Ch2_AronsonJ

toc **Constant Speed**

**__Lab: Speed of a Constant Motion Vehicle (CMV)__** **Objectives:**1. How fast does a CMV move?2. How precisely can distances be measured?3. What does a position-time graph tell you? **Materials:**Yellow CMV, spark timer*, spark tape, meter stick, masking tape//*use 10 Hz// **Hypotheses:**1. A CMV moves approximately 0.2 m/s.2. Distances can be measured within approximately 1.0 cm.3. A position-time graph measures the change in position over time. **Data:** **Class Length Range (cm)**:24.5 cm-24.8 cm //use 4 significant figures//
 * Time (s) || Position (cm) ||
 * 0 || 0 ||
 * 0.1 || 1.09 ||
 * 0.2 || 2.2 ||
 * 0.3 || 3.41 ||
 * 0.4 || 4.49 ||
 * 0.5 || 5.78 ||
 * 0.6 || 6.89 ||
 * 0.7 || 8.02 ||
 * 0.8 || 9.25 ||
 * 0.9 || 10.47 ||
 * 1.0 || 11.68 ||

**Graph Equation:**y = 11.582x + 0.99574m= Δ position/ Δ time = velocity 1. Precision is the closest measurement to the exact measurement (ex. 12.34 > 12.3). 2. Slope is equal to velocity 3. Trendline* is more precise than average. //*//r2=evaluation of fit
 * Lessons:**


 * Discussion Questions and Conclusion:**
 * 1) Why is the slope of the position-time graph equivalent to average velocity?
 * 2) The slope of the position-time graph measures the change in distance over the change in time, which is the formula for velocity.
 * 3) Why is it average velocity and not instantaneous velocity? What assumptions are we making?
 * 4) We use average velocity because it offers a more accurate representation of the data as a whole, as opposed to using one measurement. We assume that each interval may not be precisely the same velocity as that before it.
 * 5) Why was it okay to set the y-intercept equal to zero?
 * 6) We set the y-intercept equal to zero because we start at the origin, or time 0, and could never have negative time.
 * 7) What is the meaning of the R2 value?
 * 8) R2 represents the evaluation of the line of best fit, essentially telling us how accurate our measurements were and how precise the trendline fits that data.
 * 9) 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?
 * 10) That data and line of best fit would lie beneath our data and line of best fit.

Our group learned that 1) a slow, one-battery CMV moves at 11.582 cm/s; 2) distances can be measured more precisely with other instruments than meter sticks; and 3) a position-time graph tells us velocity. We know this because me used our tools to measure, as precisely as we could, the change in distance over time; acknowledged the potential for human error in measuring these; and realized that the slope of our position-time graph measured change in distance over change in time, or velocity. My first two hypotheses were not very accurate, as I overestimated the speed of the cart and underestimated the precision of our measurements, but my third one was accurate, because I wrote that a position-time graph measures the change in position over time. There were many sources of potential human error in performing the lab. One could have been that the yardstick shifted during our measurements, affecting our measurements’ precision, or that we did not collect data as precisely as we could have, affecting our line of best fit and accuracy of data. To minimize these issues we could have used a ruler or measuring tape, which are more precise tools of measurement.

= __**Notes**__ =
 * Constant speed=instantaneous speed
 * Always same value
 * V=d/t
 * Four types of motion:
 * At rest=not moving/stationary
 * Constant speed=moving at unchanging speed
 * Increasing speed=acceleration
 * Decreasing speed=negative acceleration (NOT deceleration)
 * Representing motion with diagrams
 * Velocity/Motion diagrams (Chart #1)
 * Direction
 * Arrows toward right=positive velocity (ADD signs)
 * Arrows toward left=negative velocity (ADD signs)
 * Length
 * Increasing length of arrows=increasing velocity
 * Decreasing length of arrows=decreasing velocity
 * Equal length arrows=constant speed
 * Arrows can be up, down, right or left
 * Represents acceleration/speed increase
 * At rest:
 * V=0
 * a=0 (INCLUDE in graph)
 * Shows direction
 * Ticker-tape diagrams (Chart #2)
 * Direction
 * One dot=at rest
 * Equal length between dots=constant speed
 * Increasing length between dots=increasing velocity
 * Decreasing length between dots=decreasing velocity
 * Length
 * Dots increasing space towards right=positive velocity
 * Dots decreasing space towards right=negative velocity
 * Does not show direction
 * Shows distances
 * Signs are arbitrary


 * Chart #1**


 * Chart #2**

=Graph Shapes=


 * Position versus Time Graph:**
 * Velocity versus Time Graph:**
 * Acceleration versus Time Graph:**

= Physics Classroom: Kinematics 1-D =


 * Lesson 1-a.,b., c. **
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) I read in Lesson 1-c. about the difference between distance and displacement, but I already understood that difference from our class discussion. Ms. Burns’s demonstration was very helpful: for distance, she walked 3 steps ahead, describing the distance as 3 units; and for displacement, she walked 2 units down and 3 units back up, describing the displacement as 1 unit from the origin. I also read in Lesson 1-d. about the difference between instantaneous speed and average speed, but again I understood that material from our class discussion. As an example, Ms. Burns explained to a student that his driving speed along one particular highway may have been 65 miles/hour, but the average speed at which he traveled the entire trip was closer to 50 miles/hour.
 * 3) 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.
 * 4) I read in Lesson 1-d. about the difference between speed and velocity, which helped to clarify that difference. I did not have a misconception so much as I was generally confused. Though Ms. Burns explained both concepts, I did not grasp that velocity requires a change in position to be measured, whereas speed is simply the rate of movement.
 * 5) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 6) Could someone please explain the measurement of //average// velocity? I think that I understand it, but I could use some clarification.
 * 7) What (specifically) did you read that was not gone over during class today?
 * 8) I read in Lesson 1-a. about the difference between scalars and vectors, which Ms. Burns referenced but did not divulge.

**Lesson 2-b.,c.,d.**
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) I read in Lesson 2-b. and c. about ticker tape and vector diagrams, both of which I understood from our class discussion. As Ms. Burns explained to us, ticker tape diagrams are useful because they are easier to make and show distance, though they are more difficult to read, while vector diagrams are useful because they are easier to read and show direction, though they are more difficult to make.
 * 3) 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.
 * 4) In this particular reading I understood all of the material.
 * 5) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 6) How do we make a vector diagram that accurately represents distance? I understand how ticker tape diagrams accomplish this, but not how vector diagrams do so.
 * 7) What (specifically) did you read that was not gone over during class today?
 * 8) I learned in the reading that when an object is slowing down, its direction of acceleration is opposite that of its motion.

Lesson 1-e.
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) I read about acceleration and constant acceleration, but I already understood those concepts from our class discussion. Acceleration, I learned from Ms. Burns, is the rate at which velocity changes. Constant acceleration represents a velocity that changes at the same rate over different periods of time.
 * 3) 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.
 * 4) I understand almost everything that we learned in class.
 * 5) What (specifically) did you read that you still don't understand? Please word these in the form of a question.
 * 6) How does acceleration represent direction? Though we reviewed this idea in class, I am uncertain about how acceleration represents direction.
 * 7) What (specifically) did you read that was not gone over during class today?
 * 8) I read that the direction of acceleration depends on the velocity of the object (speeding up or slowing down) and whether that object is moving in a positive or negative (away from or towards the origin) direction.


 * Lesson 3-a.,b.,c.**


 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) In Lesson 3-b. I read about the meaning of slope in an x-t graph, but I already understood that material from class. Ms. Burns taught us that the slope of an x-t graph tells us the velocity, so reading the web lesson did not further my understanding of the slope. In lesson 3-c. I read about determining the slope of an x-t graph, but again I already knew how to calculate slope. Slope is equal to the change in y-value over the change in x-value.
 * 3) 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.
 * 4) I was not confused about material from this lesson.
 * 5) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 6) I now understand everything from this lesson.
 * 7) What (specifically) did you read that was not gone over during class today?
 * 8) We reviewed all lesson material in class.


 * Lesson 4-a.,b.,c.,d.,e. **
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) In Lesson 4-b. I read about the meaning of slope in a v-t graph, but I already understood that material from class. Ms. Burns taught us that the slope of a v-t graph tells us the acceleration, so reading the web lesson did not further my understanding of the slope. In lesson 4-c. I read about relating the shape of a v-t graph to motion, but again I already knew how to determine that from our whiteboard practice in class.
 * 3) 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.
 * 4) I was a bit confused about the difference between a v-t graph with a constant, positive slope, and a v-t graph with a constant, zero slope. The lesson helped clarify that the former represents positive velocity and acceleration, while the latter represents positive velocity and no acceleration.
 * 5) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 6) I am still a bit confused about creating a v-t graph to represent data. How can I better understand how to draw velocity as slope?
 * 7) What (specifically) did you read that was not gone over during class today?
 * 8) We reviewed all lesson material in class.
 * Lesson 5-a.,b.,c.,d. and e.**

//Topic Sentences:// This lesson teaches the basics about objects in free fall. I learned: acceleration due to gravity (known as "g") occurs at 9.8m/s^2, we should ignore air resistance, the shape of an acceleration versus time graph and objects of different masses fall at the same rate (depending upon surface area).

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 **. There are two important motion characteristics that are true of free-falling objects:
 * Introduction to Free Fall **
 * 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 for //back-of-the-envelope// calculations)

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 dot diagram above depicts the acceleration of a free-falling object. The fact that the distance that the object travels every interval of time is increasing is a sure sign that the ball is speeding up as it falls downward. Recall from an [|earlier lesson], that if an object travels downward and speeds up, then its acceleration is downward. This numerical value for the acceleration of a free-falling object is such an important value that it is given a special name. It is known as the ** acceleration of gravity ** - the acceleration for any object moving under the sole influence of gravity. A matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a special symbol to denote it - the symbol ** g **. The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s/s. ** g = 9.8 m/s/s, downward ( ~ 10 m/s/s, downward) ** Recall from an [|earlier lesson] that acceleration is the rate at which an object changes its velocity over time. To accelerate at 9.8 m/s/s means to change the velocity by 9.8 m/s each second.
 * The Acceleration of Gravity **

If the velocity and time for a free-falling object being dropped from a position of rest were tabulated, then one would note the following pattern. Observe that the velocity-time data above reveal that the object's velocity is changing by 9.8 m/s each consecutive second. That is, the free-falling object has an acceleration of approximately 9.8 m/s/s.
 * ** Time (s) ** ||  ** Velocity (m/s) **  ||
 * 0 ||  0  ||
 * 1 ||  -9.8  ||
 * 2 ||  -19.6  ||
 * 3 ||  -29.4  ||
 * 4 ||  -39.2  ||
 * 5 ||  -49.0  ||

Another way to represent this acceleration of 9.8 m/s/s is to add numbers to our dot diagram that we saw [|earlier in this lesson]. The velocity of the ball is seen to increase as depicted in the diagram at the right. (NOTE: The diagram is not drawn to scale - in two seconds, the object would drop considerably further than the distance from shoulder to toes.) A position versus time graph for a free-falling object is shown below.
 * Representing Free Fall by Graphs **

Observe that the line on the graph curves. [|As learned earlier], a curved line on a position versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9.8 m/s/s), it would be expected that its position-time graph would be curved. A further look at the position-time graph reveals that the object starts with a small velocity (slow) and finishes with a large velocity (fast). Since the slope of any position vs. time graph is the velocity of the object ( [|as learned in Lesson 3] ), the small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity. Finally, the negative slope of the line indicates a negative (i.e., downward) velocity. A velocity versus time graph for a free-falling object is shown below.

Observe that the line on the graph is a straight, diagonal line. As learned earlier, a diagonal line on a velocity versus time graph signifies an accelerated motion. 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. A further look at the velocity-time graph reveals that the object starts with a zero velocity (as read from the graph) and finishes with a large, negative velocity; that is, the object is moving in the negative direction and speeding up. An object that is moving in the negative direction and speeding up is said to have a negative acceleration (if necessary, review the [|vector nature of acceleration] ). Since the slope of any velocity versus time graph is the acceleration of the object ( [|as learned in Lesson 4] ), the constant, negative slope indicates a constant, negative acceleration. This analysis of the slope on the graph is consistent with the motion of a free-falling object - an object moving with a constant acceleration of 9.8 m/s/s in the downward direction. Thus, the velocity of a free-falling object that has been dropped from a position of rest is dependent upon the time that it has fallen. The formula for determining the velocity of a falling object after a time of ** t ** seconds is  ** vf = g * t ** where ** g ** is the acceleration of gravity. The value for g on Earth is 9.8 m/s/s. The above equation can be used to calculate the velocity of the object after any given amount of time when dropped from rest. Example calculations for the velocity of a free-falling object after six and eight seconds are shown below. ** Example Calculations: ** At t = 6 s vf = (9.8 m/s2) * (6 s) = 58.8 m/s At t = 8 s vf = (9.8 m/s2) * (8 s) = 78.4 m/s The distance that a free-falling object has fallen from a position of rest is also dependent upon the time of fall. This distance can be computed by use of a formula; the distance fallen after a time of ** t ** seconds is given by the formula. ** d = 0.5 * g * t2 ** where ** g ** is the acceleration of gravity (9.8 m/s/s on Earth). Example calculations for the distance fallen by a free-falling object after one and two seconds are shown below. ** Example Calculations: ** At t = 1 s d = (0.5) * (9.8 m/s2) * (1 s)2 = 4.9 m   At t = 2 s d = (0.5) * (9.8 m/s2) * (2 s)2 = 19.6 m   The diagram below (not drawn to scale) shows the results of several distance calculations for a free-falling object dropped from a position of rest.
 * How Fast? and How Far? **


 * The Big Misconception **

The answer to the question (doesn't a more massive object accelerate at a greater rate than a less massive object?) is absolutely not! That is, absolutely not if we are considering the specific type of falling motion known as free-fall. 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. At that time, you will learn that the acceleration of an object is directly proportional to force and inversely proportional to mass. Increasing force tends to increase acceleration while increasing mass tends to decrease acceleration. Thus, the greater force on more massive objects is offset by the inverse influence of greater mass. Subsequently, all objects free fall at the same rate of acceleration, regardless of their mass.

= Kinematics Equations = a = Δv*/Δt vf = vi+aΔt (no displacement) Δd = ½(vi+vf)t Δd = vit + ½at2 vf2 = vi2+2aΔd

*(vf–vi)

= **Constant Motion** =


 * 1) How can you tell that there is no motion on a…
 * 2) position vs. time graph—Constant motionlessness is represented by a straight line along the x-axis.
 * 3) velocity vs. time graph—Ibid
 * 4) acceleration vs. time graph—Ibid


 * 1) How can you tell that your motion is steady on a…
 * 2) position vs. time graph—Steady motion is represented by a constant slope of motion.
 * 3) velocity vs. time graph—Ibid
 * 4) acceleration vs. time graph—Ibid


 * 1) How can you tell that your motion is fast vs. slow on a…
 * 2) position vs. time graph—Fast motion is represented by a greater slope, while slow motion is represented by a gradual slope.
 * 3) velocity vs. time graph—Ibid
 * 4) acceleration vs. time graph—Fast motion is represented by a steadier line, while slow motion is represented by erratic y-value behavior.


 * 1) How can you tell that you changed direction on a…
 * 2) position vs. time graph—Change in direction is represented by an inverse slope, with the positive representing movement away from the sensor and negative representing movement towards the sensor.
 * 3) velocity vs. time graph—Change in direction is represented by a stoppage in velocity, as humans pause in between strides to adjust their bodies.
 * 4) acceleration vs. time graph—Change in direction is represented by a dip in the slope, because humans have to slow down to change direction.


 * 1) What are the advantages of representing motion using a…
 * 2) position vs. time graph—Representing motion with a position graph shows the distance traveled towards or away from a starting point.
 * 3) velocity vs. time graph—Representing motion with a velocity graph shows the changes in direction.
 * 4) acceleration vs. time graph—Representing motion with an acceleration graph shows change in speed.


 * 1) What are the disadvantages of representing motion using a…
 * 2) position vs. time graph—Representing motion with a position graph does not show the specific direction in which an object turns.
 * 3) velocity vs. time graph—Representing motion with a velocity graph shows erratic behavior, which is not aesthetically pleasing or easy to read.
 * 4) acceleration vs. time graph—Representing motion with a position graph does not show the specific direction in which an object turns.


 * 1) Define the following:
 * 2) No motion—when an object does not travel any distance or change position.
 * 3) Constant speed—when an object travels at a static rate.

= **Notes** =


 * m(Position v. Time graph) = v(Velocity v. Time graph)
 * m(Velocity v. Time graph) = a(Acceleration v. Time Graph)
 * Constant slope=horizontal line for velocity
 * acceleration (a) = Δ velocity
 * m/s^2
 * Rate that velocity changes
 * velocity ONLY for average or constant speeds

= **Acceleration Graphs Lab** = //September 14, 2011// //Jake Aronson and Danielle Bonnett//

__Objectives and Hypotheses:__
 * 1) What does a position-time graph for increasing speeds look like?
 * 2) A position-time graph looks like a straight line representing a constant acceleration.
 * 3) What information can be found from the graph?
 * 4) We can determine velocity and distance traveled from the graph.

__Materials:__ Spark tape, spark timer, track, dynamics cart, ruler/meter stick/measuring tape, textbook

__Procedure__
 * 1) Set one end of the track on top of the edge of the textbook.
 * 2) Plug in the spark timer and set it to 10 Hz.
 * 3) Insert spark tape into the spark timer and connect one end of the spark tape to the tail end of the dynamics cart.
 * 4) Place the cart on the track so that the wheels fit into the grooves.
 * 5) Turn on the spark timer.
 * 6) Release the cart from the elevated end of the track and stop it once it has hit the bottom of the track.
 * 7) Measure the distance in between each dot on the spark tape, starting with the second or third dot as Point 0.
 * 8) Repeat Steps 1–7, changing Step 6 to “…from the bottom of the track…the elevated end of the track.”

__Data and Graphs__

Increasing Down Incline
 * ** Time (s) ** || ** Distance (cm) ** ||
 * 0 || 0 ||
 * 0.1 || 0.39 ||
 * 0.2 || 1.89 ||
 * 0.3 || 6.1 ||
 * 0.4 || 12.35 ||
 * 0.5 || 18.41 ||
 * 0.6 || 24.29 ||
 * 0.7 || 29.5 ||
 * 0.8 || 35.45 ||
 * 0.9 || 40.79 ||
 * 1 || 46.91 ||

Decreasing Up Incline:
 * ** Time (s) ** || ** Distance (cm) ** ||
 * 0 || 0 ||
 * 0.1 || 0.39 ||
 * 0.2 || 1.89 ||
 * 0.3 || 6.1 ||
 * 0.4 || 12.35 ||
 * 0.5 || 18.41 ||
 * 0.6 || 24.29 ||
 * 0.7 || 29.5 ||
 * 0.8 || 35.45 ||
 * 0.9 || 40.79 ||
 * 1 || 46.91 ||


 * || __Analysis:__ ||
 * 1) The equation of the increasing line can be compared to the equation Δd=½at2 +vit. By substituting our data values with the equation’s, our data tell us that our cart had an acceleration of 31.594 m/s. Our r2 value indicates that most of our data fell within 1 unit of the target point, based on the slope.
 * 2) Our instantaneous speed at halfway point is 61.37 m/s, and at the end is 46.91 m/s.
 * 3) The average speed for the entire trip is 19.64 m/s.



__Discussion Questions and Conclusion:__
 * 1) What would your graph look like if the incline had been steeper?
 * 2) If the incline had been steeper, then the curves would have been more dramatic. The graph of acceleration down the incline would have shown a more exaggerated J-curve, and the graph of acceleration up the incline would have leveled out in much less time.
 * 3) What would your graph look like if the cart had been decreasing up the incline?
 * 4) When the cart was decreasing up the incline, it showed a steep slope in the beginning, but quickly leveled off as the velocity decreased.
 * 5) Compare the instantaneous speed at the halfway point with the average speed of the entire trip.
 * 6) The instantaneous speed at halfway was much greater than the average speed of the entire trip because the first few points with much lower speeds were outliers, pulling the average down. By the halfway point the cart had gathered much more speed, so the instantaneous speed reflects a further accelerating graph, while the average speed takes into account the slow start.
 * 7) Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense?
 * 8) The instantaneous speed is the slope of the tangent line because the tangent line has a constant slope, and a midway point velocity would represent that constant slope. In addition the original line is curved, so the tangent line hitting the midway point offers a middle-of-the-road estimate of the instantaneous speed at a given point on the curved line.
 * 9) Draw a v-t graph of the motion of the cart. Be as quantitative as possible.
 * 10) (see Graph #2 above)

We learned that an object accelerating down an incline will increase its acceleration exponentially, and that the slope of the line representing acceleration data (with equation Δd=½at2 +vit) is also the acceleration. As for our hypotheses, they were not correct, but we learned from our mistakes. Initially we thought that the position-time graph would represent acceleration at a constant rate, but after performing the lab we realized that acceleration is gradually increasing, almost exponentially. This is apparent in viewing our graph, which shows the acceleration down an incline as a steadily increasing curve. Our second hypothesis was partially correct, in that we can tell distance from a position-time graph, but we actually learned the acceleration, not velocity, from the graph. The equation for our data resembles the equation Δd=½at2 +vit, which means that our slope is equal to ½ of the acceleration. There were a few sources of error that could have skewed the data. The most important one is that the end of the track may have been imbalanced atop the textbook, affecting the acceleration of the cart up and down the incline. Another source of error lay in the measurement. If the ticker tape moved or was read slightly imprecisely during measurement, then our data could have been inexact. Finally, a member of our group could have pushed the cart with too much force down the path, artificially skewing the acceleration. To minimize this issue we could use a definitely flat surface, like a low countertop, on which to perch one end of the track. We also could use tape to keep the ticker tape flat while reading it, which our group did. Finally, we could pay closer attention to how much force we put into releasing the cart down an incline, so it accelerates on its own.

= Velocity Crash Course Lab = //September 21, 2011// //By Jake Aronson, Danielle Bonnett, Michael Solimano and Jessica Smith//

We are trying to learn at what velocity does a CMV catch another, and at what velocity do two CMVs crash into each other.
 * Purpose**

Car#1—Fast, Yellow CMV (y=28.274x) Car#1—Slow, Yellow CMV (y=11.582x)


 * Data**

__CRASH RUN__ //Calculations for Time, Theoretical Distance, % Error and % Difference//
 * ** Trial ** || ** Position at Which Car #1=Car#2 (cm) ** || ** % Error ** || ** % Difference ** ||
 * 1 ||  227  ||  30.23  ||  4.22  ||
 * 2 ||  217  ||  24.49  ||  0.367  ||
 * 3 ||  218  ||  25.07  ||  0.0918  ||
 * 4 ||  215.5  ||  23.63  ||  1.06  ||
 * 5 ||  211.5  ||  21.34  ||  2.89  ||
 * Average ||  217.8  ||  24.95  ||  1.73  ||

__CATCH-UP RUN__ //Calculations for Time (t), Theoretical Value, % Error and % Difference//
 * ** Trial ** || ** Position Where Car#1>Car#2 (cm) ** || ** % Error ** || ** % Difference ** ||
 * 1 ||  80  ||  15.31  ||  21.951  ||
 * 2 ||  56  ||  19.28  ||  14.634  ||
 * 3 ||  67.5  ||  2.78  ||  2.896  ||
 * 4 ||  58.5  ||  15.68  ||  10.823  ||
 * 5 ||  66  ||  4.87  ||  0.609  ||
 * Average ||  65.6  ||  11.584  || 10.183 ||


 * Pictures #1 & #2, and Video**

media type="file" key="New Project - Mobile.m4v" width="300" height="300"


 * Discussion Questions and Conclusion**
 * 1) Where would the cars meet if their speeds were exactly equal?
 * 2) If started from opposite directions and moving towards each other at the same speed, the cars would meet directly in the middle, at x=300cm. If started from the same end, but separated 100 cm and moving in the same direction at the same speed, the cars would never meet.
 * 3) Sketch p-t graphs to represent the catching up and crashing situations. Show the point where they are at the same place at the same time.
 * 4) (See Picture #1 above)
 * 5) Sketch v-t 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?
 * 6) (See Picture #2 above). On a velocity versus time graph there is no way to find when the points are at the same place and time, as the graph does not indicate position.

Our lab group found that, started from 600 cm apart, a slow and yellow CMV would "crash" into a fast and yellow CMV after traveling an average of 217.5 cm; also, that, with the slow and yellow CMV 100 cm from the fast and yellow CMV, the faster CMV would catch up to the slower CMV at about 65.6 cm past the slower one's starting point. These numbers are slightly different from our calculated values, or theoretical values, but that difference could have been caused by many sources of error.

The primary source of error is that the CMVs did not travel in a straight line, which affected the position at which the CMVs crashed and caught up. Another source of error is imprecise measurement of the points at which the CMVs crashed and caught up to each other, which would have affected the position as well. To minimize these errors we could have attached the CMVs to a straight track, and stopped the CMVs as soon as they crashed and caught up.

= Position-Time Graphs =

=**Egg Drop Project** = //By Jake Aronson and Michael Solimano// //September 28, 2011//


 * Final Design **

Our final model was unsuccessful in protecting the egg: when our model hit the ground, the egg cracked and "scrambled." The average time from release to impact was 1.5275 seconds, which was much less time than we had hoped to achieve with our project.
 * Discussion of Result **

18.04g
 * Mass of Final Project**

d = vi(t) + (1/2)at^2 d=8.5 m vi=0 m/s t=1.5275 s a=? m/s^2 g (gravity)=9.8 m/s^2
 * Calculation of Acceleration **

8.5=(0)(1.5275)+(1/2)(a)(1.5275)^2 8.5=0+(1/2)(2.333)(a) 8.5=1.1667a 7.2855=a


 * Final Acceleration = 7.2855 m/s^2**

Our egg broke because the project design was flawed. We observed that our egg reached the ground in far less time than it should have taken, indicating that our parachute did not function properly. Because the straw cage in which we enclosed our egg and cone in which we put said cage were wider than the parachute, the parachute was blocked off from air that could buoy it upwards. We also observed that the model hit the ground on its side, rather than at the base of the cone, showing a lack of balance in the model. Because our egg was not secure and the cone was not entirely symmetric, the egg's weight must have shifted mid-drop, forcing the model to tip over and land on the egg-heavy side. Finally, the final drop was probably flawed, not releasing the model so that it could have a chance to catch air under the parachute and maintain its balance. If we had the opportunity to redo the project, then we would implement the following changes. First, we would change the design of the cone so that it was slimmer, lighter and longer, probably using newspaper rather than printer paper. Second, we would reduce the size of the straw cage so that it fit the egg securely inside, without any opportunity for the egg to move around mid-drop. Finally, we would make the top of the parachute sturdier so that the dropper could hold it from the top and allow for all parts to drop at the same time; in fact, with that revision, the parachute might catch air earlier than before. Our second design, in which we secured the egg within a taller and thinner cone, was probably better-equipped to survive the drop than was our final design.
 * Analysis **

=Falling Object Lab=

// By Jake Aronson and Danielle Bonnett // // October 5, 2011 //
 * Acceleration of a Falling Body Lab **

__ Objective: __ What is the acceleration of a falling body?

__ Hypothesis: __ The acceleration of a falling body is -980 cm/s2 due to gravity.



__ Materials: __ Ticker tape timer, timer tape, masking tape, mass, clamp, meterstick

__ Procedure: __
 * 1) Collect all materials.
 * 2) Cut a piece of ticker tape as long as the distance from which you plan to drop it.
 * 3) Slide one end of the ticker tape through the ticker tape timer and tape that end to a 100g mass.
 * 4) Drop the mass from a balcony, keeping the ticker tape straight during the drop.
 * 5) Gather ticker tape, measure distances between dots and record.

__ Data: __


 * ** Time (s) ** || ** Position (cm) ** ||
 * 0 ||  0  ||
 * 0.1 ||  1.2  ||
 * 0.2 ||  8  ||
 * 0.3 ||  21.99  ||
 * 0.4 ||  43.99  ||
 * 0.5 ||  73.99  ||
 * 0.6 ||  111.59  ||
 * 0.7 ||  157.19  ||
 * 0.8 ||  210.21  ||
 * 0.9 ||  270.1  ||
 * 1 ||  338.19  ||

__ Analysis: __

The equation of this position versus time graph is in the form, y=Ax2+Bx, but it follows the formula that we have learned in class, y=vit+½at2. That means that the slope of the trendline is equal to the acceleration of the object, so our acceleration equals 756.56 cm/s2. Considering that the R2 value is so close to 1, our data display a near-perfect fit for the polynomial trendline.
 * Graphs **

The equation of this velocity versus time graph is in the slope-intercept form, y=mx+b, but it follows the formula that we have learned in class for final velocity, vf=vi+at. That means that the slope of the trendline is equal to the acceleration of the object, so our acceleration equals 755.87 cm/s2, which is very close to the velocity predicted from the velocity versus time graph. Again, our data demonstrate a near-perfect relationship with the trendline, with an R2 value of approximately 1.


 * Sample Calculations **




 * Average % Error: 1.69
 * In calculating % Error, I chose to omit the value for x=0.1s, as its y-value skewed the average % Error, but did not disturb the graph.
 * % Difference: 10.02

__ Discussion Questions and Conclusion: __
 * 1) Does the shape of your v-t graph agree with the expected graph? Why or why not?
 * 2) No, the shape of our v-t graph does not agree with the expected graph. We expected the line of the velocity graph to slant negatively away from the origin, into the fourth quadrant of the graph, because we dropped the mass away from the starting point. However, our graph only displayed positive data, so it displayed a positive slope into the first quadrant of the graph.
 * 3) Does the shape of your x-t graph agree with the expected graph? Why or why not?
 * 4) Yes, the shape of our x-t graph agrees with the expected graph. We expected the speed to the mass to increase exponentially, displaying a J-curve on the graph, because velocity and acceleration were constantly increasing.
 * 5) How do your results compare to that of the class? (Use % Difference to discuss quantitatively.)
 * 6) We achieved a slope of 755.28, while the class achieved 839.417. Our results exhibit a % Difference of 10.02, meaning that they are approximately 10% different from the class average.
 * 7) What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?
 * 8) One factor that would have caused acceleration due to gravity to be higher could have been the movement of our hands when dropping the masses. Had our hands not been completely still, then they could have given the mass unintended momentum. A few factors that would have caused acceleration to be lower could have been friction and imprecise measurements. As the ticker tape ran through the spark timer, the pressure of the ink pressing down on the paper could have reduced the acceleration slightly, and because neither the ticker tape nor measuring tape was completely straight when we measured distances, the data that we collected could have skewed the acceleration lower or higher.

__ Conclusion: __ Our group learned that gravity causes constant acceleration to an object in free fall. We know this because the data that we measured display an exponential J-curve on the position versus time graph. Our hypothesis was not correct because we predicted a negative velocity, but the graph displayed a positive slope. The sources of error could have been the movement of our hands while dropping the masses, which would have given the masses momentum in addition to gravitational acceleration; friction caused by the ticker tape rubbing against the ticker tape timer, which would have reduced the acceleration of the mass slightly; and imprecise measurement because of non-straight measuring tape and/or ticker tape, which would have affected our recorded distances. To minimize these issues we could have placed the masses over a latch that opened on the bottom, allowing the masses to fall without any momentum to start.