Ch2_LeibowitzG

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 * __ Lesson 1- __****// Describing Motion with Words (Part I) //**


 * **Describing Motion with Words**
 * **Scalars and Vectors**
 * **Distance and Displacement**
 * **Speed and Velocity**

** 1. What specifically did you read that you already understood well from our class discussion? Describe at least 2 items fully. ** In class we already discussed the difference between distance and displacement. Distance is a quantity that refers to how much ground an object has covered during its motion. Displacement is quantity that refers to how far out of place an object is or its overall change is position. For example, if you are standing in front of a desk and move 2 steps forward and then 1 step backwards, your distance is 3 steps while your displacement is 1 step. In addition, in class we defined velocity as the rate at which an object changes in position. We also already learned the equation to calculate velocity, which is the change in distance over the change in time.

** 2. 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. ** Although I understood the concepts of distance, displacement, speed, and velocity though in class explanations and examples, I was unsure to an exact definition. Once learning the definitions through the readings, I was also unsure as to what two vocabulary words within the definitions meant- scalars and vectors. I now understand from the reading that scalars are quantities that are fully described by a magnitude or numerical value alone while vectors are quantities that are fully described by both magnitude and a direction.

** 3. What specifically did you read that you still don’t understand? Please word these in the form of a question. ** Everything was clear and I have no questions.

** 4. **** What specifically did you read that was not gone over during class today? ** The vocabulary words of mechanics and kinematics were not gone over during class today. Mechanics is the study of the motion of objects while kinematics is the science of describing the motion of objects using words, diagrams, numbers graphs, and equations.

=**__ Lesson 1- __****// Describing Motion with Words (Part II) //**=


 * **Acceleration**

** 1. What specifically did you read that you already understood well from our class discussion? Describe at least 2 items fully. ** In class we already discussed the definition of acceleration, which is the rate at which an object changes its velocity. Therefore, we knew that an object is accelerating only if it is changing its velocity, as they stressed in the reading. In addition, we already discussed the formula for acceleration, which is the change of velocity/ time, or as we defined it in class, the initial velocity minus the final velocity / time.

** 2. 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 helped me to further understand positive and negative accelerations. While I understood the graphing we did on acceleration, the book helped show me what they define as the "rule of thumb." This is that if an object is slowing down, then its acceleration is in the opposite direction of its motion. With charts the reading showed me not only this, but if the object is moving in a negative acceleration, you have to take the rule of thumb to consideration and realize that if that object is speeding up, it will therefore also have a negative velocity.

** 3. What specifically did you read that you still don’t understand? Please word these in the form of a question. ** Everything was clear and I have no questions.

** 4. **** What specifically did you read that was not gone over during class today? ** Mostly everything in the reading was discussed in class, besides for the rule of thumb, which states that if an object is slowing down, its acceleration is in the opposite direction of its motion.

=**__ Lesson 2- __****// Describing Motion with Diagrams //**=


 * **Introduction to Diagrams**
 * **Ticker Tape Diagrams**
 * **Vector Diagrams**

** 1. What specifically did you read that you already understood well from our class discussion? Describe at least 2 items fully. ** In class we already discussed the common way of analyzing the motion of objects known as a ticker tap analysis. This is when a long tape is attacked to a moving object and then threaded through a device that places a small “tick” on the tape at regular intervals of time. In class we understood that this distance between the dots on the tape represents the object’s position change over time. In addition, in class we already learned the alternative way to show motion knows as a motion diagram, or in the textbook’s case, vector diagram, which are diagrams that use arrows to show motion by either remaining the same size, increasing, or decreasing in size moving left, right, up, or down depending on the type of motion occurring.

** 2. 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 introduced the concept of “magnitude” which helped me to understand vector diagrams better. Although I understood that the size of the arrows will either increase or decrease depending on what type of motion occurring, I now know that these arrows represent the magnitude of the vector quantity.

** 3. What specifically did you read that you still don’t understand? Please word these in the form of a question. ** Everything was clear and I have no questions.

** 4. **** What specifically did you read that was not gone over during class today? ** The term “vector diagram” wasn’t mentioned in class and instead the term “motion diagram” was used. I no know that both names are interchangeable.

=**__ Class Notes (Sept. 6) //-// __ //Definitions// **=

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=**__ Class Notes (Sept. 8) //-// __ //Motion Diagrams// **=

=**__ Lab- __****// Speed of a Constant Motion Vehicle //**= September 9th, 2011- Gabby Leibowitz and Rob Kwark

**Objectives:** 1. How precisely can you measure distances with a meter stick? 2. How fast does your CMV move? 3. What information can you get from a position-time graph? **Materials:** 1. Spark timer and spark tape 2. Meterstick 3. Masking tape 4. CMV **Hypothesis:** 1. You can measure distances probably to a precision of a mm. 2. The CMV probably moves around 30 cm/s 3. From a position-time graph, you can probably get its distance in a certain time and the speed and velocity of the object **Experiment Steps:** 1. Tape a piece of the spark tape to the CMV while the rest is in the spark timer 2. Turn on the spark timer and the CMV and let the CMV pull the spark tape through the timer until you have enough dots to measure 3. Use a meterstick to measure the distance between each dot, then add up the distance per each second 4. Graph results <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;"> <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">**Discussion Questions:** <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">**1. Why is the slope of the position-time graph equivalent to average velocity?** <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">The two are equivalent because the whole purpose of the slope is to graph where the average velocity should be in relation to the data we recorded. <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">**2. Why is it average velocity and not instantaneous velocity? What assumptions are we making?** <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">Instantaneous velocity is the velocity at any given instant in time while average velocity is the average of all instantaneous speeds found by a distance/time ratio. It is average velocity because our calculations were based on the distance the CMV traveled over a certain amount of time opposed to finding how fast the CMV was moving at a given instant. This is all based on the assumption that the CMV is, in fact, moving at a constant speed, by assuming a point on the tape to start at where it looks as if it is constant. <span style="font-family: 'Times New Roman',Times,serif; font-size: medium;">**3. Why was it okay to set the y-intercept equal to zero?** <span style="font-family: 'Times New Roman',Times,serif; font-size: medium;">We assume that the CMV began at a distance of zero. The distance could not start at a measure greater than zero because we are measuring right from the time the CMV started moving and it couldn't start at a measure less than zero because that would make for impossible results. <span style="font-family: 'Times New Roman',Times,serif; font-size: medium;">**4. What is the meaning of the R^2 value?** <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">The R^2 value is a statistical term saying how good one value is at predicting another, often in a percentage. Given a set of data points, an R^2 value says how well the resulting line matches the original data points. <span style="font-family: 'Times New Roman',Times,serif; font-size: medium;">**5. If you were to add the graph of another CMV that moved more slowly on the asme aces as your graph, how would you expect it to lie relative to yours?** <span style="font-family: 'Times New Roman',Times,serif; font-size: medium;">It would be expected to lie under the line that is graphed now representing my data and have less of a slope. This is because it takes a longer amount of time to move each distance my CMV moved per second.

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">**Conclusion:** <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;"> The speed of my CMV was 31.084 cm/sec, found by measuring the slope of my line. This was pretty accurate considering my R^2 value was 0.9996 or 99.96%. My hypothesis was pretty accurate both predicting that you can measure distances to a precision of mm and that my CMV would move 30 mm/sec. My hypothesis of what a position-time graph would present held true as well, since the information on the graph allows you to find the distance of the object in a certain amount of time, which allows you to find the velocity if you wish. Many sources of error could have possible contributed to any problems that occurred within my experiment, such as the fact that we didn't measure perfectly accurate, we recorded only one second worth of data, and that we were instructed to ignore the beginning of the tape. Since our lab group decided to measure the distance between each dot on the tap separately and then add each interval together as we move from second to second, much rounding and estimation was done, in addition to the movement of the starting spot of the measuring stick each time. Also, only one second worth of data doesn't give very reliable data, for it could have changed and stabilized as more seconds went on. Finally, we were told to ignore the dots in the beginning of the tape that were more clustered together and find a more constant place to measure from. This could be misleading because we may have ignored data that was important to determining the true speed of the CMV if we began our measurements too far into the tape. In order to minimize errors, doing the trial multiple times would provide further evidence that our results were accurate. A longer measuring stick, possibly measuring tape, would benefit our experiment for we wouldn't have to continually move the stick which contributes to much of our accuracy error. Instead of measuring the intervals separately and then adding on, measuring all at once would limit the rounding and estimation done.

=**__ Class Notes (Sept. 13) //-// __ //Time Graphs// **= **__ Class Notes (Sept. 13) //-// __ //The Big Five// **

=**__ Activity- __****// Graphical Representations of Equilibrium //**=

<span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">AT REST: <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;"> <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">**The graphs above show a person at rest. For the position time graph, this is displayed by a straight, horizontal line above the x-axis at the position of the person. For a velocity time graph and acceleration time graph, a person at rest is displayed by a straight, horizontal line laying on the x-axis.**

<span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">SLOW: <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;"> <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">**The graphs above show a person moving at slow speed. For the position time graph, the red line represents a person moving away in slow speed while the orange line represents a person moving towards in slow speed. For the velocity time graph, the green line represents a person moving away in slow speed (above axis) while the purple represents someone moving towards in slow speed (under axis). For the acceleration time graph,** **the blue line represents a person moving away in slow speed while the brown represents someone moving towards in slow speed.**

<span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">FAST: <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">**The graphs above show a person moving at fast speed. For the position time graph, the red line represents a person moving away in fast speed while the orange line represents a person moving towards in fast speed. For the velocity time graph, the green line represents a person moving away in fast speed (above axis) while the purple represents someone moving towards in fast speed (under axis). For the acceleration time graph, the blue line represents a person moving away in fast speed while the brown represents someone moving towards in fast speed.**

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">1. How can you tell there is no motion on a... <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">a. Position vs. time graph: There will be a straight, horizontal line from your position with no slope <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">b. Velocity vs. time graph: There will be a straight, horizontal line from 0.0 with no slope or points of increase/decrease <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">c. Acceleration vs. time graph: There will be a straight, horizontal line from 0.0 with no slope or points of increase/decrease

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">2. How can you tell that your motion is steady on a... <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">a. Position vs. time graph: The straight line is rising or falling at a constant slope <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">b. Velocity vs. time graph: There is no jagged points of increase or decrease; It remains at either the positive line or negative line in a horizontal line <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">c. Acceleration vs. time graph: There are no jagged points of increase or decrease; It remains at 0.0 in a horizontal line

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">3. How can you tell that your motion is fast vs. slow on a... <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">a. Position vs. time graph: If your motion is fast the line will rise, remaining in a straight line <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;"> If your motion is slow the line will rise, but to a less extenet than if it was fast, remaining in a straight line <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">b. Velocity vs. time graph: If your motion is fast the line will go up and down in a more jagged way, high points representing great increase in speed; Resembles a straight line with a few points of increase; Less straight than positive vs. time graph but not rising in slope <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;"> If your motion is slow the line will also go up in down, but to a less extenet than if it was fast; Less jagged considering there are no points of great speed <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">c. Acceleration vs. time graph: If your motion is fast, the line will greatly go up and down as your speed increases or decreases, highlighting points of acceleration and remaining completely jagged; It is less of a straight line than the velocity vs. time graphs and has no consistency <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;"> If your motion is slow the line will also go up and down, but less jagged since there are fewer acceleration points

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">4. How can you tell that you changed direction on a... <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">a. Position vs. time graph: If you are walking towards the object, the line will be moving on an upwards angle <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;"> If you are walking away from the object, the line will be moving on a downwards angle <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">b. Velocity vs. time graph: If you are walking towards the object, the line will be above the 0.0 line, meaning positive velocity <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;"> If you are walking away from an object, the line will be below the 0.0 line, meaning negative velocity <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">c. Acceleration vs. time graph: You cannot tell your change in direction

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">5. What are advantages of representing motion using a... <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">a. Position vs. time graph: You can see the rising or decreasing slope in a neatly graphed straight line <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">b. Velocity vs. time graph: You can see whether the velocity is negative or positive <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">c. Acceleration vs. time graph: You can see highlighted points of a decrease or increase in acceleration

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">6. What are disadvantages of representing motion using a... <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">a. Position vs. time graph: You cannot see points of acceleration or a decrease in acceleration; You cannot see the speed or velocity of the object, only the position <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">b. Velocity vs. time graph: You cannot see a rising or falling slope: You cannot see the position of the object <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">c. Acceleration vs. time graph: You cannot see a change in direction; You cannot see a rising or falling slope; You cannot see the position of the object

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">7. Define the following: <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">a. No motion: A state that does not change position or place <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">b. Constant speed: Movement at a fixed distance per unit of time

=**__ Class Notes (Sept. 14) //-// __ //Increasing/ Decreasing Speeds// **= <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">MOTION DETECTOR AT TOP: <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;"> <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">MOTION DETECTOR AT BOTTOM: <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%; line-height: 0px; overflow-x: hidden; overflow-y: hidden;">

=**__ Lesson 3- __****// Describing Motion with Position vs. Time Graphs //**=


 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">**The Meaning of Shape for a p-t Graph**
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">**The Meaning of Slope for a p-t Graph**
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">**Determining the Slope on a p-t Graph**

** 1. What specifically did you read that you already understood well from our class discussion? Describe at least 2 items fully. ** In class we already discussed the the shape of a p-t graph if it were constant velocity. If it were moving with a constant, rightward (+) velocity, the line would look like a straight diagonal line increasing higher with a constant slope, as posted below under the letter A. We also discussed the shape of a p-t graph if were at a changing (+) velocity, seen when it accelerates. The line here would look like a curved line increasing higher with a changing slope, as posted under letter B.

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">A. B.

** 2. 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, and taking the quiz a few days ago, helped me understand that if the object is moving rightwards, it would have a positive velocity, while if it was moving leftwards, it would have a negative velocity.

** 3. What specifically did you read that you still don’t understand? Please word these in the form of a question. ** Everything was clear and I have no questions.

** 4. **** What specifically did you read that was not gone over during class today? ** Mostly everything in the reading was discussed in class, the reading just made the principle more clear that the principle about slope and velocity is that on a position time graph, the slope is equal to the velocity.

=**__ Lesson 4- __****// Describing Motion with Velocity vs. Time Graphs //**=


 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">**The Meaning of the Shape for a v-t Graph**
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">**The Meaning of the Slope for a v-t Graph**
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">**Relating the Shape to the Motion**
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">**Determining the Slope on a v-t Graph**
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">**Determining the Area on a v-t Graph**

** 1. What specifically did you read that you already understood well from our class discussion? Describe at least 2 items fully. ** <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">In class we already discussed the shape of a v-t graph with a constant velocity. If it were moving with a constant, rightward (+) velocity, the line would look like a straight horizontal line above the (0,0) at a certain position on the y-axis, as posted under letter A. We also discussed the shape of a v-t graph if were at a changing (+) velocity, seen when it accelerates. The line here would look like a straight, diagonal line, increasing with a constant slope, as posted under letter B.

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">A. B.

** 2. 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 helped me to further understand the area under the line and what that means for the graph. We briefly discussed in class that on a velocity-time graph, the area under the horizontal line, forming a box, is the displacement of the object. The reading made it clearer by describing the different shapes that can be formed on a velocity-time graph representing the displacement, not only a rectangle.

** 3. What specifically did you read that you still don’t understand? Please word these in the form of a question. ** Everything was clear and I have no questions.

** 4. **** What specifically did you read that was not gone over during class today? ** The reading taught me the meaning of the area under the line on a velocity time graph, which was not gone over in class besides for that of a constant speed, in which it resembles a rectangle. The reading told me if the velocity is decreasing or increasing, you can form triangles or trapezoids and use the appropriate equations to figure out the displacement for those graphs as well.

=**__ Class Notes (Sept. 15) //-// __ //Position-Time Graph Worksheet// **=

=**__ Class Notes (Sept. 16) //-// __ //y=Ax^2 + Bx// **=

= **__ Packet- __****// Quantitative Graph Interpretation //** =

=**__ Lab- __****// Acceleration Graphs //**= <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">September 16th, 2011- Gabby Leibowitz and Rob Kwark

**Objectives:** 1. What does a position time graph for increasing speeds look like? 2. What information can be found from the graph? **Materials:** 1. Spark timer and spark tape 2. Track 3. Dynamics cart 4. Ruler/meterstick/measuring tape **Hypothesis:** 1. A position time graph for increasing speeds will probably be pointing upwards with a curved slope 2. The graph probably shows speed, acceleration, and its position at a particular time on the graph <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">**Experiment Steps:** 1. Gather materials and set up the ramp, using a single textbook as the slope, then put the spark tape through the spark timer attaching it to the dynamics cart by tape <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">2. First, put the spark timer at the top of the ramp and let the cart travel down the ramp with the spark tape running through the timer <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">3. Set up a new spark timer with new tape and follow the exact same method, except this time, place the spark timer at the bottom of the ramp, pushing the cart upwards (then stopping it) <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">4. Graph results <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;"> <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">**Analysis:** <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">**1. Interpret the equation of the line (slope, y-intercept) and the R^2 value.** <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">The equation of the uphill line was y=-27.468x^2 + 79.134x while the equation of the downhill line was y=12.644x^2 + 16.318 x. The slope of the uphill line is negative while the slope of the downhill line is positive because the uphill graph shows the motion of the line moving TOWARDS the spark timer causing it to be positive, while the downhill graph shows the motion of the line moving AWAY from the spark timer causing it to be negative. The y-intercept for both is (0,0) because it started at 0m/s. The R^2 value of the lines shows the accuracy of the line, which for both is very accurate at a value of 0.999. <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">The instantaneous speed for the uphill graph, halfway, is around 43.4 cm/s. The instantaneous speed for the uphill graph, end, is around 6.15 cm/s. The instantaneous speed for the downhill graph, halfway, is around 42.2 cm/s. The instantaneous speed for the downhill graph, end, is around 62.8 cm/s.
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">2. Find the instantaneous speed at halfway point and at the end. **



<span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 120%; text-align: center;">CLOSE UP OF TANGENTS- <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 120%; text-align: center;">CLOSE UP OF EQUATIONS- <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">The average speed for the uphill graph is around 43.4 cm/s. The average speed for the downhill graph is around 39.1 cm/s.
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">3. Find the average speed for the entire trip. **

<span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">**Discussion Questions:** <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">**1. What would your graph look like if the incline had been steeper?** <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">If the incline had been steeper, the uphill graph would have had a slower time for it would have taken longer for the cart to make it to the top with a slope at a greater incline. Therefore, the uphill graph would look flatter and farther down, since its points would have to reach larger times for each of its positions. If the incline had been steeper on a down hill graph, it would have had a quicker time for it would have traveled down the ramp at a much faster pace. Therefore, the downhill graph would look even more curved and higher up, since its points wouldn't have to reach such a long distance because the times would be smaller and closer together. <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">**2. What would your graph look like if the cart had been decreasing up the incline?** <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">The graph would still look like a curve, but it would start at a higher speed and than decreases as it moves up the incline. Therefore, it would start higher up at a higher value and then move its values of speed down, now in the shape of a downwards curve opposed to an upwards. Also, since it is on an incline it would be flat because the slope would be less since the speeds would be slowing down as it moves up the ramp. <span style="font-family: 'Times New Roman',Times,serif; font-size: medium;">**3. Compare the instantaneous speed at the halfway point with the average speed of the entire trip.** <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">The instantaneous speed at the halfway point for the uphill graph was exactly equivalent to the average speed of the entire trip, with 42.4 cm/s. The instantaneous speed at the halfway point for the downhill graph was extremely close in value to the average speed of the entire trip, with the instantaneous speed being 42.2 and the average speed being 39.1. This shows that these values have an equivalent relationship. <span style="font-family: 'Times New Roman',Times,serif; font-size: medium;">**4. Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense?** <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">Since the uphill and downhill lines are both curved, making it have many slopes instead of just one as a linear graph would have, by drawing a linear or tangent line, going through one of the points of the graph, one can see the slope at that particular speed, in other words, the instantaneous speed. <span style="font-family: 'Times New Roman',Times,serif; font-size: medium;">**5. Draw a v-t graph of the motion of the cart. Be as quantitative as possible.** <span style="font-family: 'Times New Roman',Times,serif; font-size: medium;"> <span style="font-family: 'Times New Roman',Times,serif; font-size: medium; line-height: 24px;">

<span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">**Conclusion:** <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">The equation for my uphill graph was y=-27.468x^2 + 79.134x while the equation for my downhill graph was y=12.664x^2 + 16.318x. My hypothesis was correct in saying that a position time graph for increasing speeds will probably be pointing upwards with a curved slope. The increasing speed graph, being my downhill graph, pointed upwards (its slope being positive and its speeds increasing as time continues) and did have a curved slope. My second hypothesis was also correct in saying that my graph will show speed, acceleration, and position at a particular moment in time. Instantaneous speed, or a speed at a particular moment in time is easily found by just looking at the points plotted on the graph and reading the (x,y) values, while acceleration and average speed, and even velocity, can be found by doing their equations. Many sources of error could have occurred during the experiment, such as the fact that we pushed the cart down or up, falsely increasing its acceleration and possibly causing it to move back down the incline when we were measuring a decrease in speed. In addition, as in the lab before, we were instructed to ignore the beginning of the tape and could have ignored data that was important to determining the true speed of the cart uphill or downhill. In order to minimize these problems, one has to be sacrificed for the other. In order to be able to ignore the misleading values that could have been as a result of our touching of the cart, we'd have to begin or measurements more towards the center of the tape. However, this doesn't solve the problem of possibly starting too far into the center and ignoring valuable data. Therefore, in order to fully minimize these issues, multiple runs of the experiment can provide the ability to reach consistent results.

=** __Class Notes (Sept. 21)-__ // Practice Problems //**=

=**__ Lab- __****// A Crash Course in Velocity (Part II) //**= <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">September 16th, 2011- Gabby Leibowitz and Rob Kwark, Jonathan Itskovitch and Lauren Kostman

**Objectives:** 1. Both algebraically and graphically, solve the following two problems a. Find the position where both CMV's will meet if they start at least 600 cm apart, move towards each other, and start simultaneously b. 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 <span style="font-family: 'Times New Roman'; font-size: 16px; line-height: 0px; overflow-x: hidden; overflow-y: hidden;"> 2. Set up each situation and run trials to confirm calculations

**Materials:** 1. Constant Motion Vehicle 2. Tape measure/ metersticks 3. Masking tape 4. Stop watch 5. Spark timer 6. Spark tape **Hypothesis:** 1. The CMVs will meet at 204.84 cm in 6.59 sec (6.59, 204.84) if they start 600 cm apart, move towards each other, and start simultaneously 2. The faster CMV will catch up with the slower CMV at 207.55 cm in 3.46 sec if they start 100 m apar, move in the same direction, and start simultaneously <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">**Experiment Steps:** <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">Collision media type="file" key="Movie on 2011-09-23 at 11.21.mov" width="300" height="300" Catch-Up media type="file" key="Movie on 2011-09-23 at 11.31.mov" width="300" height="300"

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">**Results:** <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">Collision Crashed (cm) || Percent Error || Percent Difference || <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">Catch-Up Crashed (cm) || Percent Error || Percent Difference ||
 * Trial || Position
 * 1 || 200 cm || 2.4% || 2.6% ||
 * 2 || 204 cm || 0.4% || 0.7% ||
 * 3 || 208 cm || 1.5% || 1.3% ||
 * 4 || 210 cm || 2.5% || 2.2% ||
 * 5 || 205 cm || 0.1% || 0.2% ||
 * Average || 205.4 cm || 1.4% || 1.4% ||
 * Trial || Position
 * 1 || 198 cm || 4.6% || 3.9% ||
 * 2 || 210 cm || 1.2% || 1.9% ||
 * 3 || 205 cm || 1.2% || 0.5% ||
 * 4 || 211 cm || 1.7% || 2.4% ||
 * Average || 206 cm || 2.2% || 2.2% ||

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;"> Percent Error <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">Percent Difference <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">**Discussion Questions:** <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">**1. Where would the cars meet if their speeds were exactly equal?** <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">Since we set our CMVs to start exactly 600 cm apart, if they were traveling at equal speeds they would end up meeting in the middle, at 300 cm. This is because neither of the cars would be acceleration faster than the other, so each would travel a certain distance at the same amount of time. Therefore, since the distance between them is only 600, and since the CMVs are moving towards each other, at 300 cm they would have covered the same amount of distance in the same amount of time and therefore, meet in the middle. <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">**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.** <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">Collision <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">Catch-Up

<span style="font-family: 'Times New Roman',Times,serif; font-size: medium;">**3. Sketch the 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?** <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">Since it is a velocity-time graph, and not a position time graph, there is no way to find the points when they are at the same place at the same time. This is because a velocity time graph only displays velocity and time, not position, and therefore has no slope. This makes it impossible to find the equation of both lines and then do the method of substitution to find the intersection point. Without knowing this, one can merely look at the graph and realize finding the meeting point is impossible for the lines on the graph are just horizontal lines that never show intersection.

<span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">**Conclusion:** <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">The theoretical answer for the collision situation was that the CMVs will collide at 204.84 cm in 6.59 sec. The experimental answer was extremely close to our theoretical answer, in which there was only a maximum of a 2.63% error. The theoretical answer for the catch-up situation was that the faster CMV will catch up to the slower CMV at 207.55 c, in 3.46 sec. The experimental answers were also extremely close to our theoretical answer, in which there was only a maximum of a 3.88% error. Therefore, our calculations and hypothesis made before physically conducting the experiment were extremely accurate. One source of error that could have occurred during the calculation process was that we never took into account the fact that the data of either our group of our partnered group could have been miscalculated during the previous lab, therefore, throwing off the equations we formulated for each CMV and ultimately effecting our theoretical results. Many sources of error could have also occurred during the experimentation, in which the battery of the CMV could have been stronger or less strong than the time when the original calculations were taken, therefore, giving us different results. In addition, while measuring the time when the CMVs either crashed or caught up to each other, we depended on our group members to turn the CMVs on at the same time, and read the exact location of the CMVs the second they collided. Human error could have gotten in the way in which the two members turning on the CMV could have hit the switch at slightly different times, or the members recording the location they collided could have been a few seconds off, providing us with misleading, but practical, results. In order to prevent these errors, we could have re-performed the previous lab right before performing this lab, to secure accuracy. In addition, using more than one of the same color CMV could have assured us that the battery did not affect our results. Finally, performing the experiment even more times than we did, rotating group members, would have illustrated that human error didn't completely alter the data.

=**__ Project- __****// Egg Drop //**= <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">September 30th, 2011- Gabby Leibowitz and Maxx Grunfeld

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">Mass of egg drop with egg: 155.6 g <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">Mass of egg: 55.0 g <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">Mass of egg drop without egg: 100.6 g

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">Time of fall: 1.35 s

<span style="font-family: 'Times New Roman',Times,serif; font-size: 16px; line-height: 23px;">Egg was placed in a cradle of straws suspended with rubber bands wrapped around the straws attached by sewing strings to a parachute of aluminum foil.

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">After dropping our egg in its contraption from the window 8.5 m above the ground, our egg remained intact. When we calculated the acceleration, it came out to be 9.33 m/s2:

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">d = 8.5 m <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">v0 = 0 <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">t = 1.35 s <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">a = ?

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">d = v0t + 1/2at2 <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">8.5 = 1/2a(1.35)2 <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">8.5 = .91125a <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">a = 9.33 m/s2

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">Compared to the normal force of gravity (g), the acceleration we calculated is less than 9.8 m/s2 but relatively close. This means that when we dropped our egg, it came out to be close to but with slightly less of an acceleration than when an object falls only by the force of gravity. If we could redo this activity, we would try to make it lighter by possibly reducing the amount of straws or incorporating a cone shape to our final project.

=**__ Lesson 5- __****// Free Fall and the Acceleration of Gravity //**=

**__ Introduction to Free Fall __** A free falling object is an object that is falling under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of **free fall**. ** Characteristics: ** 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 distance that the object travels every interval of time is increasing (velocity and acceleration = same direction) **__ The Acceleration of Gravity __** Numerical value for the acceleration of a free-falling object (98. m/s/s) is known as the **acceleration of gravity** The acceleration for any object moving under the sole influence of gravity Symbol=g **__ Representing Free Fall by Graphs __** One means of describing the motion of objects is through the use of graphs - position vs. time and velocity vs. time graphs. A position versus time graph for a free-falling object: The line on the graph curves, signifying an accelerated motion. The object starts with a small velocity (slow) and finishes with a large velocity (fast), indicating a small initial velocity and a large final velocity. Finally, the negative slope of the line indicates a negative (i.e., downward) velocity. A velocity versus time graph for a free-falling object: The line on the graph is a straight, diagonal line, signifying an accelerated motion. 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. The constant, negative slope indicates a constant, negative acceleration. **__ How Fast? and How Far? __**
 * 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

Free-falling objects are in a state of acceleration. Specifically, they are accelerating at a rate of 9.8 m/s/s. SO the velocity of a free-falling object is changing by 9.8 m/s every second. (Ex: If dropped from a position of rest, the object will be traveling 9.8 m/s at the end of the first second, 19.6 m/s at the end of the second second, 29.4 m/s at the end of the third second, etc.) THEREFORE, 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. ** FORMULA: vf = g * t ** (g= acceleration of gravity/9.8m/s/s) Used to calculate the velocity of the object after any given amount of time when dropped from rest **__ The Big Misconception __**

9.8 m/s/s (known as 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. ** "Doesn't a more massive object accelerate at a greater rate than a less massive object?" ** NO, 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 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. **All objects free fall at the same rate of acceleration, regardless of their mass.**

= = = ** __Class Notes (Oct. 3-7)-__ // Free Fall //** =

=**__ Lab- __// Free Fall //**=
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">October 5th, 2011- Gabby Leibowitz and Rob Kwark **

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">1. What is the acceleration of a falling body? <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">1. Ticker tape timer <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">2. Timer tape <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">3. Masking tape <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">4. Mass <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">5. Clamp <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">6. Meterstick <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">1. The acceleration will always be 9.8 m/s^2 as a result of gravity and the velocity will be displayed in a diagonal line on the graph under the origin since it is being dropped. <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">**Experiment Steps**: 1. Gather materials and set up experiment, placing the timer tape in the timer and taping the mass to the timer tape <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">2. Have one partner hold the timer and the mass over the railing while the other holds the timer tape back <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">3. The partner holding the timer turns the timer on and both partners release the mass and the tape simultaneously <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">4. Measure distance between each of the dots on the timer tape <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">5. Graph results
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">Objectives: **
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">Materials: **
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">Hypothesis: **

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">Results: **
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">Analysis: **

<span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">Graphs:

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;"> <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">Discussion and Interpretation: <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">The position time graph was an increasing curve moving away from the origin. It was increasing because the mass's distance or position continued to increase as time went on, as shown on the graph. In our results, you can see that the velocity was increasing as well, and therefore, the velocity time graph illustrates a straight, diagonal line moving away (since it is above the x-axis). The equation for the position-time graph shows a parabolic equation, y= 440.32x^2 - 13.539x + 0.3736, with an R^2 value of .99, showing our results were extremely accurate. The equation for the velocity-time graph shows a linear equation, y=881.46x-13.08 (the slope being 881.46), with an R^2 value of .99, also proving accurate results. However, the slope of that line is 881.46 cm/s^2, and therefore, our group received results less than the theoretical expectation of 980 cm/s^2. This can be blamed on the friction that existed within the ticker tape timer, causing the acceleration to be slowed and all results to be less than predicted.

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">Sample Calculations: <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">Percent Error= 10.06% <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">Percent Difference= 5.69% <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;"> <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">**Discussion Questions:** <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">**1. Does the shape of your v-t graph agree with the expected graph? Why or why not?** <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">Yes, the shape of the v-t graph agrees with the expected graph because the velocity was increasing as time went on, and therefore, should be displayed in a straight, diagonal line above the x-axis moving away. It goes with the x-t graph, showing an increasing curved line, and the v-t graph produced is the accurate way to produce our x-t graph in v-t form. <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">**2. Does the shape of your x-t graph agree with the expected graph? Why or why not?** <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">Yes, the shape of the x-t graph agrees with the expected graph because the position was increasing as time went on, and therefore, should be displayed in a curved, increasing, line moving away from the origin. It goes with the v-t graph, showing a straight, diagonal line above the origin, and the x-t graph produced is the accurate way to produce our v-t graph in x-t form. <span style="font-family: 'Times New Roman',Times,serif; font-size: medium;">**3. How do your results compare to that of the class?** <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">Our percent difference was 5.69%, which isn't too far off from the rest of the class. Our group's results were the second highest in the class, at a value of 881.5cm/s^2, and therefore, we were one of the closest groups to the expected value of 980cm/s^2. There was a difference of about 47 cm/s^2 between our result and the average, leading to the conclusion that many groups received further below the expected value, as a result of the friction present. <span style="font-family: 'Times New Roman',Times,serif; font-size: medium;">**4. Did the object accelerate uniformly? How do you know?** <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">Although the R^2 value is extremely close to 100%, at 99.93%, it IS NOT exactly 100%, and therefore, our object did not accelerate uniformly. The acceleration at one point on our graph does not perfectly match up to the acceleration on another point, although it was relatively close. <span style="font-family: 'Times New Roman',Times,serif; font-size: medium;">**5. What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?** <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">The leading factor causing acceleration to be lower than it should be, which was what most groups received, is the presence of friction within the ticker tape timer. When the tape was moving through the timer, friction was created between the box and the tape, causing the acceleration speed to be less than 980cm/s^2. If different materials were used, air resistance could be a possible problem to a lower acceleration, but in this case, it didn't affect our experiment. On the other hand, any groups that received a higher acceleration, must have had some sort of error in their procedure, for there would have to be a force pulling the mass downwards to receive an acceleration higher than 980cm/s^2.

<span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">**Conclusion:** <span style="font-family: 'Times New Roman',Times,serif; font-size: 16px;">The theoretical acceleration of the mass was 980cm/s^2, the acceleration due to gravity. Our experimental acceleration of the mass was 881.46cm/s^2, slightly below the acceleration due to gravity. Therefore, although our hypothesis was not fully accurate, it came close to our experimental results. This is because we did not take into account the leading source of error that existed within almost everyone's experiments. The presence of friction that was created between the box and the tape while the tape was moving through the timer reduced the acceleration speed from what is expected. In addition, further sources of error could have came from either wrinkles in the tape, causing the results to be thrown off, or the human error of miscalculations during measurement. For example, our tape ended up ripping during the falling process, and therefore, we were left having to tape together the pieces of tape as accurately as possible to measure the distance between the dots, definitely having some impact on the accuracy of our data. In order to prevent these errors, one would have to eliminate friction entirely. Since this is almost impossible, a way to reduce the friction created is to be cautious to hold the ticker tape at an extremely high vertical angle, and therefore, the time it takes for the ticker tape to move through the timer would be reduced, ultimately causing friction to be reduced as well. In addition, performing the experiment multiple times would enable one to have more than one tape, and therefore, wrinkles or measurement issues could be reduced since one would have more than one trial to base their data from.