Ch2_SekiK

toc =Physics Classroom Notes:=

**Lesson 1 (abcd):**

 * 1) **What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully**
 * From our previous discussions in the class, I already understood the concept of vectors and scalars. Scalar quantities are only measured in size, or numeric value, while vectors take into account the direction of the object as well. Included in scalar quantities are distance and speed, while vectors include displacement and velocity.
 * 1) **What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.**
 * I was a little bit shaky about the velocity formula because we only covered very briefly, as well as being rushed for time while learning it. However, the reading cleared up the confusion by presenting the information in an easy to understand format: [[image:http://www.physicsclassroom.com/Class/1DKin/U1L1d4.gif]]
 * I now understand that velocity deals with a change in position, which would be displacement, rather than distance.
 * 1) **What (specifically) did you read that you still don’t understand? Please word these in the form of a question.**
 * I understood most of the material as it was mostly covered in class already, as well as the fact that it was relatively straightforward.
 * 1) **What (specifically) did you read that was not gone over during class today?**
 * Before reading, we did not go over the difference between constant speed and instantaneous speed. While they are relatively easy concepts to understand, the difference between the two is substantial. Constant speed is a average of all instantaneous speeds, while instantaneous speed is the speed of something at one point in time.

Lesson 1 (e)

 * 1) **What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.**
 * I already understood the formula well, as well as it derivatives. It makes logical sense that the formula is the change in velocity over the time. I also understood the concept of constant acceleration. All it is is that something is speeding up or slowing down at the same rate per second.
 * 1) **What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.**
 * I understood everything we did in class.
 * 1) **What (specifically) did you read that you still don’t understand? Please word these in the form of a question.**
 * I didn't understand some components of the Rule of Thumb:
 * [[image:http://www.physicsclassroom.com/Class/1DKin/U1L1e6.gif]]
 * It says that Example B is slowing down, but still has a positive acceleration. How is it slowing down, while maintaining a positive acceleration?
 * 1) **What (specifically) did you read that was not gone over during class today?**
 * The relationship between the time's square and the total distance it covered in its fall. We did not discuss this connection, as well as any type of free falling acceleration.

**Lesson 2 (abc)**

 * 1) **What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.**
 * From our discussion, I already understood the two types of diagrams well. Ticker tape diagrams are based on a constant space of time between each dot. As the space between the dots change, the speed changes as well. If it gets smaller, it's getting slower, and if the space is getting larger, it is accelerating. On the other hand, motion or vector diagrams depict relatively the same thing but in less precise ways. It uses arrows to show how something can have a constant speed, how something could be slowing down, and how something could be accelerating.
 * 1) **What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.**
 * Nothing stood out from the class discussion as confusing. The diagrams are straightforward.
 * 1) **What (specifically) did you read that you still don’t understand? Please word these in the form of a question.**
 * The reading as well was also straightforward and easy to understand. There were no problems.
 * 1) **What (specifically) did you read that was not gone over during class today?**
 * Everything was covered in class; in fact, we covered more in class.

Lesson 3 (abc):

 * 1) **What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.**
 * I understood all of the concepts from the reading due to our class discussion. For example, the slope on a position time graph is the velocity of the object. This is because the equation for slope is identical to the equation for velocity when considering the labels on the x and y coordinates. In addition, I already understood how the shape of the line correlates to the velocity of the object. A straight one denotes a constant velocity, while that a curved one reveals that the velocity is changing.
 * 1) **What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.**
 * I understood everything from our previous class discussion.
 * 1) **What (specifically) did you read that you still don’t understand? Please word these in the form of a question.**
 * I have no questions, as the content was discussed and understood.
 * 1) **What (specifically) did you read that was not gone over during class today?**
 * We discussed everything in class

Lesson 4 (abcde)

 * 1) **What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.**
 * I already understood all concept they discussed about velocity-time graphs. I knew that the slope was representative of acceleration because changes in velocity = changes in acceleration. In addition, I knew that the area of the shape created by the line and the x-axis was the displacement. This is because velocity multiplied by time is displacement.
 * 1) **What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.**
 * I understood everything from the class discussion
 * 1) **What (specifically) did you read that you still don’t understand? Please word these in the form of a question.**
 * I have no questions
 * 1) **What (specifically) did you read that was not gone over during class today?**
 * Everything was covered in the discussion in class.

Lesson 5 (abcde)
5a Topic Sentence: Objects that free fall are affected only by gravity, and thus, have an acceleration of -9.8m/s/s A free falling object is an object that is falling under the sole influence of gravity. There are two motion characteristics that are true of free-falling objects: Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s, a ticker tape trace or dot diagram of its motion would depict an acceleration.
 * 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

5b Topic Sentence: Free falling objects accelerate at 9.8m/s/s. This numerical value for the acceleration of a free-falling object is given a special name. It is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. This quantity has a special symbol to denote it - the symbol g. The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s/s.  Acceleration is the rate at which an object changes its velocity. It is the ratio of velocity change to time between any two points in an object's path. To accelerate at 9.8 m/s/s means to change the velocity by 9.8 m/s each second.  If the velocity and time for a free-falling object being dropped from a position of rest were tabulated, then one would note the following pattern. The free-falling object has an acceleration of approximately 9.8 m/s/s. Another way to represent this acceleration of 9.8 m/s/s is to add numbers to our dot diagram. The velocity of the ball is seen to increase as depicted in the diagram at the right.
 * ~ Time (s) ||~ Velocity (m/s) ||
 * 0 || 0 ||
 * 1 || - 9.8 ||
 * 2 || - 19.6 ||
 * 3 || - 29.4 ||
 * 4 || - 39.2 ||
 * 5 || - 49.0 ||

5c Topic Sentence: Using graphs can help to visualize the acceleration of gravity, as well as finding other information we already knew how to get. One such means of describing the motion of objects is through the use of graphs - position vs. time and velocity vs. time graphs.   A curved line on a position versus time graph signifies an accelerated motion. 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. A further look at the position-time graph reveals that the object starts with a small velocity afand finishes with a large velocity (fast). Since the slope of any position vs. time graph is the velocity of the object, the small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity. The negative slope of the line indicates a negative (i.e., downward) velocity. 

Since a free-falling object is undergoing an acceleration, its velocity-time graph would be diagonal. The object starts with a zero velocity 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. This analysis of the slope on the graph is consistent with the motion of a free-falling object - an object moving with a constant acceleration of 9.8 m/s/s in the downward direction.

5d Topic Sentence: We can find the velocity of an object with the formula final velocity = g x t Free-falling objects are in a state of acceleration. They are accelerating at a rate of 9.8 m/s/s. Thus, the velocity of a free-falling object that has been dropped from a position of rest is dependent upon the time that it has fallen. The formula for determining the velocity of a falling object after a time of t seconds is vf = g * t where g is the acceleration of gravity. The value for g on Earth is 9.8 m/s/s. The above equation can be used to calculate the velocity of the object after any given amount of time when dropped from rest.

5e Topic Sentence: Free falling objects fall at the same rate regardless of mass. Yet the questions are often asked "doesn't a more massive object accelerate at a greater rate than a less massive object?" "Wouldn't an elephant free-fall faster than a mouse?"  The answer to the question is absolutely not! That is, absolutely not if we are considering the specific type of falling motion known as free-fall. Free-fall is the motion of objects that move under the sole influence of gravity; free-falling objects do not encounter air resistance. More massive objects will only fall faster if there is an appreciable amount of air resistance present.  The actual explanation of why all objects accelerate at the same rate involves the concepts of force and mass. You will learn that the acceleration of an object is directly proportional to force and inversely proportional to mass. Increasing force tends to increase acceleration while increasing mass tends to decrease acceleration. Thus, the greater force on more massive objects is offset by the inverse influence of greater mass. Subsequently, all objects free fall at the same rate of acceleration, regardless of their mass.

=Class Notes:=

9/6
__Position__ - where something is relative to a reference object __Distance__ - how far something traveled in total __Displacement__ - how far something traveled from its original point, including distance
 * Definitions**

__Speed__ - How fast something is moving __Velocity__ - speed with direction

Displacement and velocity are vectors because they have size and direction, while distance and speed are scalar quantities.

9/8
__Average Speed__ - average of instantaneous speeds __Constant Speed__ - speed never changes
 * Definitions**

4 types of motion: Increasing/Decreasing speed, at rest, and constant speed

Motion Diagrams: When at rest, a(cceleration) = 0, and v(elocity) = 0. Constant speed = --->...--->...---> Increasing speed = --->...->...---> (acceleration is to the right, and it should be pointing in the same direction as velocity) Decreasing Speed= --->...-->...---> (acceleration is the left, and it points in the opposite direction of velocity)
 * 2 types of Diagrams**

Dot Diagrams 10Hz = between each dot is 1/10 of a second 60Hz = between each dot is 1/60 of a second The space between each dot corresponds to the speed. If the space gets larger as you read the tape, then it is accelerating because it is covering more ground in less time. If it gets smaller, than it is slowing down.

LAB: Speed of a Constant Motion Vehicle (CMV)
__Objectives__ __Hypotheses:__
 * 1) How precisely can you measure distances with a meter stick?
 * 2) How fast does my CMV move?
 * 3) What information can you get from a position-time graph?
 * 4) Spark timer
 * 5) Spark tape
 * 6) Meter stick
 * 7) Masking tape
 * 8) CMV
 * 1) I probably measure distance in centimeters up to one decimal place. This is because common ruler measure down to the millimeter, which is one decimal place in centimeters.
 * 2) It moves probably about a half a meter per second, because it moved past my laptop in one second.
 * 3) From a position-time graph, I can use the information to find the speed of the CMV. This is because we can find velocity from the change in position over the change in time.

Data Table: Position vs. Time \

Analysis: We use a linear trend line because with this, we got a .99358 for our R^2 value. This means that our points are 99.358% accurate to the actual trend line. The line is constant because our vehicle was a CMV, or constant motion vehicle. It moved at a set rate per second, which, through the equation and slope, we can deduce is about 34.951 cm per second. Discussion Questions:
 * 1) The equation for the slope is the change in y over the change in x. Because "y" is the position and "x" is the time, we can substitute those words in for the respective variables. This makes it identical to the formula for average velocity. Thus, we can say that the slope is the average velocity.
 * 2) We can say that this is a calculation of average velocity because it is an average of speeds taken at ten different points, rather than just one.
 * 3) We can set the y intercept to zero because at zero seconds, the CMV was not moving, hence a 0 in position.
 * 4) The R^2 value is the representation of how accurate our data is to the real thing. Our value, .99358, signifies a 99.358% of accuracy in our data.
 * 5) If the CMV were to move more slowly, the line would be beneath ours. This is due to its lower slope, which as discussed in question one, shows a lower average speed. Thus, our line would be getting higher at a faster pace than a slower CMV's line.

Conclusion: After using the ticker tape diagram to record the speed of our CMV, we found that it moved 34.95 centimeters per second. This was slower than our hypothesis, which stated that we thought it could move 50 cm in one second. This was found by inputting the numbers into excel and graphing it to find our slope and R constant. There were a few things that could, however, have may have skewed our results. The surface that the CMV moved on could not have been level - which would cause either an increase or decrease in speed. Our methods in measuring the distance between each dot on the tape could have been improved as well. By using a material like tape to firmly hold the ruler or the tape down, we could have had more accurate results. Furthermore, we could have used the last ten dots instead of the first ten to get rid of inaccuracies caused by the accelerating CMV.

9/12
Data Studio Intro - use setup to change graphs - in order to see them all at the same time, drag and drop them from the data box into the graph one in the displays box

- to retake a run, you could go to data and hide it, or you could go to experiment and delete the data run. DON'T USE THE DELETE BUTTON.


 * 1) How can you tell that there is no motion on a…
 * 2) position vs. time graph
 * The line would be horizontal, as demonstrated in run #3. It would not have any slope, though it has a space from the x axis. This shows that there is some distance from the sensor and I.
 * 1) velocity vs. time graph
 * The line would be horizontal, but the line would be equal to zero because there is no velocity.
 * 1) acceleration vs. time graph
 * The line would be horizontal again, as well as equaling zero. This is because we are not accelerating
 * 1) How can you tell that your motion is steady on a…
 * 2) position vs. time graph
 * We can tell if our motion is steady if there is no change in the slope of the line.
 * 1) velocity vs. time graph
 * You can tell if your motion is steady if you have a consistent velocity - or in other words, if your line stays horizontal.
 * 1) acceleration vs. time graph
 * The line would be a horizontal line equaling zero.
 * 1) How can you tell that your motion is fast vs. slow on a…
 * 2) position vs. time graph
 * The slower motion would have a shallower line, meaning it has a slope less than that of a faster line. The faster line would increase at a much quicker rate and be higher.
 * 1) velocity vs. time graph
 * While they are both horizontal lines, the faster motion should be higher than the slower motion.
 * 1) acceleration vs. time graph
 * As long as the motion was a constantly faster or slower rate, there would be no way to tell.


 * 1) How can you tell that you changed direction on a…
 * 2) position vs. time graph
 * The slope of the line would become the opposite.
 * 1) velocity vs. time graph
 * On a velocity vs. time graph, you can tell if you changed direction if the velocity becomes negative or positive.
 * 1) acceleration vs. time graph
 * There is no way to tell a change in direction


 * 1) What are the advantages of representing motion using a…
 * 2) position vs. time graph
 * Motion is easily represented. In addition, it is very easy to calculate average speed, distance, and displacement.
 * 1) velocity vs. time graph
 * Direction is most easily seen on this graph - negative means that the object is coming toward the sensor, positive means it's going away from it.
 * 1) acceleration vs. time graph
 * Any change in speed will be very easily seen


 * 1) What are the disadvantages of representing motion using a…
 * 2) position vs. time graph
 * Speed and velocity must be calculated using the formula
 * 1) velocity vs. time graph
 * Position from the origin must be calculated using the velocity formula.
 * 1) acceleration vs. time graph
 * It is impossible to find changes in direction, as well as average speed and velocity.


 * 1) Define the following:
 * 2) No motion
 * Position is constant, and both velocity and acceleration are at zero.
 * 1) Constant speed
 * While acceleration is zero and velocity is a constant, position should change at a constant rate over time.

9/15
Average speed: V = D/T V = 56/11 V = 5.1 m/s Average velocity: V = change in D/total T V = 14-10 / 11 = 4/11 = .44 m/s Acceleration is zero - whether you're at constant speed or at rest, you have no acceleration
 * segment || motion || displacement || velocity ||
 * AB || At rest || 0 || 0m/s ||
 * BC || constant speed toward origin || - 10 || -5m/s ||
 * CD || at rest || 0 || 0m/s ||
 * DE || constant speed behind origin || -16 || 32m/s ||
 * EF || at rest || 0 || 0m/s ||
 * FG || Constant speed toward origin || 16 || 16m/s ||
 * GH || constant speed away || 14 || 7m/s ||

LAB: Acceleration Graphs
Lab partner: Hella Talas

Objectives: Hypothesis Materials Procedure Data: Analysis: Increasing speed: halfway point = 31.6 m/s. end speed is 64.3 m/s Decreasing speed: halfway point = 55 m/s. end speed is 11.8 m/s Discussion Questions:
 * 1) What does a position-time graph for increasing speeds look like?
 * 2) What information can be found from the graph?
 * 1) A position time graph for increasing speed is a line whose position increases as time goes on. This is because as time goes on, speed increases. Therefore, its position gets farther and father away from the origin quicker.
 * 2) The velocity can be found from the slope, and the displacement can be calculated as well. Velocity can be found from the slope, while displacement can be calculated from the position-time graph.
 * 1) Spark tape, spark timer, track, dynamics cart, ruler
 * 1) Set the ramp on top of the Physics textbook and line up the two ends
 * 2) Place the timer onto the top of the ramp
 * 3) Put the tape through the timer, and then tape the tape to the cart.
 * 4) Turn on the timer, and then let the cart go down the ramp.
 * 5) Take the tape from the cart.
 * 6) Take the timer and place it on to the bottom of the ramp
 * 7) Thread the tape through the timer and then tape it to the cart
 * 8) Turn the timer on and push the cart up the ramp, and let it come to a full stop before stopping the timer
 * 9) With the two tapes with the marks on it, measure the distances between each and plot them into Excel.
 * 10) Make the graphs
 * 1) Interpret the equation of the line (slope, y-intercept) and the R2 value.
 * The line, being a polynomial equation, has the form y=ax^2 + bx. This directly correlates with the formula d=ViT + 1/2aT^2. Therefore, we can substitute a for T, and b for Vi. The y-intercept is at 0 because at zero seconds, it had not changed position. We use a polynomial fit line because it gave us the highest R2 value at .99995, while the linear fit gave us a value of about .93
 * 1) Find the instantaneous speed at halfway point and at the end. (You may find this easier to do on a printed copy of the graph. Just remember to take a snapshot of it and upload to wiki when you are done.)
 * 1) Find the average speed for the entire trip.
 * For increasing, the average speed is 27.55 m/s
 * For decreasing, the average speed is 46.67 m/s.
 * 1) What would your graph look like if the incline had been steeper?
 * If the incline had been steeper, the lines in our graph as a whole would be both steeper and flatter. The increasing speed would have covered more distance in less time, so the line would become much steeper. On the other hand, the decreasing speed would slow down at a much faster rate, causing the line to be flatter.
 * 1) What would your graph look like if the cart had been decreasing up the incline?
 * It would look relatively similar to the decreasing speed line on our graph because that is what it was doing: decreasing speed.
 * 1) Compare the instantaneous speed at the halfway point with the average speed of the entire trip.
 * For increasing speed, the speed at the halfway point was 31.6 m/s. The average, however, was 27.55 m/s. For decreasing speed, the speed at the halfway point was 55 m/s. The average was 46.67 m/s.
 * 1) Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense?
 * This is because at the tangent line, only one point on the curve is on the line. Thus, we can isolate the speed at that one point by finding the equation of the tangent.
 * 1) Draw a v-t graph of the motion of the cart. Be as quantitative as possible.

Conclusion: Our hypothesis stated that a position-time graph for increasing speed is "a line whose position increases as time goes on." After our experiment, we found this to be true, as even in cases where the speed is not constant, the position increases with the advancement of time. However, I did not predict the parabolic shape. The second statement said that velocity and displacement could be found from the graph. We found this to be true as well. However, there were things that could have been avoided to make this more accurate. For example, when we pushed the cart up the ramp, it was very easy to make mistakes in stopping the cart at the right time. This could have affected our data if we included the ends of the tape. The middle, however, was unaffected. It was also possible to make inaccuracies while measuring due to misreading our own data. Small errors such as making sure that the measuring tape wasn't perfectly flat against the tape could have skewed our results. This could have been avoided by making sure that our measurements were accurate. We also could have avoided the cart problem by allowing it to come to a full stop before switching the timer off.

9/21




LAB: A Crash Course in Velocity (part II)
lab partners: Hella Talas, Noah Pardes, Mike Poleway Objective: Both algebraically and graphically, solve the following 2 problems. Then set up each situation and run trials to confirm your calculations. Materials: CMV, tape measure/meter sticks, Masking tape (30 cm/group), stop watch, spark timer, and spark tape.
 * 1) Find another group with a different CMV speed. Find the position where both CMV's will meet if they start //at least// 600 cm apart, move towards each other, and start simultaneously.
 * 2) 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:

Procedure: media type="file" key="Catch Up Process.mov" width="300" height="300" media type="file" key="Crash Process.mov" width="300" height="300" Results: Crash: media type="file" key="Crash Close Up.mov" width="300" height="300" average: 370.25 Catch up: media type="file" key="Catch up 2.mov" width="300" height="300" Average: 186.33 Analysis: Discussion Questions
 * Trial || Position (that they meet at) ||
 * 1 || 369 cm ||
 * 2 || 380 cm ||
 * 3 || 372 cm ||
 * 4 || 360 cm ||
 * Trial || Position (that they meet at) ||
 * 1 || 192 cm ||
 * 2 || 180 cm ||
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 17.6px;">3 || <span style="font-family: 'Times New Roman',Times,serif; font-size: 17.6px;">187 cm ||
 * 1) Where would the cars meet if their speeds were exactly equal?
 * If the cars were exactly equal, they would meet at the halfway point; that is, 300 cm. However, in the catching up situation, the cars would never meet, because of their equal speeds. The only reason why the blue car was able to "catch up" was because its speed was higher than that of the yellow vehicle.
 * 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?
 * No, you cannot find the points where they intersect.

Conclusion: After our experiments, we found that our initial calculations were very accurate. We predicted that in the crash test, they would meet at 362.57 cm away from the faster car, while the slower car would travel 237.43 cm. With a percent error of -2.12%, we can be sure that our results are accurate. In the Catching up test, we predicted that it would take 189.78 cm for the blue car to catch up to the slower one. With a percent error of 1.82%, we can be sure, like in the crash test, that our results are good. However, we could have been more accurate in our testing. For example, turning each of the vehicles simultaneously on is close to impossible. This skewed our results, because one would have slightly more time to travel. In addition, our blue vehicle always ended up curving from its original path. This made our results inaccurate. Timing could be fixed by doing more trials so that as we get more results, the average result becomes more pronounced than the exception. In that way, we can fix any error with timing. We could have fixed our curving car by leaning it against a wall with no holes so that it would reliably go straight. This means that we would get better results because our car wouldn't waste time by curving and not going straight. This would ensure that we get more accurate results more often, making our graph as a whole better.

Lab: EGG DROP PROJECT
Partner: Max Llewellyn (sp?) Egg Status: Completely Intact The egg was placed in the interior of the contraption - the part right in between the wings. Our cone shape distributed the force from the landing in the best way possible so as to not harm the egg, while the wings helped in the case that it landed on its side (which it did). We found straws to be the most efficient material, because it was hollow, helped to cushion the egg, and was overall, very light. The parachute in the back helped to slow the egg down so that it would hit the ground at the maximum force. This was a very efficient design as well, weighing in at about 23.11 grams.

Calculations:

Conclusion: As we made each separate prototype, our designs got better each and every time. Our first prototype was too light - we made it completely out of paper, and therefore, the egg cracked. The second one was very similar to our final, but we found that often, the egg would pop out and break. Therefore, on the third, we added much more reinforcement to keep this from happening. What we could have done better is to have made the parachute more symmetrical - we found that often, the contraption would spiral down very dangerously, though the extra protection in the wings often saved it. Making the parachute better could have made for a safer fall by letting it land on the tip of the cone, the safest point.

LAB: What is acceleration due to gravity?
Purpose:
 * 1) What is acceleration due to gravity?
 * 2) What does a v-t graph of a free falling object look like?
 * 3) How are you going to find the acceleration?

Hypothesis:
 * 1) Acceleration due to gravity is equal to -9.8 m/s/s.
 * 2) A v-t graph of a free falling object is a diagonal line in the negative region. It is diagonal because it is acceleration, and it's negative because it is going away from the origin in the negative direction.
 * 3) The slope of the velocity-time graph is the acceleration.

Materials:
 * 1) Spark Timer
 * 2) Spark Tape
 * 3) Mass (>100)

Procedure: Data: time/position data table Velocity table: Graph:
 * 1) Tear off a long piece of spark tape.
 * 2) Thread it through the timer, and tape the mass over the end
 * 3) One person should drop, while another should hold the timer and maker sure the tape passes through freely.
 * 4) Go to Balcony, turn on the timer, and drop the mass

Analysis: Our slope on our v-t graph was equal to 767.02 cm/s. Compared to what it should be, 980 cm/s, our velocity is significant lower than it should be. This is due to friction between the tape and the timer, which slowed down the freefall of our object. Another very small factor is air resistance, though the contribution it made is miniscule and probably insignificant. Our percent error is equal to: (Theoretical - Experimental)/Theoretical X 100 (980-767.02)/980 X 100 = 21.7% This percent error is largely due to the force of friction acting upon our experiment. Because free fall is the absence of any other force but gravity, adding friction is predictably going to skew our results off. Our percent difference is: ((average experimental value - individual experimental value) / average experimental value) x 100 834.03 - 767.02 / 834.03 x 100 = 8.03% Because friction was acting upon everyone's experiments, our percent difference from the rest of the class is relatively closer. The same force affected everyone, so we should be getting closer results. The y-intercept on our graph is not set to 0 because it is impossible to perfectly coordinate the turning on the timer and the drop time. Class results: Measurements in cm/s Discussion Questions: Conclusion: Our acceleration due to gravity was 767.02 cm/s, which is considerably slower than the actual acceleration, which is 980 cm/s. This is explained by friction between the tape and the timer, which slowed the object considerably. Our v-t graph's shape is right on, though the slope is not. It is as we expected, a straight diagonal line that increases due to the acceleration. Likewise, our x-t graph's shape is correc tin the parabolic line, though it's slope probably isn't. There are many errors within this lab that we could try to fix next time. The first things is to make sure that our tape is completely flat, which allows for the most clear markings in our tape. Any bends or folds could create inaccuracies, because if a mark is made after a fold, measurements would become skewed. This is because after unraveling the fold, the tape would actually become longer just in that one spot, making it inaccurate. Another problem we could fix is inspecting the marks better after dropping. This, in our case, led to a huge inaccuracy that skewed our results. We can avoid this next time by saving many of our trials and then picking the best one.
 * v-t Graph Class Data ||
 * 853.72 ||
 * 861.69 ||
 * 805 ||
 * 708.97 ||
 * 767 ||
 * 864 ||
 * 881.5 ||
 * 887.79 ||
 * 876.56 ||
 * Average: 834.03 ||
 * 1) Does the shape of your v-t graph agree with the expected graph? Why or Why not?
 * It agrees with the shape of our v-t graph because it is an object in freefall. That means that it is accelerating at 9.8 m/s, which would make the graph a straight diagonal line above the origin, going upward. Our graph does that.
 * 1) Does the shape of your x-t graph agree with the expected graph? Why or why not?
 * Again, it agrees because an accelerating object will make a curve with the slope becoming steeper as the object accelerates. Our graph does the same.
 * 1) How do your results compare to that of the class? Use percent different to discuss quantitatively
 * We had the second slowest velocity in the class at 767.02 cm/s. When calculated, our percent difference from the class was 8.03%. Even though we are not close to 980 cm/s, we are closer to the results that the class gained.
 * 1) Did the object accelerate uniformly? How do you know?
 * It very clearly does not accelerate uniformly on our v-t graph. There are two incredibly off points, which skewed our results considerably. We double checked our numbers, and they were correct. Something odd happened in the intervals of .3-.4 seconds, and .4-.5 seconds. This caused our object to clearly accelerate erratically.
 * 1) What factor(s) would cause acceleration due to gravity to be higher that it should be? Lower than it should be?
 * If someone were to pull on the object or to throw it at the ground, then the acceleration due to gravity would be much higher. Otherwise, nothing would make it higher than it should be besides the Earth changing in some way to change it. However, there are many things that can make the gravity seem to be lower than it should be. One reason is that gravity isn't the only force acting on the object. According to our definition of freefall, gravity is the only force. Clearly, it wasn't freefalling of there was friction between the tape and the timer. Air resistance is also a possibility, but would be very insignificant in our experiment.