Ch2_Ordover

=toc Constant Speed=

Lesson 1 (b ,c, and d)
Basic Terms
 * 1) I had already understood vectors and scalars well. Scalars are described by solely using numeric values. However, vectors are described with numeric values and direction.
 * 2) I did not understand the difference between distance and displacement when we discussed it in class today. I did not get that displacement did not count the total distance moved. I now understand that it is just the distance from a starting point.
 * 3) I had no questions, everything was clear.
 * 4) Average speed and average velocity was not gone over in class.

Class Notes: Constant Speed
average speed- average speed over a period of time constant speed- speed does not change --> --> --> a=0 instantaneous speed- speed at any given time increasing speed- speeding up -> --> ---> > ->.........--> (a) decreasing speed- slowing down -> > ---> --> ->.........<-- (a) at rest- no motion v=0 a=0

Lesson 2 (a, b, and c)
Graphing
 * 1) I understood the ticker tape diagram well. Dots are put on the paper at regular intervals. If they are farther apart then the paper was moving faster. If they are closer together, then the paper was moving slower. If the dots are at a constant distance from each other, then that is constant speed. If they are at different distances from each other, then there is acceleration.
 * 2) I did not understand that a change in arrow size represented acceleration on a vector diagram. Now I understand that increasing and decreasing arrow sizes refer to acceleration.
 * 3) I understood everything from the reading.
 * 4) Everything was gone over in class.

Activity: Graphing Acceleration (Graphical Representations of Equilibrium)
Slow

Fast

At Rest
 * 1) How can you tell that there is no motion on a…
 * 2) position vs. time graph - horizontal line
 * 3) velocity vs. time graph - horizontal line at 0
 * 4) acceleration vs. time graph - horizontal line at 0


 * 1) How can you tell that your motion is steady on a…
 * 2) position vs. time graph - it is a straight line, slope=speed
 * 3) velocity vs. time graph - it would be a straight line at the speed you are moving, slope=acceleration, area=displacement
 * 4) acceleration vs. time graph - horizontal line at 0


 * 1) How can you tell that your motion is fast vs. slow on a…
 * 2) position vs. time graph - the slope would get steeper the faster that you moved
 * 3) velocity vs. time graph - the slope would get steeper the faster that you moved
 * 4) acceleration vs. time graph - can not tell


 * 1) How can you tell that you changed direction on a…
 * 2) position vs. time graph - the slope would become negative
 * 3) velocity vs. time graph - the velocity would be negative
 * 4) acceleration vs. time graph - should stay the same
 * 5) What are the advantages of representing motion using a...
 * 6) position vs. time graph - you know the position at a given time and can determine speed
 * 7) velocity vs. time graph - you know the velocity at a given time and can determine position
 * 8) acceleration vs. time graph - you know the acceleration at a given time


 * 1) What are the disadvantages of representing motion using a…
 * 2) position vs. time graph - can not determine acceleration or instantaneous speed
 * 3) velocity vs. time graph - can not determine position easily
 * 4) acceleration vs. time graph - can not determine velocity or position


 * 1) Define the following:
 * 2) No motion - not moving
 * 3) Constant speed - moving at the same speed for a period of time

Lab: Speed of a CMV
9/9/11

Purpose: In this lab we are attempting to graph the constant motion of a vehicle on a position vs. time graph. We will then analyze the data to see if it is accurate and what we can take out of it.

Objectives: 1) How precisely can you measure distances with a meterstick? 2) How fast does your CMV move? 3) What information can you get from a position-time graph?

Materials: spark timer, spark tape, meterstick, masking tape, cmv

Hypotheses: 1) I can measure distances to the nearest mm. 2) I believe the cmv will move about about 45 cm/s. 3) A position-time graph will show where the cmv is at a specific time.

Data table: Position vs. Time Graph: Position vs. Time Analysis: The slope is rise over run and the rise is displacement while the run is seconds. Therefore the slope is displacement over seconds which is the same as velocity. Hence, the slope of the line is the same as the average velocity. The R^2 value shows us how accurate the data was and my data was extremely accurate with an R^2 value of .998.

Conclusion: The average speed of my cmv was 62.423 cm/sec. This was significantly higher than my estimation of about 45 cm/sec. However, the R^2 of my data was .998 which meant my data was very accurate. On the contrary, my first hypothesis was correct as we were only able to measure to the nearest mm with the meter stick or the ruler. My third hypothesis was also true as I was able to determine the position of the cmv at any given time. I was also able to determine the velocity of the object by calculating the slope of the line. There were a few possibilities for why there might have been inaccuracies in the data. The first is that there might have been slight errors when I measured the ticker tape. Even if they were only off by less than a mm, it could have affected the data. Another reason could have been that the table was not perfectly flat. There could have been a slight slope that affected the data. The final reason was that there could have been any kind of minuscule objects such as dirt that slightly slowed down the cmv.
 * 1) Why is the slope of the position-time graph equivalent to average velocity? The slope of this graph is equivalent to average velocity because slope is y over x. In this graph y is distance and x is time. Distance over time is equal to average velocity. Therefore the slope of this graph is average velocity.
 * 2) Why is it average velocity and not instantaneous velocity? What assumptions are we making? This is average because it is over a period of time and not a one specific time. We are assuming the speed is constant.
 * 3) Why was it okay to set the y-intercept equal to zero? We can set the y-intercept equal to 0 because at 0 seconds the cmv has not moved yet. Therefore its position and time are 0.
 * 4) What is the meaning of the R2 value? The R^2 value signifies how accurate the data actually was.
 * 5) If you were to add the graph of another CMV that moved more slowly on the same axes as your current graph, how would you expect it to lie relative to yours? It would be below the current line on the current graph as it would have a smaller slope due to its lower speed.

Lesson 1 (e)
Acceleration
 * 1) Acceleration is the rate at which an object changes velocity. An object is not accelerating if it is moving at constant speed. Even if an object is moving fast, it is not necessarily accelerating.
 * 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. I did not understand what type of units were used for acceleration. I now know that it is a speed unit per a time unit, such as m/s/s.
 * 3) I understood everything.
 * 4) Average acceleration was not gone over in class.

Lesson 3

 * 1) I understood the meaning of the slope of a P-T graph. The slope is the average speed because it is distance over time. Also, a positive slope means the object is moving away from the camera, while negative slope means it is coming towards the camera.
 * 2) I did not understand what it meant if the slope changed. I now understand that a change in slope represents acceleration, either positive or negative.
 * 3) I understood everything.
 * 4) How to determine slope, however I already knew.

Lesson 4

 * 1) I understand what the shape means on a V-T graph. A straight line represents either constant or no motion. If the line has a slope that is not 0, then it means there was acceleration. The velocity is positive if it is on top of the x-axis and negative if it is below.
 * 2) I did not fully understand how the shape related to the motion. I now understand that a downward slope means the object has leftward acceleration and upward slope means the object has rightward acceleration. If the line is positive it is rightward velocity, if the line is negative it is leftward velocity.
 * 3) I understood everything.
 * 4) We did not learn how to determine the area of a V-T graph.

Lab: Acceleration Graphs
Purpose: We are attempting to graph negative and positive acceleration on a position vs. time graph. We will then analyze the data and graph.


 * Objectives:**
 * What does a position-time graph for increasing speeds look like? The slope will increase in steepness as the time gets higher. This will happen because slope is equal to velocity and the velocity becomes greater as the speed increases.
 * What information can be found from the graph? Position, acceleration, and velocity can be found from the graphs.

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

Procedure: First the metal ramp was set up by being put on one textbook. The spark timer was set up at the top of the ramp with the car next to it. The spark tape was put through the spark timer and taped to the car. We turned on the spark timer and let the car roll down the ramp to get increasing speed. Then we did the same thing, but put the spark timer at the bottom of the ramp and pushed the car up the ramp to get decreasing speed.

Data table: Position vs. Time a) Interpret the equation of the line (slope, y-intercept) and the R2 value. b) Find the instantaneous speed at halfway point and at the end. (You may find this easier to do on a printed copy of the graph. Just remember to take a snapshot of it and upload to wiki when you are done.) c) Find the average speed for the entire trip.
 * Analysis:**
 * 1) The slope of the increasing speed line was y=20.274x^2 + 14.455x. This was positive because it was increasing in speed. The slope of the decreasing speed line was y=-29.35x^2 + 63.793x. This slope was negative because it was decreasing speed. The accuracy of both was great because the R^2 for both was .999.
 * 1) Increasing speed half way – 36.5 cm/s end – 66.67 cm/s
 * 2) Decreasing speed half way – 36.67 cm/s end – 4 cm/s
 * 1) Increasing speed – 35 cm/s
 * 2) Decreasing speed – 34.3 cm/s

1) What would your graph look like if the incline had been steeper? Slope would have been steeper for negative and positive acceleration. 2) What would your graph look like if the cart had been decreasing up the incline? The slope would decrease in steepness as time passed. It would start off being steep and it would get less steep as time passed. 3) Compare the instantaneous speed at the halfway point with the average speed of the entire trip. For both the uphill and downhill graphs, the instantaneous speed at halfway was only slightly larger than the average speed. 4) Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense? Slope must be found over a certain amount of time and since this graph is always changing its slope, a tangent must be drawn to find instantaneous speed. The tangent represents the speed of just that point because it does not touch any other points on the graph. 5) Draw a v-t graph of the motion of the cart. Be as quantitative as possible.
 * Discussion Questions:**

Conclusion: My hypothesis for the shape of an increasing speed graph was correct. It did increase in slope or steepness as the graph went on. My second hypothesis was correct as well. We were able to find position, velocity, and acceleration. Position can be found just by looking at the graph. Velocity can be found at specific point by drawing a tangent line, and average velocity can be found by doing displacement over time. Finally, acceleration can be found by finding change in velocity over change in time. There are a couple of places where errors could have been made. One is when measuring the dots. This is hard to avoid, but it helps to double check and make sure you measured correctly. In addition, we were only able to measure to the nearest mm. Another spot of error could have been in using the dots at the very beginning of the tape which are not always accurate. They may not be accurate because it is impossible to push the cart at the exact time that the spark timer makes a spark. One can avoid this by not using the the very beginning of the tape.

HW: Interpreting Position - Time Graphs (D and E)
D E

Lab: Crash Course
9/23/11 Lab Partner: Maddie Margulies

Purpose: We want to find where two constant motion vehicles will crash if they are put 6m apart and move towards each other at different speeds. We also want to find out where a faster cmv will catch up to a slower cmv if they are placed 1m apart and facing the same direction.

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

Calculations (page on the left, part one on bottom left, part two on bottom right):

Procedure:

Part 2 media type="file" key="crash lab part 1.mov" width="300" height="300"

Part 1 media type="file" key="crash lab part 2.mov" width="300" height="300"

Results:

Discussion Questions: 3. Sketch velocity-time graphs to represent the catching up situation. Is there any way to find the points when they are at the same place at the same time?
 * 1) Where would the cars meet if their speeds were exactly equal? They would meet exactly in the middle for part 1 because they would cover the same distance if they were the same speed. For part 2, they would never catch up and would always be 100 cm apart.
 * 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.

Analysis: Percent Error Percent Difference

Conclusion: For part one we put two different constant motion vehicles of different speeds 6m apart and let them move towards each other. We tried to find where they crashed. My average results for the part 1 were 154.2 cm. This was 17.2 cm different than my theoretical yield of 137 cm. This yielded a percent error of 12.55%. The percent difference for this part ranged from 4.41% to 1.43%. For part two we put the faster cmv 1m behind the slower cmv and saw where it would catch up to the slower one. In part 2 my theoretical yield was 141.78 cm. My average result was 147.2 cm, which yielded a percent error of 3.82%. The percent difference ranged from .815% to .136%. These results were very consistent. There are a few possible reasons for error. One is that the cars did not travel in perfectly straight lines. One of them veered off to the side. This caused it to travel a farther distance. We could fix this next time by using different cars or using a track. In addition, we had to approximate the crash spot because they did not actually crash. Next time we could have used a camera and used that to get a more accurate crash spot. The final source of error is that the cars might not have started moving at the same time. This could be fixed by using a machine to start them at the same exact time.

Egg Drop Project
Matt Ordover and Mike Poleway

Final Structure did not work. It was 27.03 grams. Our design was a box of plastic straws that was held together by tape. The box had two layers and shredded paper inside to cushion the egg. After the egg was put in, more paper was put in and the top was taped down.

Analysis Our acceleration was 8.67 m/s^2. This is lower than the maximum possible acceleration of 9.8 m/s^2. This makes sense because our device created some wind resistance. However, our acceleration was higher than most other groups because we had a small surface area and no parachute. Our high acceleration is probably one of the main reasons that our egg broke.

Conclusion: Our first design was more creative than our final design as it used wood and rubber bands. It did not work because there was very little surface area and not much cushion for the egg. We decided to use straws for the second design as they are lighter and cushion the egg more. We thought this design did not work because it did not have enough cushion. On our final design we tried improving on the second design by using straws, but adding more cushion. We ended up keeping it simple by making a box out of straws and using cushioning it with shredded paper. Our design was good, but the high acceleration of 8.67 m/s^2 caused it to hit the ground at a high speed and the cushioning was not enough. If I did this again I would have get my same design except I would have added a parachute to it. Many other groups did this and it lowered their acceleration greatly. I believe my design would work very well with a parachute attached to it.

Lesson 5
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:
 * A) 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

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. Recall from an [|earlier lesson], that if an object travels downward and speeds up, then its acceleration is downward. B) ** The Acceleration of Gravity ** 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 **.  ** g = 9.8 m/s/s, downward ( ~ 10 m/s/s, downward) **   C) ** Representing Free Fall by Graphs ** One way of describing the motion of objects is through the use of position vs. time and velocity vs. time graphs. A position versus time graph for a free-falling object is shown below. Observe that the line on the graph curves. A curved line on a position versus time graph shows accelerated motion. Since a free-falling object is undergoing acceleration, its position-time graph should 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). The small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity. The negative slope of the line indicates a negative/downward velocity. V-T The graph is a straight, diagonal line. A diagonal line on a velocity versus time graph signifies acceleration. Since a free-falling object is accelerating its velocity-time graph should be diagonal. The object starts with a zero velocity and finishes with a large, negative velocity; the object is moving in the negative direction and speeding up. This is known as negative acceleration. Since slope of a velocity-time graph means acceleration, the constant negative slope indicates a constant, negative acceleration. D) The formula for determining the velocity of a falling object after a time of ** t ** seconds is ** v **** f **** = g * t **, where ** g ** is the acceleration of gravity. The value for g on Earth is 9.8 m/s/ E) The questions are often asked "doesn't a more massive object accelerate at a greater rate than a less massive object?" The answer to the question is absolutely not! That is, absolutely not if we are considering the specific type of falling motion known as free-fall. Free-fall is the motion of objects that move under the sole influence of gravity; free-falling objects do not encounter air resistance.

Freefall Lab
Lab partner: Maddie Margulies

Purpose: What is acceleration due to gravity?

Hypothesis: I predict that the acceleration due to gravity will be 9.8 m/s^2. I predict that the velocity-time graph will look like a diagonal line that starts at the origin and moves up to the right. We will find the acceleration from that graph by finding the difference in velocity from one tenth of a second to another.


 * Materials ** : Ticker Tape Timer, Timer tape, Masking tape, Mass, clamp, meterstick.

Data:

Graphs Position-Time: Velocity-Time:

Analysis Percent error: ((Theoretical-experimental)/(Theoretical))x100 ((980 cm/s^2-861.69cm/s^2)/(980 cm/s^2))x100=12.07%

Percent Difference: ((Average experimental value-Individual experimental value)/(Average experimental value))x100 ((834.03 cm/s^2-861.69 cm/s^2)/(834.03 cm/s^2))x100 3.32% difference

Class Results: Average = 834.03

It does agree with the expected graph as it is linear with a positive slope. It is linear because it is the v-t graph of a parabolic x-t graph. It also agrees with the expected graph. It is a parabolic shape because it has a constant acceleration. We got a result of 861.69 cm/s^2. The average class result was 834.03 cm/s^2. This yielded a percent difference of 3.32%, which means our results were very similar to the rest of the class. This means that our results were pretty accurate. The objects did accelerate uniformly. I know because the velocity-time graph was linear. This means that the acceleration was constant because it is the same as the slope. Air resistance or higher altitude could make the acceleration due to gravity lower than it should be. Another factor would be the friction between the spark tape and the spark timer. There are not many reasons as to why the acceleration would be higher. The only possible reasons would be a lower altitude or air pressure.
 * 1) Does the shape of your v-t graph agree with the expected graph? Why or why not?
 * 1) Does the shape of your x-t graph agree with the expected graph? Why or why not?
 * 1) How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.)
 * 1) Did the object accelerate uniformly? How do you know?
 * 1) What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?

Conclusion: My hypothesis was correct in that that v-t graph was linear with a positive slope. The x-t graph was a parabolic shape because the velocity was not constant, but the acceleration was. Our results showed that the mass fell with an acceleration of 861.69 cm/s^2. This was not too far off from the theoretical value of 980 cm/s^2. This gave us a percent error of 12.07%. The average class result was 834.03 cm/s^2. This gave us a percent difference of only 3.32%. This means our results were very similar to the rest of the class. It also means they are probably accurate. There are a couple possible sources for error. One is that there could have been friction between the spark tape and the spark timer. This could not be completely solved, but it could be fixed a little by making sure that the tape is straight and it is cleanly being fed through the spark timer. Another source of error could have been error in measuring the dots on the spark tape. They could have been measured incorrectly or if the measuring tape or meter stick moved the results would have been off. This could have been fixed by taping the spark tape down and taping one side of the measuring tape down. These were the main sources of error in this experiment.