 6) It is okay to set the y-intercept equal to zero because when no time has passed (0 s) the car hasn’t began to move so it has moved 0 cm/s.
 * 7) What is the meaning of the R2 value?
 * 8) The R2 value is whether the plotted data’s line of best fit fits the data set closely.
 * 9) 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) Being that the CMV moved slower it would go less cm/s, so the slope of the line would be slighter, falling underneath the the original line.

Homework- Describing Motion with Words
vs.


 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) I understood the difference between distance and displacement, distance being the total amount of space traveled and displacement being the amount of space something is from where it started. I also understand the difference between speed and velocity, speed being scalar and only relating to size only, instead of size and direction like velocity does.
 * 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 didn't really understand the difference between scalar and vector, but the reading made it clear that scalar has to deal with speed because it only deals with size, while vector deals with size and direction and is generally used in velocity.
 * 5) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 6) I didn't find that i was really confused or unclear on any of the subjects form the reading.
 * 7) What (specifically) did you read that was not gone over during class today?
 * 8) Everything in the reading was basically covered in class, however the reading went into a little more depth about the difference between speed and velocity and scalar and vector then we had in class.

Motion Diagrams and Ticker Tape-

 * Average Speed- The total distance over the total time
 * Constant Speed- The same speed at all times. (Your instantaneous speed will always be the same as the constant speed)
 * Instantaneous Speed-the speed your going now. The change in speed because you arent going the same speed the whole time.
 * Eqution used for all of these-V=change in d/change in t


 * Types of Motion-
 * Motion Diagrams
 * Atrest- V=0, a=0
 * Constant-
 * ---v--> v---> ---v--->
 * a=0
 * --v--> -v> ---v->
 * a--->
 * acceleration points in same direction as velocity
 * v--> v---> -v->
 * <--a
 * acceleration points in the opposite direction of velocity


 * Signs are arbitrary, they follow the x-axis y-axis graphs.
 * For example if it goes down its negative, and it it goes left its negative.
 * For example if it goes up and right its positive.
 * If a sign is tilted then the graph tilts with it.
 * Ticker tape Diagrams/ Spark Diagrams/ Oil Drop Diagrams
 * Increase- . . . . ..
 * Decrease- . . . . ..

Interpreting Position- Time Graphs




Homework 9/9/11:
First- Constant Second and Third- Accelerating vs.


 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) I understood both the usage of ticker tape and motion diagrams. I know that when the dots on a ticker tape are the same distance apart the speed remained constant through however if the dots started out closer together and grew farther apart there was an acceleration or if they came closer together there was a deceleration. I also understand the accelerations are represented by arrows getting longer and decelerations are represented by arrows getting shorter in motion diagrams, and if there was not change in speed or velocity the arrows would remain the same size.
 * 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) There was nothing that i truly didn't understand in the reading because everything was well explained in class, and if something did appear a little blurry I could reference my class notes.
 * 5) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 6) I didn't find that i was really confused or unclear on any of the subjects form the reading.
 * 7) What (specifically) did you read that was not gone over during class today?
 * 8) Everything in the reading was covered in class, however the reading went into a little more depth about the ticker tape and the what the distance between the dots represented.

Graphing Acceleration Activity-
No Motion/At Rest
 * Graphs- **

Constant Motion (Fast and Slow) Constant Motion (Towards and Away)
 * 1) How can you tell that there is no motion on a…
 * 2) position vs. time graph-
 * 3) velocity vs. time graph
 * 4) acceleration vs. time graph
 * 5) For all of the the above- On the at rest graph we can tell that there is no motion because when we look at the graph the line has very few fluctuations that go above or below zero. The line is nearly horizontal with as light positive slope.


 * 1) How can you tell that your motion is steady on a…
 * 2) position vs. time graph- The line rose at a gradual rate and than evened out and continued for a short time at the same pace.
 * 3) velocity vs. time graph- There is a very small in line that leads to a pretty steady line slightly above 0.
 * 4) acceleration vs. time graph- For the most part the line remains exactly on 0. It does fluctuate slightly above or below 0 on a few occasions.


 * 1) How can you tell that your motion is fast vs. slow on a…
 * 2) position vs. time graph- The faster the motion is, the shorter it takes to go a greater distance or have a larger change in position. The slower a motion is, the longer it takes for something to change or move positions.
 * 3) velocity vs. time graph- The faster motion will have a faster/higher velocity then the slower motion, so the faster motions line should be slightly above that of the slow motion line.
 * 4) acceleration vs. time graph- Because during the experiment the fast and slow walk were constant, the acceleration remains close to 0 throughout.


 * 1) How can you tell that you changed direction on a…
 * 2) position vs. time graph- When the line, or position, starts at a high number and then gets lower as time goes on, the motion that is being preformed is moving towards/closer the sensor, but when the position starts at a low number and gets higher, the motion being performed is moving farther away from the sensor.
 * 3) velocity vs. time graph- the velocity will change because velocity equals displacement over time, and when your displacement will be different depending on where you are, close or far away from the sensor.
 * 4) acceleration vs. time graph- Because there is no acceleration during either of these motions, you couldn’t tell that there was a change in direction because the acceleration would remain around 0 in both ways.


 * 1) What are the advantages of representing motion using a…
 * 2) position vs. time graph- The movement of an object staying still will not change its position. This means that the graph has a straight line representing no change in its position. It can also show if there is a large change in position indicating that an object is moving faster than another. Also it can show the displacement depending on how far an object moves forward and back. It will show how much distance is covered at a certain speed.
 * 3) velocity vs. time graph- It can show that an object is moving towards a certain point and back by the negative and positive intervals of the graph. Negative would be moving closer to the original position and positive would be moving away from the original position.
 * 4) acceleration vs. time graph- It shows whether an object is increasing its speed, decreasing its speed, or staying at a constant rate for a certain amount of time over the course of a distance.


 * 1) What are the disadvantages of representing motion using a…
 * 2) position vs. time graph- Occasionally, one of the plotted points will be on the zero or another number, which represents either change in the moving position or no change at all from the starting point.
 * 3) velocity vs. time graph- A possible disadvantage for representing our data using a velocity graph would be that there are some peaks and valleys in the beginning and end of the data. It is also difficult to see the data because of its closeness unless you adjust the scale.
 * 4) acceleration vs. time graph- A possible disadvantage for representing our data using an acceleration graph would just be because the acceleration for this specific lab was 0 its very hard to see any truly outstanding differences between the data.


 * 1) Define the following:
 * 2) No motion- When an object is at rest and is not changing its position. Speed, velocity, and acceleration equal 0.
 * 3) Constant speed- When an object is going at the same speed for the entire time that it is moving from one position to another.

**Acceleration Cart Lab-**
Julie, Remzi, Joey

**Objectives:** **Hypothesis:** - The position time-graph will have an upward incline with a positive slope - The graph will show us how fast an object travels a certain distance. Its rate of acceleration and velocity.
 * What does a position-time graph for increasing speeds look like?
 * What information can be found from the graph?

**Procedure:** - Step 1- Get materials - Step 2- Lean Ramp on textbook and put cart at the top of the ramp - Step 3- Feed ticker tape through timer and attach to cart - Step 4- Turn on ticker time and release car - Turn off ticker timer and interpret results

Data-

**Analysis:** a) Interpret the equation of the line (slope, y-intercept) and the R2 value. b) Find the instantaneous speed at halfway point and at the end. (You may find this easier to do on a printed copy of the graph. Just remember to take a snapshot of it and upload to wiki when you are done.) > i. Slope of tangent line= (8-2)/(.77-.37)= 6/.4= 15 cm/s > i. Slope of tangent line= (8-0)/(.79-.55)= 8/.24 = about 33.3 c) Find the average speed for the entire trip. speed = total distance/ total time
 * 1) When we constructed our line of best fit we tried two different lines before we came to the correct one (a polynomial). When we compare the two lines we find that the polynomial is closer to 1.0 (0.99979) than that of the linear, which was only about 0.934… This means that the polynomial line contains almost 100% of our points compared to only about 93.4. Our polynomial line has a slope of 13.67 and a y intercept of 0.7452.
 * 1) These are the tangent lines for the halfway point speed and the end point speed. Using these I used the points, (.37,2) and (.77, 8) from the tangent line for the slope of the instantaneous speed at the halfway point. I also used the coordinates (.79,8) and (.55,0) from the tangent of the end point line to find the slope/ instantaneous speed at the end.
 * 2) The instantaneous speed is about 15 cm/s
 * 1) The instantaneous speed at the end is about 33.3 cm/s
 * 1) Average speed= 14.48 cm/ 1 s
 * 2) The average speed is 14.48 cm/s

**Discussion Questions:** 4. Draw a v-t graph of the motion of the cart. Be as quantitative as possible. After starting the lab we measured our results for 1.0 second and came up with decent results. When we placed a polynomial line of best fit we found that our equation for the line was: (y=13.67X^2+0.7452x) and our R^2 value was extremely close to being 1 it came out as (0.99979). For this laboratory we created two individual hypotheses, the first being: An upward incline with a positive slope and the second being: The graph will show us how fast an object travels a certain distance. It will display the distance at which our cart travels in 1.0 second. Both of our hypotheses seemed to be accurate with the data we received. Both of our hypothesizes were proven on the position time. Some sources of error that could have affected our results are the fact that the ticker timer starts and a person may not release the car at a perfect time, which creates a few extraneous dots. Another error that could have occurred was in the measuring of the distance between one dot and the next on the ticker tape, which could have led to slightly off results. In order to minimize these errors there are a few things that can be done. The first thing that I would recommend doing is starting on the ticker tape a few dots in to assure that it was recording the time when the cart was moving and not just the wait time in between the launch and timer start process. The second way we could minimize error is to use other methods of measuring the space in between the ticker tape dots. We could have used different measure devices like a flat ruler or a tape measure. Another option is to have two people measure it and see if the results they produced were comparable.
 * 1) What would your graph look like if the incline had been steeper?
 * 2) [[image:Screen_shot_2011-09-14_at_11.19.47_AM.png]]
 * 3) What would your graph look like if the cart had been decreasing up the inclineCompare the instantaneous speed at the halfway point with the average speed of the entire trip.
 * 4) [[image:Screen_shot_2011-09-14_at_1.08.04_PM.png]]
 * 5) Compare the instantaneous speed at the halfway point with the average speed of the entire trip.
 * 6) Avg. Speed = ∆d/∆t = 14.48 cm/1 sec
 * 7) i. Avg. Speed: 14.48
 * 8) Inst. Speed = 15cm/s
 * 9) Because the graph doesn’t have a constant speed, the instantaneous speed is the slope of the tangent line.
 * 10) - If a line is tangent to a curve it touch at only 1point. If you know the slope of the tangent line you know the slope of thepoint that you chose on the curve which is included on the tangent line.

Homework 9/14:

 * What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 1) I understood that the equations of acceleration and how one would go about trying to find it. I also understood the difference between a constant acceleration and a constant velocity.
 * 2) Average Velocity Equation-
 * 3) [[image:Screen_shot_2011-09-29_at_6.33.43_PM.png]]
 * 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.
 * 1) I wasn’t really confused on anything that we went over in class, but the reading just went into a lot more depth in the topic of acceleration then we did in class.
 * What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 1) How do you find an acceleration if you are given both a velocity and a direction?
 * What (specifically) did you read that was not gone over during class today?
 * 1) This reading went into a lot more detail about free fall and the direction of an accelerator vector

Homework 9/15-
Lesson 3- Describing Motion with Position Time Graphs Lesson 4- Describing Motion with Velocity Time Graphs-
 * What (specifically) did you read that you already understood from our class discussion? Describe at least 2 items fully.
 * 1) For the most part I understood the difference between the graphs at a constant speed and the graphs of acceleration. I also already understood the differences between graphs when the acceleration was positive and when the acceleration is negative.
 * 2) ﻿[[image:Screen_shot_2011-09-25_at_4.45.59_PM.png width="521" height="196"]] vs [[image:Screen_shot_2011-09-25_at_4.46.10_PM.png width="494" height="166"]]
 * 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.
 * 1) Nothing that I read really confused me because our class notes were very detailed.
 * What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 1) I found that there was nothing that I truly didn’t understand that I couldn’t figure out by resorting back to our class notes.
 * What (specifically) did you read that was not gone over during class today?
 * 1) The reading talked about the importance of slope in position time graphs, which we only briefly touched upon in class.
 * What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 1) I understood the differences between graphs of constant velocity and graphs with an accelerating velocity. I also understood when you would make the velocity graph line negative and when it would be positive.
 * 2) ﻿[[image:Screen_shot_2011-09-25_at_4.48.55_PM.png]] and [[image:Screen_shot_2011-09-25_at_4.49.04_PM.png]]
 * 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.
 * 1) Nothing that I read really confused me because our class notes were very detailed.
 * What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 1) When and why would you need to find the area of a v-t graph?
 * What (specifically) did you read that was not gone over during class today?
 * 1) We didn’t go over finding the area of a V-T graph in class.

A Crash Course in Velocity Lab-
Julie, Remzi, Maddi, Ben S



Purpose- In this lab we are attempting to see if our predicted calculations, of the distance it takes for two CMV's to crash and how long it takes for one to catch up to the other, match up with our actual results from the trials.

Calculations- A. B.

Videos- Crashing CMV's media type="file" key="Catch up Vid.mov" width="300" height="300"

media type="file" key="crashingcorrect1.mov" width="300" height="300"

@https://www.box.net/shared/static/5gv7rx9k4gsrsoy83n2f.mp4 media type="custom" key="10761916"

Catching Up

Crashing

Percent Error-
 * Crashing-
 * Run 2-
 * ( l theoretical-experimental l / theoretical ) X 100
 * ( l 361.4- 427 l / 361.4) X 100
 * Percent Error- 18.1%
 * Catching Up-
 * Run 2-
 * ( l theoretical-experimental l / theoretical ) X 100
 * ( l 294.23-240 l / 294.23) X 100
 * Percent Error- 18.4 %

Percent Difference- >>>
 * Crashing-
 * Run 1-
 * ( l average experiment- individual experimental l / average experimental ) X 100
 * ( l 186.5- 173 i / 186.5 ) X 100
 * Percent Difference= 7.24%
 * Catching Up-
 * Run 1-
 * ( l average experiment- individual experimental l / average experimental ) X 100
 * ( l 236.3 - 230 l / 236.3 ) X 100
 * Percent Difference- 2.67%

Discussion Questions-
 * 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.
 * 4) 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

3. 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? On the Velocity - Time graph, no. It shows that one CMV has a higher initial velocity. They are at different positions on the graph so you can't tell if they were at the same position at a certain time just by looking at the graph. It is better to use a position-time graph to see this.

Conclusion- For the most part when we ran trials with our CMV's, the data matched relatively closely with our predicted hypothesis/solutions. Being that we were unsure if the CMV we used in these trials was the same CMV we had previously recorded the data for, that could have lead to error in our results. That also could have lead to the large percent errors between our experimental vs. theoretical values in the Catching Up trial. For the distances, because the blue car was faster for the crashing run, it had to travel a larger distance of 361.4 cm in 7.45 s as apposed to the mere 283.6 cm covered by the yellow CMV. For the catching up experiment the blue CMV once again had to travel a larger distance of 294.23 cm in about 6.0 s trying to catch up to the 194.23 cm traveled by the yellow CMV. Other error in this experiment could have been a result of not letting the CMVs go at the same time, old batteries, or having one CMV run on a surface that was not flat. I would redo the first part of the lab and make sure that the CMV we used, was the CMV we had recorded the data for in the prior lab. Also, I would have ran a few more trials to make sure that are trial times were accurate.

Egg Drop Project-
mass of project- 46.1 g time of fall- 1.19s acceleration- 12 m/s
 * t-average=1.19s
 * d=8.5m
 * v-initial-0
 * a=?
 * d=vit+0.5at^2
 * 8.5=0+0.5(1.19^2)a
 * 8.5=.70805a
 * a=12 m/s^2

This is my Egg Drop Experiment Project. The straws on the outside triangle were supposed to absorb the impact so that the egg, located in the inside triangle, wouldn't crack. During the testing of prototypes, it worked one out of two times. However on the final drop, the outside triangle didn't absorb the impact and the egg cracked. For the value of "a", my results were wrong because the times that were taken during the drop were inaccurate. Because there was no parachute located on the prototype, the drop would have been quick and "a" should have been close to 9.8m/s squared, instead of the 12.0 m/s squared that I calculated. If I were to redo this project, I would have attached a parachute to it because it would slow down the fall and lessen the blow. Also, instead of the structure being a triangle, like my final project was, I would make it a cone shape because the tip too would lessen the impact of the fall.

Interpreting Position- Time Graphs
Graph A. Graph E. Graph F. Graph G.

Homework 10/3:
--
 * A free falling object falls under ONLY the influence of gravity and these objects acted upon only by the force of gravity are in a state of free fall.
 * The important motion characteristics true for free-falling objects:
 * They do not encounter air resistance.
 * They all (on Earth) accelerate downwards at a rate of 9.8 m/s/s
 * A ticker tape trace or dot diagram of a free falling motion would depict an acceleration.
 * [[image:U1L5a1.gif]]
 * This numerical value for the acceleration of a free-falling object is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity represented by the symbol g.
 * One way of describing the motion of objects is through the use of graphs -__ [|position versus time] __ and __ [|velocity vs. time] __ graphs. This is an example of a position versus time graph for a free-falling object.
 * 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. The position-time graph shows that the object starts with a small velocity (slow) and finishes with a large velocity (fast). The small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity and 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.


 * A diagonal line on a velocity versus time graph signifies an accelerated motion, this graph reveals that the 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.
 * Since the slope of any velocity versus time graph is the acceleration of the object the constant, negative slope indicates a constant, negative acceleration.
 * If dropped from a position of rest, a free falling 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. . 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 for the velocity of a free-falling object after six and eight seconds are shown below.

Free Fall Lab-
**Objective**: What is the acceleration of a falling body?
 * Hypothesis**- The acceleration due to gravity should be around 9.8m/s2. The V-T graph should start at the origin and go down in a straight line with a negative slope of -9.8m/s2. You can tell the acceleration due to gravity by looking and finding the slope of the line from the data set.


 * Data-**

Velocity= displacement/time V=

In this position-time graph, the line equation y=Ax^2 + Bx is equal to the equation d=v1t +1/2 at^2. In the case of this graph, A is half of the acceleration, while B is the initial velocity. In this case, half of the acceleration in 4.59m or 459 cm, while the initial velocity is 44.78 cm/s or .4478 m/s. y=Ax^2 + Bx d=v1t +1/2 at^2 A= 1/2 acceleration B= initial velocity
 * Position- Time Graph**

In this graph, the line portrays the equation y=mx=b, or vf= vi + at. Because slope= velocity/time, it has the same value as the acceleration. This graphs shows us that our acceleration is 907.4 cm/s or 9.07 m/s and that the y-intercept is -39.9.
 * Velocity-Time Graph**


 * Percent Error-**
 * ( l theoretical-experimental l / theoretical ) X 100
 * ( l 9.8 - 9.07 l / 9.8 ) X 100
 * Percent Error- 7.45 %


 * Percent Difference-**
 * ( l average experiment- individual experimental l / average experimental ) X 100
 * ( l 805.9 - 907.4 l / 805.9 ) X 100
 * Percent Difference- 12.6%


 * Discussion Questions-**
 * 1) Does the shape of your v-t graph agree with the expected graph? Why or why not?
 * 2) Not the shape of our graph does not agree with the expected V-T graph. This is because the expected V-T graph the line would have a negative slope, however since in our experiment we made down have a positive velocity, the V-T graph has a positive slope.
 * 3) Does the shape of your x-t graph agree with the expected graph? Why or why not?
 * 4) Once again our graph does not match that of the expected one, because we made our falling positions positive instead of negative. This is because we made our starting point zero, and the distances that followed it to be positive, instead of measuring how high we are, using that as the starting point, and recording all the points that lie below it to be negative.


 * 1) How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.)
 * 2) Our result were closer to the actual speed of acceleration then that of the class. We found ourself to have a 12.6% difference from the class, while only having a 7.45% difference form the actual free fall acceleration.


 * 1) Did the object accelerate uniformly? How do you know?
 * 2) No, our object did not accelerate completely uniformly because it had a smaller rate of acceleration then a normal free fall object. This is because we had our free fall object attached to ticker tape going through a ticker tape timer, which created friction, slowing down the rate at which the object fell.


 * 1) What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?
 * 2) Factors like friction would cause the acceleration due to gravity to be lower then it should be. Factors like incorrect measure of the tickertape could cause the acceleration due to gravity to appear to be faster then it actually is.

Our hypothesis was not correct due to many different variables. The acceleration due to gravity should have been 9.8 m/s^2 as we predicted, however there were factors besides gravity that affected our results. The tape going through the spark timer affected the acceleration because of the friction it created, slowing down/altering the acceleration due to gravity. Also, our V-T graph didn't come out as expected because we assumed the line would have a negative slope. However, we measured the values to be positive instead of negative,making down have a positive position and velocity, so the line had a positive slope. We had minimal error in this lab, resulting in a percent error of about 7.45%. This percent error was cause by the friction between the ticker tape and the timer, and the fact that our first dot wasn't made precisely as the free fall object began its drop. Also, as usual error could have occurred through our measuring, for example our measurement couldn't have been as accurate as we thought they were. If I could change this lab to address error, I would try and minimize friction between the ticker tape and the timer as best as possible by making sure it gets fed through without any resistance. Also, I would use a much more detailed measuring tool that allowed us the ability to be as accurate as possible.
 * Conclusion-**