Chapter+2

=**__Chapter 2 Notes__**=
 * Average speed, constant speed, and instantaneous speed are all solved by V = change in direction/change in time
 * Four types of motion are: At rest (V=0), Constant speed, increasing speed, and decreasing speed
 * Motion Diagram
 * V = 0, a = 0
 * To show increasing speed, make the arrows bigger
 * accel points in same direction as velocity
 * To show decreasing speed, make the arrows smaller
 * accel points in opposite direction as velocity
 * To show decreasing speed, make the arrows smaller
 * accel points in opposite direction as velocity


 * V = change d/change t (constant and average speed only)
 * V = vi +vf/2 )average speed)
 * change d/change t = Vi +Vf/2
 * **change d = 1/2(Vi + Vf)change t**
 * Acceleration is rate of change in speed
 * a = Vf - Vi/change t
 * **Vf = Vi +at**

__Free fall__: Any object only acted on by the force of gravity. toc =**__Chapter 2 HW__**= 1) What we learned in class that I understood well were distance, displacement, and speed. These were easy concepts to understand in class, as the reading only fortified my understanding of these topics. Distance is the total amount that an object moved, while displacement is an objects position relative to where it began. Speed is how quickly an object moved from one position to another. 2) What I was a little shaky on from class was velocity, but after reading, it now makes sense. I thought that velocity and speed were the same except that velocity had a direction. After reading, I now realize that speed measures the total distance traveled, while the velocity measures the level of displacement. 3) Thankfully, there was nothing that I still do not understand, as the reading has helped me learn all of the subjects. 4) The things we did not cover in class but I now understand were Scalars, Vectors, and Acceleration. 1) What I read that we had already learned in class was both of the main topics in this reading, which were Ticker Tape Diagrams and Vector Diagrams. The ticker tape is designed to track an objects motion in a given time, and shows whether that object was at a constant velocity or if it accelerated or decelerated. The vector diagrams are used to have a physical representation of somethings movement, and whether it is at constant velocity, accelerating, or decelerating. 2) There was nothing that I was shaky on, but now understand. The reading was quite clear on what it meant to teach. 3) There is one thing that I am confused about, but I think it will be shown in time. I am still wondering: How can the vector diagrams show things such as force and momentum? However, I think we will learn about this relatively soon. 4) We went over everything in the reading in class. 1) Acceleration: The rate of an object's change in velocity over a certain period of time. It does not matter what the object's speed is, only its velocity. Also, since it is a vector quantity, it has a direction associated with it as well. The direction depends on if the object is speeding up or slowing down and if the object is moving in a positive or negative direction. Constant Acceleration: When an object's velocity is changing by a constant amount each second. This is different than constant velocity. In this case, the object's velocity changes constantly, while in constant velocity, the velocity is staying the same 2) During class, I did not think that acceleration was a vector quantity, since it seemed to just change when velocity changed. Now I understand that acceleration also depends on the direction the object is moving as well as if it is speeding up or slowing down. For example, when an object is speeding up in a negative direction and its acceleration is in the same direction of the velocity, its acceleration will be negative. 3) How would acceleration be affected if an object was moving in either upwards or downwards, instead of in a compass direction? 4) In the reading, we read about free-falling objects, which experience constant acceleration, and continuously gain a constant amount to their velocity. 1) Slope of a p-t graph: Slope is found by the change in the y-coordinates divided by the change in x-coordinates. Since the y-coordinates is position and the x's are time, this formula would be the change in position over the time, which is also the equation for velocity. Therefore, the slope of the line equals the object's velocity. Meaning of the Shape of a p-t Graph: If the line is straight, then the velocity is constant since the slope is not changing. Also, if the line is pointing upwards, then the object has a positive velocity, and vice versa. Furthermore, if the line is curved, an object's velocity is changing, and it takes a different amount of time for an object to travel a certain distance. If an object has a curve to the right before going up, it is gaining speed, while if the line curves up and then moves right, it is losing speed. 2) There is nothing that I am still shaky on. 3) How would a p-t graph look like if an object moved upwards or downwards vertically? 4) We went over everything in class.
 * Acceleration = -9.8 m/s^2
 * Negative because it is moving downward
 * **Physics Classroom Lesson 1 Summary (9/7/11)**
 * **Physics Classroom Lesson 2 Summary (9/8/11)**
 * **Physics Classroom Lesson 1e Summary (9/13/11)**
 * **Physics Classroom Lesson 3 & 4 (9/14/11)**
 * __Lesson 3__ **

1) Plotting points on a v-t graph: Since the y-axis is velocity and the x-axis is time, a point is plotted at the object's velocity at that time period. If the object's velocity is increasing, the plots will be on a line with a positive slope, and vice versa. Also, if an object stops moving, the velocity remains constant at whichever y-value (velocity) it was last at. Increasing and Decreasing Velocity on V-T Graphs: If the line is moving away from the origin left to right, than the object is speeding up since is moving further away from where it started. Contrastingly, if the line is moving towards the origin left to right, it is slowing down, because it is returning to the spot where it started moving. 2) During class, I was confused on how the acceleration was found using the line on the v-t graph. However, after reading, I know see that since the line's slope is the change in velocity of change in time, it has the same formula that is used for acceleration 3) How would you find the amount of displacement on a v-t graph is the line was curved? 4) Using Velocity-Time Graphs to find the amount of displacement. This is done by finding the area of a shape expanding from the line of velocity to the x-axis.
 * __Lesson 4__ **

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.
 * __Lesson 5__**
 * 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.

=__Lab: Speed of a CMV__= 1) You can not measure very precisely with a meter stick. Only to about the cm. 2) The CMV moves at about 50 cm/s 3) A position-time graph shows you the acceleration of an object.
 * Objectives:**
 * How precisely can you measure distances with a meter stick?
 * How fast does my CMV move?
 * What information can you get from a position-time graph?
 * Materials:**
 * Spark timer & spark tape
 * Meterstick
 * Masking tape
 * CMV
 * Hypothesis 1, 2, and 3:**

Length of laptop (cm) = 32.95
 * Observations:**


 * Experiement:**

We use a linear trend line because we got an R^2 value of 0.98, which is almost perfect. This means that our points on the graph are 98% on the trend line. Our line was constant because the vehicle we used is a CMV, which means "constant motion vehicle". Since it moved at a set rate per second, we can deduce that the slope is equal to this rate, which for our CMV was 58.365 cm/s.
 * Analysis:**

**Discussion Questions:** __1) Why is the slope of the position-time-graph equivalent to the average velocity?__ The slope of the position-time-graph is equivalent to the average velocity because they are one in the same. You find average velocity by dividing the amount of displacement by how long it took to get there. This is the same as how the slope is written, making them the same number. __2) Why is it average velocity and not instantaneous velocity? What assumptions are we making?__ It is average velocity because the graph is an average of 10 points, while Instantaneous velocity is the velocity at a single point. We are assuming that the CMV isnot changing speeds drastically and moving at a constant speed. __3) Why was it okay to set the y-intercept equal to zero?__ It was okay to set the y-intercept to equal zero because we started recording the time at 0 seconds, and when the time was at 0 seconds, the CMV had moved 0 cm. __4) What is the meaning of the R 2 value?__ The R 2 value represents how accurate our line of best fit is. The closer to 1 the value is, the more accurate our line is. __5) If you were to add the graph of another CMV that moved more slowly on the same axes as your current graph, how would you expect it to lie relative to yours?__ Since the CMV is moving more slowly, that means it would have a lower velocity, and by extension, a lower slope. This would result in the line lying lower than my own. Since the average velocity of the other CMV at any given time will be slower than mine, the location should be farther behind my CMV at any same time.

**Conclusion:** After recording, we found that the speed of our CMV was 58.365 cm/s. This was slightly faster than our hypothesis, which stated that we thought it would move at a rate of 50 cm/s. We found our speed by putting the points recorded on thje ticker tape into excel and graphing it to find our slope and R constant. However, there were a few things that may have skewed our results. First, the surface that the CMV moved on may not have been level. This would cause either an increase or decrease in speed. The way we measured the distance between each dot on the tape could have been improved as well. For more accurate results, we could have laid out the entire strip and measured it with a tape measure instead of a ruler. This would have eliminated the room for movement, thus making our measurements more accurate.

=__Activity: Graphical Representations of Equilibrium__=
 * Objectives:**
 * What is the difference between static and dynamic equilibrium?
 * How is “at rest” represented on a position vs. time graph? On a velocity vs. time graph? On an acceleration vs. time graph?
 * How is constant speed represented on a position vs. time graph? On a velocity vs. time graph? On an acceleration vs. time graph?
 * How are changes in direction represented on a position vs. time graph? On a velocity vs. time graph? On an acceleration vs. time graph?
 * Materials:**
 * Motion Detector with USB link
 * Experiment:**

1) __How can you tell that there is no motion on a...__ 1. Position vs. Time Graph The line would be horizontal, as demonstrated in run #1. It has no slope, although it is still shown on the x axis. This showed that there is some distance between the sensor and myself. 2. Velocity vs. Time Graph The line would be slightly horizontal, but the line would be equal to zero because there is no velocity. 3. Acceleration vs. Time Graph The line would be slightly horizontal again, as well as equaling zero. This is because we are not accelerating.
 * Disscusion Question:**

2) __How can you tell that your motion is steady on a...__ 1. Position vs. Time Graph We can tell if our motion is steady if there is no change in the slope of the line. 2. 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. 3. Acceleration vs. Time Graph The line would be a horizontal line equaling zero.

3) __How can you tell that your motion is fast vs. slow on a...__ 1. 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. 2. Velocity vs. Time Graph While they are both horizontal lines, the faster motion should be higher than the slower motion. 3. Acceleration vs. Time Graph As long as the motion was a constantly faster or slower rate, there would be no way to tell.

4) __How can you tell that you changed direction on a...__ 1. Position vs. Time Graph The slope of the line would become the opposite. 2. Velocity vs. Time Graph On a velocity vs. time graph, you can tell if you changed direction if the velocity becomes negative or positive. 3. Acceleration vs. Time Graph There is no way to tell a change in direction

5) __What are the advantages of representing motion using a...__ 1. Position vs. Time Graph Motion is easily represented. In addition, it is very easy to calculate average speed, distance, and displacement. 2. 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 . 3. Acceleration vs. Time Graph Any change in speed will be very easily seen

6) __What are the disadvantages of representing motion using a...__ 1. Position vs. Time Graph Speed and velocity must be calculated using the formula 2. Velocity vs. Time Graph Position from the origin must be calculated using the velocity formula. 3. Acceleration vs. Time Graph It is impossible to find changes in direction, as well as average speed and velocity.

7) __Define the following:__ 1. No motion There is no change in position, and both acceleration and velocity are at zero. 2. Constant speed The position should be changing at a constant rate over time, despite acceleration being at zero and velocity being constant.

=__Lab: Acceleration Graph__=
 * Hypothesis:**
 * The position-time graph of an increasing speed would have a line with a positive slope
 * It will show the acceleration of the object


 * Procedure:**
 * 1) Gather required materials
 * 2) Set up ramp and place ticker and cart on ramp
 * 3) Feed ticker tape through ticker and attach to cart
 * 4) Run experiment
 * 5) Record Observations
 * 6) Analyze


 * Experiment:**


 * Analysis:**

__1) What would your graph look like if the incline had been steeper?__ If the incline had been steeper, the lines on our graph would be flatter and steeper. This is because since the cart would be increasing speed, it could be covering the distance in much less time. __2) What would your graph look like if the cart had been decreasing up the incline?__ Since it is decreasing speed, it would look very similar to the decreasing speed line on our graph. __3) Compare the instantaneous speed at the halfway point with the average speed of the entire trip.__
 * Discussion Questions:**

__4) Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense?__

__5) Draw a v-t graph on the motion of the cart. Be as quantitative as possible.__

Our hypothesis stated that a position-time graph for increasing speed is "a line with a positive slope." 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. On the other hand, I did not anticipate the parabolic slope. Our second statement said that acceleration could be found. This was also true. However, there were things we could have done to make it 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. If we included the ends of the tape, this would have affected out data. The middle, however, was unaffected.
 * Conclusion:**

=**__Lab: Crash Course in Velocity__**= __Partners: Kouske Seki, Michale Poleway, Hela Tallas__

1) Find another group with a different CMV speed. Find 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.
 * Objective:**


 * Materials:**
 * CMV
 * Tape measure/meter stick
 * Masking tape, stop watch
 * Spark timer/tape

__Crash Calculations__
 * Calculations:**

__Catch Up Calculations__

__Crash Process:__ media type="file" key="Crash Process.mov" width="300" height="300"
 * Process:**

__Catch Up Process:__ media type="file" key="Catch Up Process.mov" width="300" height="300"

__Crash Results:__ media type="file" key="Crash 1.mov" width="300" height="300" Average Position: 370.25 cm
 * Results:**
 * Trial # || Position that they collided (cm) ||
 * 1 || 369 ||
 * 2 || 380 ||
 * 3 || 372 ||
 * 4 || 360 ||

__Catch Up Results:__ media type="file" key="Catch up 2.mov" width="300" height="300" Average Position: 186.33
 * Trial # || Position they met at (cm) ||
 * 1 || 192 ||
 * 2 || 180 ||
 * 3 || 187 ||


 * Analysis:**

__1) Where would the cars meet if their speeds were exactly eqaul?__ For the crash test, the CMVs would meet up exactly half way, at 300cm. In the catch up test however, if they had the same speed, they would never catch up because they are moving forward at the same rate.
 * Discussion Questions:**

__2) Sketch position-time graphs to represent 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?__

After our experiments, we found that our pre-experiment calculations were quite accurate. In the crash test, we predicted that the slower car would travel 237.43 cm, while the faster car would travel 362.57 cm. They would also meet up at 362.57 cm. With a percent error within 2.12%, we are certain of the accuracy of our test results. 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. As in the crash test, with such a small percent error margin of 1.82%, we can safely say our measurements were accurate. However, our testing methods could have been more accurate. For example, our blue vehicle always ended up curving from its original path. This made our results inaccurate. Also, turning both CMVs on simultaneously is near impossible. This messed up our results because one would travel for slightly more time. In addition, While timing can be fixed by multiple trials, there is no surefire way to have perfect results each time. We could have fixed our curving car by leaning it against a wall, although when we tried this, we found this also ineffective because there were holes and screws in the wall.
 * Conclusion:**

=__Egg Drop Experiment__= Group: Noah Pardes and Robert Kwark Final Design was a succes, egg was completely intact Weight of design: 26.9g

d = 8.5 m, t = 1.52 s, V(initial) = 0, a=? d = vt + 1/2 a(t^2) 8.5 = 0t + 1/2 x (1.52)^2 x a 8.5 = 1.1552a a = 7.36 m/s^2
 * Calculations:**

Our resulting acceleration was lower than the ideal acceleration of an object in free fall, which is 9.8 m/s^2. This was to be expected as there were other forces acting on the design, mainly air resistance. However, our result was higher than the result of most other groups in our class that succeeded. It would appear that our acceleration was the highest of any successful group.

We caught a nice break in this project, in that our first prototype worked extremely well. Even so, we attempted to improve on our design by reducing the weight, but the second prototype failed. This was most likely because there was not enough support underneath the egg. The second prototype would almost always, as it was falling, hit against the wall of school, spin erratically, and land almost right on the egg. In the end, we went with the first design since it had worked well to begin with, and it was no exception when the final drop happened. The main reason it was a success was because we created such a wide surface area which caught more air resistance underneath, slowing the descent enough to leave the egg intact. Also, although it was more of an accident through design, we had created, within our cradle, a smaller cradle of tape so the egg would not have direct force against the ground, but be held back by the tape. The acceleration of our device was 7.36 m/s^2, which was lower than the maximum acceleration 9/8 m/s^2. On the other hand, our acceleration was one of, if not the fastest, acceleration out of all the groups whose eggs survived. If I could do this again, I would not change a thing, since ours was a success. Our design weighed very little, as well as the fact that it performed its task completely. The one flaw that I saw was that if there was a particularly strong wind, it would hit the wall and flip end over end, most likely ending in faliure. Luckily this did not happen on the final drop.
 * Conclusion:**

=__Free-Falling Lab__= //**Purpose:**// //**Hypothesis:**// //**Procedure:**// 1) Gather materials, which include: 2) Feed ticker tape through ticker device 3) Attach ticker tape to 100g weight 4) Drop weight over balcony and record
 * What will the ticker tape diagram look like from a falling object?
 * What is the acceleration due to gravity?
 * How can you answer the second question using a v-t graph of the dropped object?
 * It will look like the object is increasing speed. This means there will be more ticks when it ends than when it begins.
 * The acceleration due to gravity is a constant -9.8m/s^2
 * The slope of the v-t graph is the acceleration.
 * Ticker device
 * Ticker Tape
 * Mass object (100g weight)
 * Masking Tape
 * Timer
 * Power strip

//**Calculations:**// __Position-Time Graph Data__ __Velocity Time Graph Data__

x-t graphs:
 * //Analysis://**
 * __x-t and v-t graphs__**

__**Discussion and interpretations of the graphs**__ The slope on our v-t graph was 876.56 cm/s. Comparatively, where the ideal velocity was 981 cm/s, ours is somewhat lower than it should be. The main reason this happened is because of friction created when the ticker tape is sliding through the ticker, which slowed down our object during free fall. Another facotr was air resistance. If our weighted object had a much larger surface area, this could have been a major factor, but since it was only the size of a large coin, it was a very small, almost insignificant factor.
 * __v-t graph__**

__Percent Error:__ ((Theoretical - Experimental)/Theoretical) * 100 (980-876.56)/980 * 100 = 10.5% The main reason for percent error is due to the friction slowing down our free fall. This messes with the experiment because total free fall is the absence of any force other than gravity, and clearly there were forces of friction and air resistance to skew our results. __Percent Difference:__ Class Results (in cm) ((Average Experimental - Individual Experimental) / Average Experimental) * 100 ((834.03 - 876.56) / 834.03) * 100 = -5.10% Since friction was acting upon everyone's experiments, our percent difference compared to the rest of the class is much closer than compared to the theoretical result. The same force of friction affected everyone, so everyone should be getting closer results. The reason our y-intercept on our graph is not set to 0 is because it is impossible to perfectly coordinate when the ticker was turned on to when the weight began to fall.
 * __% Error and Percent Difference__**
 * 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?__ Our v-t graph agrees with the expected graph because it is an object in freefall. Idealy, an object in free fall is accelerating at 9.8 m/s, which makes the graph a straight diagonal moving away from the origin. This is what our graph looks like. __2) Does the shape of your x-t graph agree with the expected graph? Why or why not?__ Similarly to the v-t graph, it agrees because as the object accelerates, on a graph it will make a curve with the slope becoming steeper. Our graph does that. __3) How does your result compare to that of the class? (Use Percent difference to discuss quantitatively.)__ My group had the second fastest fall in the class, at 876.56 cm/s. As shown above, our percent difference to the class average of 834.03cm/s was -5.10%. Despite the fact that we are quite far off from the ideal 981 cm/s, we were still within about 5% of the class results, which is very close. __4) Did the object accelerate uniformly? How do you know?__ As clearly seen in my graph, the object did not fall uniformly. Besides the fact that there is a moderate variety throughout, the most noticeable change is between 0.3 and 0.4. Something happened in this time span which caused our object to fall erratically. __5) What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?__ If someone were to throw the object towards the ground or pull on it from the landing height, then the acceleration due to gravity would be much higher because there is force behind the acceleration of 9.8m/s that is already there. On the other hand, there are many things that can make the gravity seem to be lower than it should be, such as forces acting on an object. According to our definition of freefall, gravity is the only force acting upon an object. Therefore, the object wasn't free falling since there was friction between the ticker and the tape.
 * //Discussion Questions://**

Our acceleration due to gravity was 876.56 cm/s, which is a lot slower than the actual acceleration of an object in free fall, which is 981 cm/s. This is explained by friction between the ticker and the tape, which slowed the object considerably. Although the slope of our v-t graph is very off, our shape is exactly as it should be. The shape was as we expected, a straight diagonal line that increases due to the acceleration. Similarly, our x-t graph's slope is probably wrong, but our shape is correct in correct in the parabolic line. There are many errors within this lab that we could try to fix next time. The first things is to make sure that we have the ticker on the correct setting. Since we had it set to 60hz and not 10hz, we got a much larger quantity of data, which could have thrown off our results. Also, if our tape was completely flat, it would allow for the most clear markings on our tape. Because our tape was not totally flat, a few folds and bends in the tape. These folds and bends in the tape caused the distance of our measurement to be shorter than it should be, skewing our results. . Another problem we could fix is inspecting the marks more closely after dropping. This, in our case, led to a very unnecessary amount of data that could have skewed our results. We can avoid this next time by making sure it is on the right setting before we drop it, and inspect it and compare to our classmates after we drop it.
 * //Conclusion://**