Ch2_FavaA

toc =**__Class Notes__**=

Constant Speed Notes

 * constant speed - speed is the same the entire time; slope will be the same; therefore instantaneous speed is also the same
 * instantaneous speed - speed at a certain time; single point
 * average speed - the average of all the speeds therefore total distance divided by total time

Motion Diagrams Notes

 * position-time graph (x-t)
 * tells where you are located
 * this will change for constant speed, though stays same for at rest
 * steeper slope = faster
 * shallower slope = slower
 * negative slope = moving towards origin
 * velocity-time graph (v-t)
 * tells how fast you are going
 * velocity=0 when at rest
 * horizontal line not at 0 = constant speed
 * higher up = faster
 * lower down = slower
 * negative = moving in backwards direction which is towards the origin
 * take absolute value for speed
 * positive = moving forward which away from the origin
 * area = displacement
 * acceleration-time graph (a-t)
 * acceleration at constant speed and at rest are both = 0
 * velocity = area "under" graph
 * acceleration is the slope/derivative of velocity, and velocity is the slope/derivative of position
 * types of motion
 * at rest
 * [[image:motion1d_fig1.png width="640" height="142"]]
 * constant speed
 * [[image:motion1d_fig2.png width="640" height="140"]]
 * increasing speed/speeding up
 * can do this at constant rate
 * decreasing speed/slowing down
 * can do this at constant rate
 * acceleration is whenever you are changing your speed, so it is both increasing and decreasing speed
 * [[image:Screen_shot_2011-09-19_at_10.12.26_PM.png]]
 * [[image:Screen_shot_2011-09-19_at_10.12.40_PM.png]]

Kinematics Notes

 * [[image:kinematics_notes.png]]

Increasing and Decreasing Speed

 * increasing speed
 * slope of position-time gets steeper
 * positive slope = away
 * negative slope = towards
 * velocity is straight line, constant, starting low ending high
 * positive y-value = away
 * negative y-value = towards
 * acceleration isn't sloped
 * positive = away
 * negative = towards
 * [[image:03356.png width="485" height="274"]]
 * decreasing speed
 * slope of position-time graph is getting shallower
 * positive slope = away
 * negative slope = towards
 * [[image:Screen_shot_2011-09-20_at_9.56.47_PM.png width="212" height="153"]]
 * velocity is straight line, starting high ending low
 * positive y-value = away
 * negative y-value = towards
 * acceleration isn't sloped
 * positive = towards
 * negative = away

Acceleration of Gravity and Free Fall Notes

 * object free fall only impacted by gravity
 * acceleration of gravity (g) = 9.8 m/s/s, can be rounded to 10 m/s/s
 * free fall doesn't have air resistance
 * instant dropped until time hits ground is considered being in free fall
 * all free-falling objects accelerate at the same rate
 * speed at any point depends on time of fall
 * at max height, a=9.8 not 0
 * misconception: bigger mass has same acceleration as smaller mass
 * Vf=gt or a=v/t
 * d=(gt2)/2
 * slope of free fall v-t should be -.98 m/s/s
 * dot diagram can be used to show free fall
 * initial velocity = 0 when dropping
 * sample math problem:
 * [[image:Photo_on_2011-10-12_at_14.25.jpg]]

=__Kinematics Notes (Textbook)__=


 * Kinematics: Describing Motion With Words (Physics Classroom HW):**
 * 1) What (specifically) did you read that you already understood well from class discussion? Describe at least two items fully.
 * The difference between distance and displacement is something I understood. Distance is scalar and therefore doesn't include direction, or is the overall amount of space covered. Displacement is vector, therefore including direction, and refers to the amount of space moved from the starting point (position); moving the same amount forward and backward would thus result in 0 displacement, though there would be distance.
 * I had also previously understood the difference between average and instantaneous speed. Instantaneous speed is the speed at any one moment, whereas average speed is the average of all the speeds. Average speed can therefore be calculated by dividing the total distance traveled by the total amount of time. Instantaneous speed is therefore that exact speed of the object, whereas average speed is an average and therefore the speed fluctuates during that period of time. Average speed can be taken between two speeds. Constant speed means that the speed isn't changing, so the instantaneous speed and average speed would be the same for the entire time (usually a straight line). There is only one equation for all of these but the information you are putting in is what varies depending upon what you are finding.
 * 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 confused about the difference between speed and velocity. I knew that they were very similar, but not exactly the same. I though velocity may have include acceleration whereas speed did not. Though I know realize that speed is a scalar quantity and velocity is a vector quantity; therefore velocity incorporates direction. Velocity is how fast an object is changing its position, which is why average velocity is found using displacement (which also refers to change in position); therefore, velocity can be negative if its changing position backwardly. Speed is just how fast an object is moving, which is why average speed is found by total distance over time, which is why its always positive.
 * 1) What (specifically) did you read that you still don't understand? Please word these in the form of a question.
 * I think I understand everything that I read fully and feel comfortable with the material.
 * 1) What (specifically) did you read that was not gone over during class today?
 * The definitions of scalar (magnitude) and vector (magnitude and direction) quantities weren't gone over in class, as well as whether each of the quantites was scalar or vector. However, it was an easy, simple topic to understand.


 * Kinematics: Describing Motion With Words (Physics Classroom HW 9/13):**
 * 1) What (specifically) did you read that you already understood well from class discussion? Describe at least two items fully.
 * I had understood that acceleration is a vector quantity and therefore includes direction. It is also the rate that an object is changing its velocity, therefore acceleration is both speeding up and slowing down. Constant acceleration is when an object is changing its velocity at a constant rate
 * I also understood the concept of average acceleration and how to find it. It is essentially found by dividing the change in velocity over the total time. The change in velocity is the final velocity - the initial 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 shaky about the difference between positive and negative acceleration. I thought positive acceleration meant speeding up while negative meant slowing down. Though the positive and negative actually refer to direction. If an object is moving in the negative direction, regardless of whether its velocity is negative or positive. This is referred to as the "Rule of Thumb".
 * 1) What (specifically) did you read that you still don't understand? Please word these in the form of a question.
 * I understood everything that I read.
 * 1) What (specifically) did you read that was not gone over during class today?
 * Free-falling motion was not gone over in class. When an object is free-falling, its speed is increasing, and it is therefore accelerating. If the object is accelerating with a constant increase in speed, then the distance is proportional to the time. The amount of time is squared, and that number is then multiplied to the distance for the first interval to get the total distance traveled.


 * Kinematics: Describing Motion With Diagrams (Physics Classroom HW 9/9):**
 * 1) What (specifically) did you read that you already understood well from class discussion? Describe at least two items fully.
 * Ticker tape diagrams are a means of visually representing data. Tape is placed inside a ticker, such as a spark timer, which makes a dot on every regular interval of time (which is pre-determined, such as 1.0 second intervals). The distance between each dot is the change in position of the object for that amount of time; the bigger the distance, the faster the object is moving. These diagrams can also show acceleration, which is change in velocity and therefore a change in the distance between dots. Constant velocity is when the dots are evenly spaced.
 * Vector diagrams are also used as a means of visually representing data. They are arrows that show both magnitude and direction. The size of the arrow therefore determines the velocity. If all the arrows in a diagram are the same, then it is at constant velocity. If they change size throughout the diagram, then it is showing acceleration.
 * 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.
 * The reading did not really help to clarify any of the information we were taught in class. It was more of a review of the material for me and didn't address my areas of confusion.
 * 1) What (specifically) did you read that you still don't understand? Please word these in the form of a question.
 * What exactly is magnitude/what exactly does it represent?
 * How can acceleration be negative if velocity is in positive direction?
 * 1) What (specifically) did you read that was not gone over during class today?
 * Everything that I read was gone over in class.


 * Kinematics: Describing Motion with Position vs. Time Graphs (Physics Classroom HW 9/15):**
 * 1) What (specifically) did you read that you already understood well from class discussion? Describe at least two items fully.
 * Position-time graphs show the position of a moving object, as well as information about its velocity and acceleration. The slope of a position-time graph is directly relevant to velocity. If the slope is positive, then velocity is positive, and vice versa. Constant velocity is shown by constant slope. An accelerating moving object will create a curved line on a position-time graph since the velocity, and therefore slope, are changing. If the slope is getting larger, then the speed is increasing. If the slope is getting smaller, then the speed is decreasing. Also, the steeper the slope, the faster the velocity, and vice versa. Slope is equal to velocity.
 * To calculate slope, two points are to be found on the line and the difference between the y-value and x-values are to be found. The difference in y-value is divided by the difference in x-value, therefore the rise is divided by the run.
 * 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.
 * The reading didn't really help to increase my understanding in any of the topics.
 * 1) What (specifically) did you read that you still don't understand? Please word these in the form of a question.
 * How exactly do you depict between negative and positive acceleration?
 * 1) What (specifically) did you read that was not gone over during class today?
 * We didn't go over how to calculate slope but it was something I had already known.


 * Kinematics: Describing Motion with Velocity vs. Time Graphs (Physics Classroom HW 9/15):**
 * 1) What (specifically) did you read that you already understood well from class discussion? Describe at least two items fully.
 * Velocity-time graphs show us about velocity, direction, and acceleration. If the velocity is positive and above the x-axis than it is moving in the positive direction. In addition, if the velocity is negative and below the x-axis than it is moving in the negative direction. If the velocity is a horizontal line, then the object is moving with constant velocity. The slope of a v-t graph is a numerical value for the acceleration. If the slope is negative, then the acceleration is negative, regardless of whether it is above or below the x-axis.
 * 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 now further understand the difference between speeding up and slowing down on a velocity-time graph. If the line is moving towards the origin, then the velocity is decreasing and the object is slowing down (regardless of whether it's negative or positive since refers to magnitude). If the line is moving away from the origin, then the velocity is increasing and the object is speeding up.
 * 1) What (specifically) did you read that you still don't understand? Please word these in the form of a question.
 * I understood everything I read.
 * 1) What (specifically) did you read that was not gone over during class today?
 * We did not go over how to determine area on a velocity-time graph. The area created by the line and the axes shows the total displacement of the moving object over a certain interval of time. The area of the shape is then found using the normal equations for that shape (i.e 1/2bh for triangle, bh for rectangle, and 1/2b(h1+h2) for trapezoid.

=__Free-Fall Notes (Textbook)__= topic sentence: An object affected solely by gravity and no air resistance is free falling with an acceleration of 9.8m/s/s downward. Introduction to Free Fall A free falling object is an object that is falling under the sole influence of gravity; said to be in a state of free fall : 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. 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-falling objects do not encounter air resistance.
 * All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s (often approximated as 10 m/s/s)

The Acceleration of Gravity 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 - denoted by the symbol g. There are slight variations in this value that are dependent primarily upon on altitude.

Representing Free Fall by Graphs One means of describing the motion of objects is through the use of graphs - position versus time and velocity versus time graphs.

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

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. 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 velocity is also downward/negative.

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

Since a free-falling object is undergoing an acceleration (g = 9,8 m/s/s, downward), it would be expected that its velocity-time graph would be diagonal. The object is moving in the negative direction and speeding up, therefore a negative acceleration.

How Fast? and How Far? The velocity of a free-falling object is changing by 9.8 m/s every second. 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 Example Calculation: At t = 6 s vf = (9.8 m/s2) * (6 s) = 58.8 m/s

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

Example Calculations: At t = 2 s d = (0.5) * (9.8 m/s2) * (2 s)2 = 19.6 m

The Big Misconception The acceleration of gravity is the same for all free-falling objects regardless of how long they have been falling, or whether they were initially dropped from rest or thrown up into the air. 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. 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.

=__Interpreting Graphs - Constant Speed__=
 * [[image:Screen_shot_2011-09-20_at_10.12.25_PM.png]][[image:Screen_shot_2011-09-20_at_10.11.46_PM.png]]

=__Interpreting Position-Time Graphs (E,F,G)__= E. F. G

=__**Activity: Graphical Representations of Equilibrium**__=

//Data://
 * The above graphs show the velocity, position, and acceleration for an object at rest, therefore no movement in front of motion detector
 * The above graphs show the velocity, position, and acceleration for an object at constant speed, where run #5 represents away from the motion detector and run #6 represents towards the motion detector.
 * The above graphs show the velocity, position, and acceleration for an object at a faster constant speed, where run #1 represents away from the motion detector and run #2 represents towards the motion detector.
 * The above graphs show the velocity, position, and acceleration for an object at a slower constant speed, where run #3 represents away from the motion detector and run #4 represents towards the motion detector.

//Discussion Questions://
 * 1) How can you tell that there is no motion on a…
 * position vs. time graph
 * steady horizontal line at starting point
 * velocity vs. time graph
 * horizontal line at zero
 * acceleration vs. time graph
 * horizontal line at zero


 * 1) How can you tell that your motion is steady on a…
 * position vs. time graph
 * line with a constant slope
 * velocity vs. time graph
 * steady horizontal line that is not on 0
 * acceleration vs. time graph
 * steady horizontal line that is not on 0


 * 1) How can you tell that your motion is fast vs. slow on a…
 * position vs. time graph
 * steeper slope means faster moving
 * velocity vs. time graph
 * y-value of line is higher for faster moving than for slower moving
 * acceleration vs. time graph
 * horizontal line because still constant speed so acceleration=0, therefore doesn't distinguish between fast and slow


 * 1) How can you tell that you changed direction on a…
 * position vs. time graph
 * when the slope changes between positive and negative
 * positive slope is away from detector and negative is towards
 * velocity vs. time graph
 * when line crosses the x-axis between negative and positive, therefore changing its sign
 * positive is away from detector and negative is towards
 * acceleration vs. time graph
 * when line crosses the x-axis between negative and positive


 * 1) What are the advantages of representing motion using a…
 * position vs. time graph
 * clearly shows slope which is the average speed; can also determine exact position
 * velocity vs. time graph
 * can determine when there is a change in direction and the speed, therefore shows both motion and speed; can also find acceleration
 * acceleration vs. time graph
 * can clearly see change in speed, such as speeding up and slowing down


 * 1) What are the disadvantages of representing motion using a…
 * position vs. time graph
 * doesn’t differentiate as clearly between fast and slow speeds and change in direction
 * velocity vs. time graph
 * don't see actual position
 * acceleration vs. time graph
 * doesn't differentiate between at rest and constant velocity


 * 1) Define the following:
 * No motion
 * object is not moving/at rest, so both velocity and acceleration are 0.
 * Constant speed
 * object is moving at a constant rate where the same distance is covered in the same amount of time, therefore acceleration is 0

= __**Lab: A Crash Course in Velocity (Part 1)**__ = September 7, 2011; Andrew Chung

//Objectives//: //Hypotheses//: //Data:// Position-Time x(t) //Graph:// //Analysis:// //Discussion questions:// //Conclusion:// In essay format, answer the following questions: What results did you get? Was your hypothesis accurate? (Be specific, using data from the lab to support claims) What sources of error may have contributed to inaccuracies? What could you do to minimize these issues if you had to redo this lab? The equation of the trendline for the speed of the CMV was y=64.629x, therefore the CMV had an average velocity of 64.629 centimeters per second. My hypothesis was therefore inaccurate as I hypothesized the average velocity to be 40 centimeters per second, significantly below the actual velocity. In addition, distance is measured by a hundredth of a centimeter, which was more precise than I hypothesized with a tenth of a centimeter. Lastly, a position-time graph shows the relationship between a data point’s position and time, as I hypothesized, which is the velocity of that point. There were multiple sources of error that could have contributed to inaccuracies. Estimating the last significant figure (hundredths place) when using the meter stick was one source of error as it is not entirely accurate. When measuring the distance with the meter stick, the stick could have shifted also changing the measurement. There is also an issue of point of view when using the meter stick since it is ½ centimeter above the spark tape that contains the dot being measured. To minimize these issues when redoing the lab, different measurement tools could be used. A ruler or measuring tape would minimize the point of view issue, as they are both on the same level as the spark tape. A different tool other than the spark timer could also be used to measure time, one with more precision and accuracy. Multiple trials could also be conducted to ensure results are consistent.
 * 1) How fast does a CMV move?
 * 2) How precisely can distances be measured?
 * 3) What does a position-time graph tell you?
 * 1) The CMV will move at 40 centimeters per second.
 * 2) Distance can be measured to the tenth of a centimeter.
 * 3) A position-time graph shows a data point's position relative to time.
 * The position-time graph above shows the position of the CMV at the respective time. The slope of the linear trendline estimates the actual speed of the CMV. The closer the R2 value is to 1, the more accurate the trendline is. The equation of the line gives the slope as the coefficient of x, therefore making that coefficient equal to the average velocity of the trendline. The CMV is moving at constant speed since the trendline is a straight line, and not curved.
 * 1) Why is the slope of the position-time graph equivalent to average velocity?
 * Slope is defined as the change in the y-axis divided by the change in the x-axis. Since the y-axis is position and the x-axis is time, the slope is therefore equal to the change in position divided by the change in time. Average velocity is defined as the change in distance over the change in time. Since position and distance are the same, the slope is, thus, equivalent to the average velocity.
 * 1) Why is it average velocity and not instantaneous velocity? What assumptions are we making?
 * Average velocity is the average of all the velocities over a period of time, whereas instantaneous velocity is the velocity at one certain point. We are assuming that there are no outliers in the velocity to throw off the average.
 * 1) Why was it okay to set the y-intercept equal to zero?
 * It is okay to set the y-intercept to zero because at zero time the position is also zero, as the CMV has yet to move.
 * 1) What is the meaning of the R2 value?
 * The R2 value refers to the accuracy of the trendline in comparison to all of the actual data points. It is more accurate the closer it is to 1, and therefore less accurate the farther it is from 1.
 * 1) If you were to add the graph of another CMV that moved more slowly on the same axes as your current graph, how would you expect it to lie relative to yours?
 * The graph of the slower CMV would lie lower than mine. This is because, although both cars would start at the origin and have the same time values, the slower CMV would have a smaller average velocity and therefore a smaller slope, resulting in a less steep, lower trendline.

=__**Lab: Acceleration Graphs**__= September 14, 2011; Andrew Chung

//Objectives://
 * 1) What does a position-time graph for increasing speeds look like?
 * 2) What information can be found from the graph?

//Hypothesis://
 * 1) The slope will continue to get steeper as the speed increases and the position will get higher
 * 2) Velocity, acceleration, position, and direction can be found from the position-time graph

//Materials://
 * spark tape, spark timer, track, dynamics carts, measuring tape

//Procedure://
 * 1) Put ramp on top of Physics textbook to create an incline.
 * 2) Put ticker tape through spark timer at the highest point of the ramp and attach it to the cart.
 * 3) Turn the spark timer on and place the cart on the ramp.
 * 4) Let the cart run down the ramp for downward incline.
 * 5) Remove the ticker tape and, using measuring tape, measure the distance between each dot for decline.
 * 6) Put ticker tape through spark timer at lowest point of the ramp and attach it to the cart.
 * 7) Turn the spark timer on and place the cart at the bottom of the ramp.
 * 8) Push the cart up the ramp for incline.
 * 9) Repeat Step 5.
 * 10) Record the distance at each time and graph on a position-time graph for both incline and decline.
 * 11) Analyze results.

//Data://

//Graph://

//Analysis://
 * 1) The equation of the increasing down incline trendline was y=10.73x2+14.262x. The A value is equal to 1/2 the acceleration, therefore making the acceleration 21.46 cm/s2. The B value is equal to the initial velocity. Our B value was very high at 14.262, suggesting that we started measuring the dots on the ticker tape too late, when the cart was already greatly moving. The R2 value for this equation was 0.99996, which is very close to 1 and therefore implying that the trendline was very accurate to our data points. The trendline is also a polynomial opposed to linear, because the R2 value for linear was 0.96822 which was much lower than the R2 value for the polynomial. The equation of the decreasing up incline trendline was y=1.9911x2+38.908x. The A value is once again equal to half the acceleration, making the accleration 3.9822 cm/s2. The B value for initial velocity was once again very high. The R2 value was 0.9982, which is also very close to 1 suggesting accuracy of trendline. I had adjusted the data points of the decreasing up incline because our original starting point had created a graph with a clear curve up, suggesting that we had began measuring too early and had inaccurate velocity, as the slope of the line is supposed to become less steep.
 * 2) Increasing Down Incline: instantaneous speed at halfway=27.14 cm/s; instantaneous speed at end=40.01 cm/s. Decreasing Up Incline: instantaneous speed at halfway=40.70 cm/s; instantaneous speed at end=42.49 cm/s [[image:down_instant.png width="400" height="300"]][[image:up_instant.png width="400" height="241"]]
 * 3) Increasing Down Incline: average speed=27.01 cm/s. Decreasing Up Incline: average speed=39.84 cm/s.[[image:average_speeds_up_down.png width="480" height="159"]]

//Discussion Questions://
 * 1) What would your graph look like if the incline had been steeper?
 * 2) If the incline had been steeper, the cart would have accelerated faster, therefore increasing its speed between each interval even more. This would cause the graph to have a greater, steeper curve since slope would be steeper.
 * 3) What would your graph look like if the chart had been decreasing up the incline?
 * 4) If the car had been decreasing up, as shown by one of the trendlines, the graph's curve would become more shallow/flatten out as time increased. The slope would therefore be decreasing opposed to increasing/steeper.
 * 5) Compare the instantaneous speed at the halfway point with the average speed of the entire trip.
 * 6) For both increasing down incline and decreasing up incline, the instantaneous speed at the halfway point and average speed were very close to each other. Instantaneous for increasing down was 27.14 cm/s and the average was 27.01 cm/s. Instantaneous for decreasing up was 40.70 cm/s and the average was 39.84 m/s. The instantaneous speed for the halfway point is the speed halfway through the acceleration process, therefore should be the middle. The average speed is the average of all the speeds at each time interval, which should therefore also represent the middle. Thus, the two calculations were very close.
 * 7) Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense?
 * 8) The tangent line is a line that crosses through one point on the curve with a constant slope. Therefore the slope of the tangent line will be equal to the slope of the one point on the curve that it crosses. Since velocity is the slope of the position-time graph, the instantaneous velocity/speed can be found using this slope.
 * 9) Draw a v-t graph of the motion of the cart. Be as quantitative as possible.
 * 10) [[image:Screen_shot_2011-09-15_at_8.14.10_AM.png]]

//Conclusion:// After doing the lab, our hypothesis was shown to be correct. We hypothesized that the slope would continue to get steeper and the position higher as the speed increased. Indeed, the graph shows the slope of the increasing down incline to be getting steeper as the cart continues to accelerate down the ramp incline. The equation of this graph was y=10.73x2+14.262x, a polynomial since the R2 value was higher for polynomial than linear. A definite source of error in our experiment was that we started measuring the distances of the dots on the ticker tape too late, since our initial velocity was so high (B value); therefore meaning that the cart had already been accelerating previous to when we started measuring. Other sources of error could include the reaction time between starting the spark timer and releasing the cart and the estimating of the last significant figure in the position measurements. To minimize these errors, a different measuring tool could be used instead of the spark timer to remove the human reaction time error possibility from releasing cart and starting timer, as well as to provide more accuracy and precision. In addition, a different measuring tool could be used for the distance so estimation of the last significant figure would be unnecessary.

=Lab: A Crash Course in Velocity (Part II)= September 21, 2011; Andrew Chung, Sammy Caspert, John Chiavelli

//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.
 * 2) Find the position where the faster CMV will catch up with the slower CMV if they start at least 1 m apart, move int he same direction, and start simultaneously.

Procedure:
 * 1) For Crashing Experiment (Objective 1)
 * media type="file" key="good vid crash.mov" width="300" height="300"
 * 1) For Catching Up Experiment (Objective 2)
 * media type="file" key="good catch up.mov" width="300" height="300"

Calculations:
 * 1) [[image:a.png width="800" height="271"]]
 * 2) [[image:b.png width="800" height="440"]]

//Data:// Crashing Table
 * trial ||> distance(cm) ||
 * 1 ||> 184.15 ||
 * 2 ||> 183.36 ||
 * 3 ||> 182.87 ||
 * 4 ||> 177.67 ||
 * 5 ||> 181.52 ||

Catching Up Table
 * trial ||> distance(cm) ||
 * 1 ||> 181.23 ||
 * 2 ||> 180.17 ||
 * 3 ||> 180.56 ||
 * 4 ||> 181.23 ||
 * 5 ||> 185.37 ||

Analysis:
 * Multiple trials were done to ensure that our data was not random. Since our data was pretty consistent, we can assume that it is representative of the real distances, and not just a random mistake/outlier. Though, there is obviously some error. This error is calculated using both the percent error and the percent difference. For our percent error, we used the average of all the trials since we didn't have any outliers. Percent error calculates how close our experimental values were to the value we got using mathematical calculation. The percent difference is found for each individual trial and compares each trial to the average of all the trials.
 * Below is the mathematical analysis for the crashing experiment. Since the percent difference for each trial compared to the average was very low (none higher than 2%), we used the average to calculate our percent error. The percent error came out to be 2.19% which is also very low.
 * [[image:Screen_shot_2011-09-22_at_10.33.44_PM.png]]
 * Below is the mathematical analysis for the catching up experiment, which also had a very lower percent error of 0.0055%, which used the average experimental, which was found to be valid since all the trails had very low percent differences from the average.
 * [[image:Screen_shot_2011-10-06_at_7.47.09_PM.png]]

//Discussion Questions://
 * 1) Where would the cars meet if their speeds were exactly equal?
 * If the speeds were equal, then the two cars would be covering the same distance in the same time. Therefore, if they started 600 centimeters away from each other, they would meet at 300 centimeters (the middle). This is because as time passes, they travel the same amount of centimeters. If the speeds were equal, and one car started 1 meter in front of the other, then the two cars would never meet. Since they are covering the same distance in the same amount of time, one car would always be 1 meter in front of the other.
 * 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.
 * [[image:x(t)_crashing.png]]
 * [[image:x(t)_catching_up.png]]
 * 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?
 * There is no way to find the points when they are at the same place because velocity-time graphs do not show position.
 * [[image:v(t).png]]

//Conclusion//: As obtained from the previous experiment, the velocity for the blue, faster car was 64.629 cm/s and the velocity for the yellow, slower car was 29.06 cm/s. Our theoretical results, which were obtained through mathematical calculations, were significantly close to our experimental results in both experiments. For the crashing problem, we calculated the distance to be 185.98 centimeters, and the average of all of our trials was 181.91 centimeters. The percent differences allow us to conclude that the average experimental value can be used since not very far off from each trial, and our percent error of 2.19% shows that there was not much error in the lab. Though there were some sources of error in this lab. One source of error could be in human reaction time when taking the distance. The reaction time between when the cars actually crashed or caught up to each other and when the person reacted could throw off the measurements. This could be fixed by using a tool that records the exact distance when the cars collide or catch up. Another source of error would be in estimating the final significant figure for the measurements when using measuring tape. Another possible source of error could be the switching of batteries. The batteries in the slower, yellow car had died and were replaced, therefore the original velocity previously found with the old batteries could have been slightly different from that with the current batteries. To fix these errors, a different tool could be used to find the intersection point, one that could mark the exact point when the cars crash or catch up. In addition, a different measuring tool could be used; a tool that provides further precision and would eliminate the estimation of the hundredths place when using the measuring tape. Also, completing the first part of the lab where the velocity of the cars is found could be done right before this part of the lab, to ensure that the velocities are correct and not effected by battery life.

=Lab: Egg Drop= September 28, 2011; Julia Sellman

//Final Device:// Mass of egg: 58.51 g; Mass of device and egg: 132.23 g

//Calculation for Acceleration//:

//Results//: Our egg had a small crack that was bleeding. The parachute and paper padding inside aided in protecting the egg from the crash.

//Analysis//: Our two prototypes had an entirely different design, where the egg was closed in a tight box with projections coming out of every side. The egg popped out of this prototype the first time and the second hadn't protected it at all. We decided to change the design to a bigger box with a parachute. This did a better job of protecting the egg, especially since it was suspended in the center with paper padding both below and above it. Though, the egg still wasn't able to completely slow its acceleration before coming to a full stop. The parachute also helped to slow the acceleration as the egg dropped.

//Improvements//: To improve our design, we would definitely make the parachute bigger. Since the parachute was about as wide as the actual box, it couldn't fully function in slowing the acceleration. We would probably just change our design entirely though. A cone would be the best design since it protects the egg, while allowing it to slow its acceleration by slowly sliding down the cone towards the tip. The tip then absorbs the majority of the impact and evenly distributes the pressure around the egg to prevent it from cracking.

=__Lab: Falling Object__=

//Objective//:
 * What is the acceleration of a falling body?

//Materials//:
 * ticker tape timer, timer tape, masking tape, mass, clamp, meterstick

//Hypothesis//:
 * The acceleration of a falling body should be equal to 9.8 m/s/s, as this is the acceleration of gravity, unaffected by air resistance. The position-time graph should be a curved line (J-shaped) and the velocity-time graph should be a straight diagonal line with a negative slope and in the negative section (below the x-axis).

//Procedure//:
 * 1) Place ticker tape through the ticker tape timer.
 * 2) Tape 100 g piece to the end of the ticker tape that had been placed through the timer.
 * 3) Hold the tape and timer out over the edge.
 * 4) Turn on the timer and drop the piece.
 * 5) Using measuring tape, measure the distance between each dot on the ticker tape and record the data.
 * 6) Analyze results.

//Data//:

//Graphs//:

//Analysis//:
 * Our velocity-time graph had a high R2 value of .97937 though it could have definitely been higher. Our experimental acceleration, which is the slope of the velocity, was found to be 851.07 cm/s2, when the theoretical is 981 cm/s2. The slope is found from the equation of the linear trendline which is y=851.07x+2.525, following the form y=mx+b. B, or 2.525 cm/s, is the initial velocity, so the smaller the better. The y=mx+b equation is derived from Vf=at+Vi, which is how b=Vi. It should be zero but it isn't because the timer and when the item was dropped may not have been the exact same moment. The position-time graph has a polynomial trendline with a very high R2 value of .99939, therefore indication our line to be very accurate. The polynomial equation y=Ax2+Bx is derived from the equation d=1/2at2+vit, which is why A=1/2acceleration and B=initial velocity. Therefore according to the position-time graph, our acceleration was 837.22 cm/s2 which is relatively close to the acceleration found from the v-t graph. Our initial velocity according to the x-t graph was 1.6964 cm/s, which was also relatively close to the 2.525 cm/s found using the v-t graph.
 * [[image:sample_calc_instant_veloc.png]]
 * [[image:%_diff_error_free_fall.png]]

//Discussion Questions//:
 * 1) Does the shape of your v-t graph agree with the expected graph? Why or why not?
 * No, I had predicted the v-t graph to have a negative slope, though our data shows a positive slope. This is because we measured the distance on the ticker tape as positive.
 * 1) Does the shape of your x-t graph agree with the expected graph? Why or why not?
 * Yes, I had predicted the x-t graph to have increasing position as time increased, therefore increasing velocity, which our data showed.
 * 1) How do your results compare to that of the class? (Use percent difference to discuss quantitatively)
 * Our acceleration was equal to 851.07 m/s/s, whereas the class average was equal to 839.417 m/s/s. The percent difference between ours and the class average was only 1.39%, which indicates that the two were very close since this is very well within the 20% range.
 * 1) Did the object accelerate uniformly? How do you know?
 * Yes, the object did, for the most part, accelerate uniformly since the slope of the v-t graph was almost constant. The slope of the v-t graph is equal to the acceleration. The R2 value was equal to .97937, which is very close to 1, indicating the linear trendline to be closely accurate to the data points.
 * 1) What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?
 * Latitude and altitude are two factors that can cause the acceleration due to gravity to be different. In addition, the friction caused by the ticker tape rubbing on the timer could have slowed down the acceleration since it slowed down the velocity.

//Conclusion//: I had hypothesized that the acceleration would be negative, therefore the velocity-time graph would have a negative slope. Our data showed the v-t graph to have a positive slope since we did not take the direction of the object falling into consideration when measuring the distance on the ticker tape. My hypothesis for the position-time graph was accurate since the slope of the line and position increased as time passed. The trendline for our v-t graph was linear, therefore indicating that the majority of the data points had uniform slope. Our acceleration was found to be 851.07 m/s/s, where the theoretical is 981 m/s/s; thus, our data was off by 14.43% which is relatively high though still within a 20% range. One major source of error in this lab is the friction created by the ticker tape rubbing through and against the timer. We held our timer vertically so this friction was lower than if we had held it horizontally, though there was still much friction produced. Another source of error could have been starting the spark timer before dropping the object so that the time started before the object was falling. Other sources of error are rooted in the measuring of the ticker tape. When using measuring tape, we have to estimate the hundredths value so that could have thrown off the accuracy. In addition, the ticker tape and measuring tape were relatively long and could have easily moved slightly during the measuring process, therefore throwing off the data. To fix these problems, a different tool could be used to measure the time and distance that the object is falling, one where friction wouldn't be an issue. In addition, this information could be recorded electronically, therefore taking away the entire issue of measuring long ticker tape and it possibly moving.