Ch2_SolimanoM

= = =toc Class notes: Velocity= = 1-D Kinematics, Lesson 1 Summary (method 2a) =

1. The discussed topics that I understood most fully from the class were the ideas of distance, [|displacement], and [|velocity]. I understood that distance is the total “ground” on which an object moves. For example, if an object moves five feet North, and then moves five feet South, its total distance is ten feet, as this is the total “ground” which it has covered. If the object were to follow this same pattern, its displacement would be different. //The Physics Classroom// describes displacement as an object’s overall change in position, which is what I understood it to be. Displacement also takes into account an object’s direction, which distance does not. For example, if an object followed the same pattern above, its displacement would be zero, because it did not have a change in position when it finished moving five feet North and then five feet South. Velocity is the rate at which an object changes position. Velocity accounts for direction, so if the object does not change its position, it has no velocity. But, if an object moves in a direction and has a positive or negative gain in position it will have a velocity.

2. I was originally not sure of the difference between speed and velocity. While I understood velocity and its calculations, I did not understand speed and their differences. From reading, I now understand that speed is calculated by an object’s distance traveled, over its time traveled. I now understand that this differs from velocity, as speed does not take into account direction, while velocity does, as velocity is calculated by an object’s displacement over its time.

3. Is average speed an acceptable measure of a long period travel time, as it does not exhibit the intervals of the travel, but rather just one average speed for the trip?

4. In class, we did not learn about scalars and vectors. These are two categories in which the mathematical quantities that describe the motion of objects are placed. [|Scalars] are the quantities that are fully described by only a magnitude, or numerical value. Vectors differ, in that they are described by a magnitude as well as a direction. I also learned that distance is a scalar quantity, while displacement is a vector quantity. We also distinguished instantaneous speed and average speed in the reading, instantaneous is taken at one point, while average is the average of speeds taken over time.

=Lab: "Constant Speed of a CMV"=

**Michael Solimano**, Jess Smith 9/7/2011 Period 2
 * Objective: ** What is the speed of a Constant Motion Vehicle (CMV)?

The slope of our graph (the coefficient of x) represents the average velocity of our ten data points. This is because the slope is equal to change in y/ change in x. In this graph, this equals change in position over change in time, which is average velocity. The r squared value was high at .9993, meaning our trend line had a high correlation to the data points given. This graph shows constant velocity, as the slope does not largely change on the graph at any point.
 * Hypotheses: **
 * 1) How fast does a CMV move?
 * 2) A CMV moves about ½ foot per second.
 * 3) How precisely can distances be measured?
 * 4) Distances can be measured to the nearest tenth with our metric rulers.
 * 5) What does a position-time graph tell you?
 * 6) A position-time graph shows an object's position on a plane as time progresses.
 * Graphs:**
 * Position Time Data for CMV**
 * Analysis:**

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

1. Why is the slope of the position-time graph equivalent to average velocity? The slope of the position-time graph was graphed on a graph where the x-axis was time and the y-axis was position in centimeters. Because slope is measured as the change in y (position) over the change of x (time), the slope actually measured the change of position over the change in time throughout our graph. Change in position over change in time is equal to the velocity, so our slope measured the average velocity of our graph.
 * Discussion questions **

2. Why is it average velocity and not instantaneous velocity? What assumptions are we making? Our measurement was an average velocity because it took into account ten distinct points of distance taken per second when determining the velocity. An instantaneous velocity would have been taken at one point, and not accounted for the series of points on the sparktape. The average velocity was a more reliable answer, and assumed that our CMV was at a constant speed when being measured, and was not fluctuating in speed. This also assumes that the instantaneous velocity may not have been an accurate representation of the CMV velocity, as the CMV may have been starting up and had not yet reached a constant velocity.

3. Why was it okay to set the y-intercept equal to zero? We were able to set the y-intercept equal to zero because our car started at time 0, where its position was 0. This was the starting point from which our average velocity was graphed.

4. What is the meaning of the R2 value? The R squared value is a measure of how close the trend line was to our given points on the graph. A higher R squared value indicates that our slope/trend line had an accurate correlation with our data points, and was an accurate model of constant velocity.

5. If you were to add the graph of another CMV that moved more slowly on the same axes as your current graph, how would you expect it to lie relative to yours? I would expect the graph of the slower CMV to lie below our graph. This is because the slower CMV will not travel as many centimeters in the given second, and thus will not be as high on the y-axis, which was the CMV’s position. This will make the slope, or average velocity, lower than our CMV as well.


 * Conclusion: **

The average velocity of our CMV was 28.274 cm/s. This was calculated by the slope of our graph, shown above, which was a measure of average velocity because it measured the change in position over the change in time. Our graph had a 99.93% R squared value, which indicates that our trend line was an accurate representation of our data points, and that our data accurately showed a constant velocity. I hypothesized that a CMV moves an average of ½ foot per second. When converted to centimeters, my hypothesis was about 15 centimeters per second. This was close to half of the average velocity of our CMV, and thus was not close to our actual results. There were many possible sources of error in this lab, many of which revolved around measuring the sparktape. We measured the sparktape markings with a metric ruler. This ruler only measured to a tenth of a centimeter, so the experimenters were forced to estimate to the 100th of a cm, which could have led to errors in our data. There were also difficulties in perspective when reading the sparktape, as the metric ruler could not be placed directly over the sparktape. This could have also led to errors in our data. Also, the surface on which the CMV moved may not have been entirely flat. Fluctuations in the surface could have had adverse affects on the CMV’s velocity, and could have created fluctuations in the CMV’s position reading. This would thus have a negative effect when trying to calculate a realistic average velocity. The second problem could have been solved by using a transparent ruler, as this would alleviate the problems with perspective when reading the sparktape. Also, a more precise tool could have been used to measure the position of the sparktape markings, although no such tool was available. Also, ensuring that the testing surface for the CMV was flat, and had no obstructions would have limited the chances of adversely effecting the CMV’s velocity.


 * Class Notes on Speed of a CMV Lab:**

=Class notes: Graph shapes, at Rest and Constant Speed=



=1-D Kinematics: Lesson 2 Summary (method 2a)=
 * 1) One subject that was clear in class and on the Physics Classroom was [|ticker tape] and its applications. I understood that the ticker tape is a piece of tape, on which a moving object is attached. Dots are placed on the tape at uniform intervals, and thus the distance between the dots can be used to analyze the motion of an object. If the distance is small between dots, the object was moving at a slower pace than an object that made dots with a larger space between them. Also if the dots are a uniform distance from each other, then the object is at constant velocity, while if the dots are becoming larger and larger it shows acceleration, and if they are becoming smaller and smaller it shows negative acceleration. I also understood the difference between ticker tape and vector diagrams. Both are capable of showing constant velocity, acceleration, and negative acceleration, while ticker tape allows us to analyze velocity with numbers but without direction, and vector diagrams show us direction, but not quantifiable numbers. The vector diagram also can show motion in any direction with the arrows.
 * 2) One thing that I was unsure of in class was the relationship between negative acceleration and a [|vector diagram]. After reading about vector diagrams, I now understand that when the arrows become shorter and shorter in one direction, it represents a negative acceleration, while if they are becoming larger and larger in one direction, it depicts acceleration.
 * 3) I understand all of the covered material.
 * 4) Everything that I read was gone over in class.

= Class Notes: Motion Diagrams and Ticker Tape Diagrams = -Constant speed will be the same as the instantaneous speed, and the speed will always be the same value. -An average speed will have varying instantaneous speeds. For all of these speeds the equation is V=d/t Types of motion -Four types: -Rest- Not moving, in the same place. -Constant speed, moving where speed does not change. -Increasing speed (acceleration)- -Decreasing speed (acceleration)- Acceleration -Acceleration is any time you are changing speed, increasing and decreasing speed are examples. There is no deceleration in Physics. -Ways of representing these are motion diagrams, which show direction of velocity and ticker tape diagrams. Motion diagram The larger arrows represent acceleration or a gain in speed. At rest v=0 and a=0 For the last one, the acceleration is (-) so the arrows become smaller and smaller. The first is at constant speed, so the arrows are equal. The arrows can be in any direction including up and down. The signs (+) or (-) should be added to the velocity arrows and the acceleration, based on their direction. Example third has negative acceleration and positive velocity. Ticker Tape Diagrams

At rest there is only one dot. At constant speed the dots are equal length apart. If increasing there will be more and more space between each one. Slowing down there will be progressing less dots. Does not show direction, but shows numerical readings. The motion diagram shows direction, but not values. Direction -Signs are arbitrary and subjective, or made up by someone.

="The Big 5 Notes"= == = = =Class Activity: Graphing Acceleration=


 * 1) How can you tell that there is no motion on a…
 * 2) position vs. time graph- At no motion the graph registers no data, as there is no change in position.
 * 3) velocity vs. time graph- At no motion the graph forms a constant line along 0 as time passes, depicting no velocity.
 * 4) acceleration vs. time graph- At no motion the acceleration graph is constant at 0 as time passes, as there is no change in velocity.
 * 1) How can you tell that your motion is steady on a…
 * 2) position vs. time graph- At a steady pace, moving away from the detector, the position will gradually become greater at the same rate, as the subject is moving farther away from the motion detector at the same velocity.
 * 3) velocity vs. time graph- At a steady pace the velocity should be constant as time progresses, as the subject is maintain a constant, or steady, velocity.
 * 4) acceleration vs. time graph- At a steady pace, moving away from the detector, the acceleration should move along 0 as time progresses, as the subject is not changing his or her velocity.

Slow Fast >> Slow:
 * 1) How can you tell that your motion is fast vs. slow on a…
 * 2) position vs. time graph- When moving fast, the graph will move higher on the y axis (position), in a shorter amount of time than the slower graph, which will take a longer time to reach the maximum distance from the detector.
 * 3) velocity vs. time graph- Both graphs will maintain a constant line, while the faster moving subject will register a higher velocity on the y axis than the slower moving subject.
 * 4) acceleration vs. time graph- Both graphs should maintain a line at 0 because although they are moving fast and slow, they are not changing their speeds.
 * 1) How can you tell that you changed direction on a…
 * 2) position vs. time graph- If moving toward instead of away from the detector, the distance becomes smaller and smaller as you move towards the motion detector.
 * 3) velocity vs. time graph- If moving toward instead of away from the detector the velocity will read negatively, as the position is now moving towards the motion detector.
 * 4) acceleration vs. time graph- If moving toward instead of away from the detector the acceleration will be at 0.



Fast:
 * 1) [[image:fast_toward.png width="800" height="516"]]
 * 2) What are the advantages of representing motion using a…
 * 3) position vs. time graph- The observer can see the subject’s gains in position, as well as the increasing, decreasing, or uniform rate at which they are doing so. The graph can also show an object’s change of direction.
 * 4) velocity vs. time graph- The observer can see the subject’s velocity only, and analyze the subjects gains or losses in velocity without having to worry about the position. Velocity vs time can also show the observer when an object is at constant velocity.
 * 5) acceleration vs. time graph-The acceleration vs time graph is the most direct means of analyzing a change in the object’s speed and acceleration, and is also a way to identify constant velocity.


 * 1) What are the disadvantages of representing motion using a…
 * 2) position vs. time graph- The graph is too indirect to be used to analyze velocity and acceleration precisely, as the observer needs to worry about the object’s position.
 * 3) velocity vs. time graph- This graph does not show position, only direction as well as velocity.
 * 4) acceleration vs. time graph- When moving at constant speed, the graph shows nothing that can be analyzed. Also, the graph offers no data on position.


 * 1) Define the following:
 * 2) No motion: when an object is at rest and gains no speed in any direction.
 * 3) Constant speed: When an object moves at a uniform velocity.

=Class Notes: Increasing and Decreasing Speed Graphs=

=1-D Kinematics: Lesson 1E (method 2a)=

= 1-D Kinematics: Lesson 3 Summary (method 2a) =
 * 1) I understand the definition of [|acceleration] to be the rate at which an object changes its velocity, and when an object is moving at a constant speed, it has zero acceleration. I also understood the equation of average acceleration to be (vf – vi)/t. The equation of acceleration is (change of velocity)/time.
 * 2) One thing that I was unclear on that I now understand was the relationship of direction and acceleration. I understand that when an object is slowing down, its acceleration is in the opposite direction of its motion. So, when an object speeds up it has a positive direction, and when it has a negative acceleration it is slowing down, or is accelerating in the negative direction.
 * 3) I understood all material.
 * 4) We did not learn in class that acceleration is a vector quantity, or one that is described by a magnitude and a direction. We also did not learn the definition of constant acceleration, which is when an object changes its velocity by the same amount each second. An object changing its velocity at a non-constant rate is called non-constant acceleration. A free falling object shows non-constant acceleration.
 * 1) I understood the concepts of the [|position time] graph. I understood that it is a measure of an object’s distance from the origin at given time points. At constant speed, the graph will be a line with the same slope at each interval, as the object is moving the same distance toward or away from the origin at a given point. If accelerating, the graph will form a curve, showing the change in velocity that the object has. I also understood that the slope in a position time graph is very important, as the degree of the slope is important in measuring the velocity of the given object. A small slope has a small velocity, while a large slope has a greater velocity.
 * 2) One thing that the reading cleared up for me was the concept of acceleration towards the origin. In the reading, I learned that a positive acceleration towards the origin is negative, while a negative acceleration towards the origin is positive. Both are still represented by curves on a position time graph.
 * 3) I understood all of the material that I read.
 * 4) Everything read was gone over in class, except for the equation for [|slope], but this was a prerequisite knowledge.

=1-D Kinematics: Lesson 4 Summary (method 2a)=
 * 1) One of the things that I was always clear on was the graphing on a [|v-t graph]. I understood that these graphs show an object’s velocity as time progresses. If an object has a constant velocity, it will register as a line parallel to the x-axis, because the object has that one velocity. If an object’s velocity is uniformly increasing, it will be a line with one slope, representing the increasing velocity of the object. If an object is speeding up, its value becomes greater, while if it is slowing down, its value will decrease. I was also comfortable in determining the slope of a v-t graph. The slope is found by finding the rise over run, or the basic slope formula. The slope of the graph can tell you many things about the velocity and acceleration of the object.
 * 2) One thing that the reading cleared up for me was finding the area on a v-t graph. I learned that the area of the v-t graph is the object’s displacement, and can be found using the area formulas for rectangles, triangles, and trapezoids.
 * 3) I understood all of the topics that I read.
 * 4) One thing that was not fully gone over in class was the different [|areas] of a v-t graph. As stated above, the reading fully covered this topic, and showed how to calculate the area for graphs that form rectangles, triangles, and trapezoids.

=Lab: Acceleration on an Incline=
 * Michael Solimano, Jess Smith 9/14/2011**

The graph will curve upwards gaining a steeper slope, similar to an exponential graph. You can see the change in slope over a given interval to see the changes in velocity, and any acceleration if it is there.
 * Objectives:**
 * Hypotheses**
 * What does a position-time graph for increasing speeds look like?
 * What information can be found from the graph?

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

media type="file" key="My First Project.m4v" width="300" height="300"
 * Procedure:**

a) Interpret the equation of the line (slope, y-intercept) and the R squared value. The equation of our line was y=15.262x² + 3.3921x for our car moving down the incline. This follows the equation y=Ax² + Bx, which can be substituted for, and will yield the equation: change distance= 1/2at + Vit. This means that our A coefficient, which was 15.262x² is equal to ½ of our acceleration, thus the acceleration was 30.524 cm/second. Also, the B coefficient, ours was 3.3291x, is equal to the initial velocity of our car. This is accurate, as our car was close to a 0 velocity when measurements started. The R squared value for our graph was .99995, meaning that the polynomial line graphed onto our points was a very accurate representation of our data points. For the car moving up the incline, our equation was y= -21.698x² + 57.435x. Using the above equations and relationships, -21.698 x² was our A coefficient, and thus ½ of the car’s acceleration. Meaning the acceleration was -43.396 cm/second. This makes sense because our car was slowing down, and thus had a negative acceleration. Also, 57.435x was our B coefficient, and thus was our initial velocity, which makes sense for this graph, as we started measuring as the car was in motion, in order to highlight its slowing velocity. The R squared value for this curve was .9913, meaning the curve was an accurate representation of our data points.
 * Analysis:**

b) Find the instantaneous speed at halfway point and at the end. Instantaneous speed (halfway)= 16.67 cm/second Instantaneous speed (end)= 45.00 cm/second
 * Car accelerating down incline**

Instantaneous speed (halfway)= 29.63 cm/second Instantaneous speed (end)= 15.00 cm/second
 * Car accelerating up incline**

c) Find the average speed for the entire trip. The average speed for the car moving down the incline is found by the equation: (change distance)/(total time). This meant our average speed was 18.59 cm/ second. For the car moving up the incline the average speed was 34.27 cm/ second. The second measure was greater, as the experimenter’s push of the car propelled the car a greater distance in the same amount of time.

The graph showing acceleration would have had a greater curve, and the slopes would have been greater as time progresses. This is because the steeper slope would have increased the acceleration of the cart, which means it is gaining velocity over each interval, and would have traveled a greater distance (y axis) in a shorter amount of time. If the car was traveling up the ramp, it would have slowed at a greater rate, and the graph would have leveled out quicker on the top part of the graph as the car is slowing down.
 * Discussion Questions:**
 * 1) What would your graph look like if the incline had been steeper?

2. What would your graph look like if the cart had been decreasing up the incline? If the cart had been decreasing up the incline, it would have leveled out on our graph as it approaches a rest or zero velocity, meaning its velocity would have slowed, and it would have covered less distance, depicted on the y-axis, in the given time interval.

3. Compare the instantaneous speed at the halfway point with the average speed of the entire trip. The instantaneous speed at the halfway point of the car moving down the graph is 5 cm/.3 second or 16.67 cm/second. This is close to our average speed for the trip, which is 18.59 cm/ second. Although close, the instantaneous speed is often not an accurate representation of the average speed, because it is taken at only one point. Our instantaneous speed was probably close to our average speed because at the halfway point the car had been accelerating significantly, and had a steep slope. For the car moving up the incline the instantaneous speed halfway was 29.63 cm/second, and the average speed was 34.27 cm/ second. This meant that at the halfway point, the car was losing velocity as it moved up the incline, and registered a lesser speed than the average speed, which included the great speed in the beginning of the run when the experimenter pushed the car.

4. Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense? The instantaneous speed is the speed of the car at one point in the graph. This one point can only follow one slope, and thus will form a straight line. Because our graph is a curve, representing multiple points and changing slopes, the instantaneous speed can only hit the curved line at one point, the tangent line. The slope of this line is equal to the car’s speed.

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

The results for our car moving down an incline were represented by a curve on our graph. This, our table, and the curved velocity-time graph, showed that as the car was moving down the ramp, it was accelerating. The equation of this graph was y=15.262x² + 3.3921x. This equation is derived from the equation y=Ax² + Bx, which when substituted for, will yield the equation: change distance= 1/2a(t squared) + Vit. The A coefficient (15.262x²) is equal to ½ of our acceleration, thus the car’s acceleration was 30.524 cm/second. The B coefficient (3.3921x) is the initial velocity of our car on the graph, which is a reasonable number, as our initial velocity should be close to zero. The R squared value for this graph was .99995, meaning that our graphed line was an accurate representation of our data points. Our second graph represented the car moving up the incline, and slowing down. This graph was also a curve, but curved the other way, representing the slowing velocity of the car as it neared zero velocity. The equation for this curve was y= -21.698x² + 57.435x. The A value (-21.698 x²) was again ½ of the acceleration, -43.396 cm/second, which makes sense as the car was slowing down, and will have a negative acceleration. The B coefficient (57.435x) was the initial velocity, as the car was in motion when we started reading. The R squared value for this graph was .9913, again showing that the graphed line had a close correlation with our data points. We also took instantaneous speeds and average speeds for our graphs. The average speed for the car going down the incline was 18.59 cm/second, and the car going up the ramp’s average speed was 34.27 cm/second. The halfway speeds for both were 16.67 cm/second (down ramp) and 29.63 cm/second (up incline). Also, we took end point instantaneous speeds, which were 45.00 cm/second (down ramp) and 15.00 cm/second (up ramp). These showed the different speeds of the cars at different points in their respective accelerations, and showed the differences inherent in instantaneous speeds and the average speeds of the cars, which is an average of all of the instantaneous speeds. Our hypotheses were close to our data. We hypothesized that a graph for increasing speed will be a curve, similar to an exponential graph, and ours’ was a curve on both our position-time graph and on the velocity-time graph. We also hypothesized that on the graph we would be able to see the changes in slope, and thus the changes in velocity over the time intervals on the graph. This was accurate, as our curved graphs do allow us to find slopes and velocities of the data at given points or for a series of points. Error could have come from many sources. One source of error could have been from the observer’s perspective when reading the ticker tape. The ruler was not flush against the ticker tape, and thus the observer had to try and match up the points with the ruler. Also, our rulers only measured to a tenth of a centimeter, which made the observer estimate the measurement to the nearest hundredth. If this were to be conducted again a translucent ruler would have allowed us to have a better perspective when reading, and a more precise ruler could have been used if available.
 * Conclusion:**


 * Class Notes on Lab**

=Lab: A Crash Course= Michael Solimano, Jess Smith, Jake Aronson, Danielle Bonnett 9/21/2011
 * Objectives ** :

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


 * 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. [[image:aaab.png]]
 * 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.



media type="file" key="Procedure.m4v" width="300" height="300"
 * Procedure:**

Car #1 Fast, Yellow CMV equation: y=28.274x Car #2 Slow, Yellow CMV equation: y=11.582x
 * Data/Analytical Hypotheses: **

Problem #1 Problem: Time calculations (t) 28.274 (cm/s)t + 11.582 (cm/s)t = 600 cm 39.856 (cm/s)t = 600 cm
 * t= 15.05s **


 * Velocity x Time = Distance**

11.582 (cm/s) x 15.05s = **174.31 cm**

Trials: Sample calculations shown below

% Error = ((|Theoretical Value – Individual Experimental Value|) / (Theoretical Value)) x 100 % Error = ((|174.31 cm – 227.00 cm|)/174.31 cm) x 100 % Error = % 30.23
 * Sample Calculations**

Sample %Difference Calculation: %Difference = ((|average experimental value - ind. experimental|)/(Average experimental)) x 100 %Difference = ((|217.80 cm - 227.00 cm|)/(217.80)) x 100 = %4.22

This value for A was higher than expected, because %24.95 is a large percent of error. This is probably because our car did not move in a straight line of motion during our tests. In doing this, the actual points where the cars met were probably skewed.

Problem TIme calculations (t) 11.582 (cm/s)t + 100 cm = 28.274 (cm/s)t 100 cm = 16.692 (cm/s)t 5.99s = time
 * Problem #2 **

Velocity x time = Distance

11.582 (cm/s) x 5.99s = **69.376 cm**

Trials: Sample Calculations shown below

% Error = ((|Theoretical Value – Individual Experimental Value|) / (Theoretical Value)) x 100 % Error = ((|69.376 cm – 80 cm|) / (69.376 cm)) x 100 % Error = % 15.31
 * Problem 2 **
 * Sample Error Calculation **

%Difference = ((|average experimental value - ind. experimental|)/(Average experimental)) x 100 %Difference = ((|65.60 cm - 80 cm|)/(65.60 cm)) x 100 %Difference = %21.951
 * Sample %Difference Calculation:**

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

If the speed of the cars were equal, they would meet at the center of our track, or 300 cm. This is because the two cars moving in opposing directions with the same velocities will reach the same point, 300 cm, at the same time and meet. If traveling in the same direction, the cars will never meet, as they have the same speeds and the trailing car will never reach the car ahead of it.
 * Discussion questions **
 * 1) Where would the cars meet if their speeds were exactly equal?

2. 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? Yes, when the two cars meet, and presumably hit as they did in our experiment, their velocities will become near 0. Although the velocity-time graph does not show position, it does show the time at which they will hit, which can then be used to find the position. If the cars do not hit in the experiment, and rather pass each other, the graph would not show their collision, and would only show their velocities.

Theoretical data for our lab was generated by setting the speed equations of the CMV’s equal to the distance traveled, which gave us the seconds at which the CMV’s would either collide or catch up to each other, which we then plugged in to generate a theoretical position at which they would collide or catch up with each other. These calculations showed us that the CMV’s would collide at 174.31 cm and would catch up to each other at 69.376 cm. Our theoretical data was relatively close to our actual data, based on the percent errors of % 24.95 and % 5.44. We also calculated the percents of difference in the lab, which were %1.73 (problem 1) and %10.183 (problem 2). These were measures of the difference of the individual points of data from our average values. These numbers showed us that in problem #1 our data points were relatively close to our average, while in problem 2 the data points were on average much farther from the average. The error could have arisen from many places in this lab. One source was that one of our CMV’s did not move in a straight line, which would have made precise calculations of position less likely. Also, the points of catching up and collision of the CMV’s was estimated by the experimenter, and exact calculations were not possible with the metric tape. The metric tape also was not a precise tool to measure our data as it did not offer measures to the nearest hundredth of a cm. Fluctuations or imperfections in the surface that we tested the CMV's on also could have effected their velocities and our results. If we were to do this lab again, we could have used a car that moved straighter, or possibly used a straight track to test the CMV's, and used more precise tools for measuring if available. Also, we could have made sure that our testing area had no imperfections, and would be appropriate for testing constant velocity.
 * Conclusion: **


 * Class Notes on Lab:**

=Class Notes on Acceleration Problems= =Quantitative Graph Interpretation=

=Egg Drop Project= Our egg drop project was not successful, as our our egg cracked after being dropped from the window. Our project involved using a parachute, a basket for the egg made of straw, and a cone in which the basket was put in. The contraption fell at a very quick rate, which was not inhibited by the parachute, and thus the egg cracked when it hit the ground at a great speed. The contraption's main flaw was the parachute. The parachute was not successful, as it was probably too small when compared to the egg basket and the cone. If I were to make the project again, I would have made the parachute larger than the objects below it, which would have hopefully slowed the speed of the egg in the contraption. The basket and cone were not designed to sustain falling and hitting the ground at a great speed, and with a better parachute I believe they would have protected the egg from cracking. The basket also appeared to fall on its side, which was not as protected with straw, when it hit the ground. Thus, I would also make the sides of the basket more protective with straws if I were to remake the project.

Mass of Project: 18.04 Time of fall: 1.5275s Acceleration Calculations:

= =

=Physics Classroom Lesson 5 Summary (Method 1)= Lesson 5 Summary Method 1 Free falling object is one that is falls under the influence of gravity, and is in a state of ** free fall **.
 * Free-falling objects do not encounter air resistance.
 * All free-falling objects accelerate downwards at 9.8 m/s/s.

A [|ticker tape trace] or dot diagram of its motion would depict an acceleration. The dot diagram at the right depicts the acceleration of a free-falling object. The fact that the distance that the object travels every interval of time is increasing is a sure sign that the ball is speeding up as it falls downward. Free-fall acceleration is often witnessed in a strobe light demonstration. The room is darkened and a jug full of water is connected by a tube to a medicine dropper. The dropper drips water and the strobe illuminates the falling droplets at a regular rate - say once every 0.2 seconds. Instead of seeing a stream of water free-falling from the medicine dropper, several consecutive drops with increasing separation distance are seen. The pattern of drops resembles a dot diagram.

A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth). This numerical value is known as the ** acceleration of gravity ** ,and has a special symbol to denote it - the symbol ** g **.
 * The Acceleration of Gravity **
 * g = 9.8 m/s/s, downward ( ~ 10 m/s/s, downward) **

The value of the acceleration of gravity (**g**) is different in different gravitational environments.

There are a variety of means of describing the motion of objects. One is through the use of graphs - [|position versus time] and [|velocity vs. time] graph. A position versus time graph for a free-falling object s.

A curved line on a position versus time graph signifies an accelerated motion. 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. Finally, the negative slope of the line indicates a negative 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. The object starts with a zero velocity (as read from the graph) and finishes with a large, negative velocity. 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. The velocity of a free-falling object is changing by 9.8 m/s every second. The formula for determining the velocity of a falling object after a time of ** t ** seconds is ** v **** f **** = g * t ** where ** g ** is the acceleration of gravity. The value for g on Earth is 9.8 m/s/s. ** Example Calculations: ** At t = 6 s v f (9.8 m/s 2 ) * (6 s) 58.8 m/  This distance can be computed by use of a formula; the distance fallen after a time of ** t ** seconds is given by the formul file:///Users/solimami/Desktop/Picture%20clipping.pictClipping a.   ** Example Calculations: ** At t = 1 s d = (0.5) * (9.8 m/s 2 ) * (1 s) 2 = 4.9 m
 * d = 0.5 * g * t **** 2 **

The acceleration of gravity is the same for all free-falling objects. The answer to the question (doesn't a more massive object accelerate at a greater rate than a less massive object?) 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.

= Freefall Lab = Michael Solimano 10/3/11

Hypotheses: What is the force of gravity on acceleration? The force of gravity is -9.8 m/(s squared). This will be found from the slope of our v-t graph.

Hypothesized Graphs:

Data:

Graphs: Position-Time
 * Velocity-Time for Freefall**


 * Class Data ||  ||
 * || ** Period 2 ** ||
 * || 754.43  ||
 * || 856.73  ||
 * || 851.07  ||
 * || 891.38  ||
 * || 891.12  ||
 * || 798.13  ||
 * || 710.65  ||
 * || 755.87  ||
 * || **// 659.39 //**  ||
 * || **// 1225.4 //**  ||
 * Average || ** 839.417 **  ||

Sample Calculations

Sample Percent Difference:

Percent Error

Analysis: The equation for our velocity-time graph was 798.13x - 63.215. In this equation, the slope (798.13) is equal to the acceleration of our weight that we dropped. Thus, according to this equation, the acceleration was 798.13 cm/s. The y intercept, or -63.215, is equal to the initial velocity of our weight. These results are taken from the equation V(final) = at + V(initial), which is another form of y=mx+b. The equation for our position-time graph was 391.86 (x squared) - 60.111x. In this equation, the A coefficient, according to the equation y = 1/2at(squared) + v(initial)t, is equal to 1/2 of the acceleration. Thus, the acceleration was 783.72 cm/s. This equation also tells us that the B coefficient, -60.111, was our initial velocity. These large accelerations are relatively close to the force of gravity on objects, 9.8 m/(s squared).

Discussion Questions: 1. Our actual graph did not agree with our hypothesized velocity time graph. My hypotheses was innacurate, as my graph was not a straight line, as a thought that the rates of acceleration would become greater as the weight fell. This was not accurate, as our graph was nearly a straight line with nearly uniform rates of acceleration, which it should have been given the freefall of the weight with the force of gravity.

2. Our actual graph does agree with my expected graph. In both graphs a curve upwards is formed, representing acceleration on a position-time graph. This represented the acceleration of our weight as it fell in freefall.

3. Our results were below the average of the class, but only 4.92% below. This was a relatively close number, and could have been caused vy the friction of our spark tape moving through the spark timer or by air resistance.

4. Our object did accelerate close to uniformly. We knew this from our velocity-time graph. On this graph, our data points and trend line had a very high r squared value. This is important, as it means our data points nearly formed a straight line with one slope, which represents one uniform acceleration for all of the points.

5. Acceleration due to gravity is typically 9.8 m/(s squared). However, this number is subject to change on different parts of the Earth, and a slight variation could have made our values greater or lesser than expected. Also, the experimenters were expected to keep the weight still before dropping it, and if they had a slight move downward as they dropped the weight, it could have thrown off the acceleration. Also, the friction caused by the sparktape and the air resistance encountered by the weight could have both served to slow the acceleration of the weight while it fall.

Conclusion: The data collected from our weight that was in freefall was indicative of freefall acceleration. This was indicated by both our position time graph and velocity time graph. Our position time graph formed a steep curve upwards, showing the increasing velocities as the weight fell and accelerated. Our velocity time graph was best represented by a line with one slope. This was accurate, as our weight was constantly accelerating as it fell, and its acceleration could be found by the slope of our v-t graph. Both graphs had high r squared values, .99932 and .99117 respectively, indicating that our data points were accurately represented by the trend lines. This also shows us that our data points on the v-t graph were very close to linear, and thus had a uniform rate of acceleration. Our percent difference and error were also relatively low, %4.92 and values all less than %.01 respectively. Also, based on the equations found in the analysis, we were able to find two examples of our actual weight's acceleration, 798.13 cm/s and 783.72 cm/s. These are below the expected 980 cm/s, but are valid accelerations, given the sources of error in the lab discussed below. These readings did confirm our original hypothesis. Although our acceleration was not exactly 980 cm/s, it was relatively close, and error discussed below show why it was not exactly 980 cm/s. Our graphs, including a curved x-t graph and a straight v-t graph, also showed acceleration of our weight. As discussed above, our hypothesized x-t graph mirrored our actual results, while a fundamental error in my hypothesized v-t graph did not match our actual and correct v-t graph. Many groups had values differing than the expected 980cm/s rate of acceleration. This could have been caused by many sources in this lab. Two of the biggest causes of differences in the lab were the friction of the spark tape moving through the spark timer and the air resistance that was encountered. Both of these factors would have decreased the acceleration of the weight. Also, when reading our spark tape, it was hard to keep the spark tape and metric tape lined up for proper reading, and the metric tape forced the reader to round the nearest hundredth of a cm. If we were to conduct this lab again, a differently shaped weight could have reduced air resistance, and different means of marking the position and time of the weight and measuring the spark tape could have been used if available. A different means of measuring position and time would have reduced the friction of the spark tape.

=Freefall Notes=