Chapter 2: One Dimensional Kinematics

Class Notes: Constant Speed

Position: where an object is located in reference to its surroundings
Distance: how far an object has traveled, regardless of direction
Displacement: how far an object is from its starting location; must have a reference point and include a direction
Velocity: rate of change of position (how fast an object is going). This can be negative. Speed is the same thing, but can't be negative. Velocity must reference direction. (Measures displacement, not distance)
Vectors: size and direction
Scalar: size only

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Lesson 1: Describing Motion With Words
B,C,D

  1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
    • Since these readings are about topics that have been covered in class, I was familiar with most of them. Two topics that I was already familiar with (from class) were the definitions of distance and displacement. Displacement refers to an object's overall change in position, and distance refers to how much an object has traveled overall. The example used with the physics teacher only reaffirmed my knowledge.
  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.
    • A concept I had trouble with during class was the definition of scalar vs. the definition of vector. I did not understand if a quantity could be both a scalar and a vector, instead of just one or the other. For example, are all scalars also vectors, but not all vectors are scalars? (This is sort of like the statement "All squares are rectangles, but not all rectangles are squares") The reading helped me see that it is not possible for a quantity to be both a scalar and a vector. It can only be one.
  3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
    • I do not have any questions, as the questions I had from class were answered in the readings.
  4. What (specifically) did you read that was not gone over during class today?
    • Instantaneous speed was not covered in class today. It is "the speed in any given instant in time". It is compared to the speed on the speedometer of a car. It is different from average speed because objects do not travel at consistent speeds all the time.

E

  1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
    • I already knew from our discussion in class that the formula for acceleration is (final velocity - initial velocity) / time. This is a different formula from velocity. In class, we also talked about constant acceleration. This is when an object moves faster or slower at a consistent pace. This is unlike constant speed because the object is either getting faster or slower. The object stays in motion, but does not move at the same rate for a set distance. Instead, the velocity either increases or decreases by a constant amount each second.
  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 have any questions from class that the reading helped to clarify.
  3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
    • I do not have any questions, as the questions.
  4. What (specifically) did you read that was not gone over during class today?
    • We didn't go over the free-falling object in class. A free falling object is the acceleration of an object when it is dropped from above ground. A free falling object does not have constant acceleration. Instead, it goes faster and faster, so its acceleration increases as it falls.


Speed of a Constant Motion Vehicle Lab: September 9Partner: Nicole Tomasofsky

Purpose:The purpose of this lab is to find the speed of a CMV, a constant motion vehicle, and see what information can be gleaned from position time graphs. Also, from this lab, I am trying to find how precisely you can measure distances with a meter stick.

Objectives:

  1. How precisely can you measure distances with a meter stick?
  2. How fast does the CMV move?
  3. What information can you get from a position time graph?



Materials:

  1. spark timer
  2. spark tape
  3. meter stick
  4. masking tape
  5. CMV


Hypothesis:

  1. How precisely can you measure distances with a meter stick? You can measure distances to the nearest millimeter using the meter stick. This is my hypothesis because I have used meter sticks in the past and know that there are millimeter marks on the meter stick, and they are the smallest unit of measurement.
  2. How fast does the CMV move? 75 cm/s. This is my hypothesis because
  3. What information you get from a position time graph? You can get the instantaneous speed and average velocity of an object.

Data:

Length of laptop = 30.0 cm
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= 45.12 cm/s

The CMV moves 45.12 cm/s





Discussion questions

    1. Why is the slope of the position-time graph equivalent to average velocity?
      • Slope is nothing more than rise over run, or y/x. The formula for average velocity is displacement over time elapsed. In my graph, position is the y axis, and time is the x axis. So in this case, y/x = displacement/time. This makes the slope equal the average velocity of the CMV.
    2. Why is it average velocity and not instantaneous velocity? What assumptions are we making?
      • Average velocity is more accurate than instantaneous velocity because instead of just looking at one point on the graph, all our data is taken into consideration. The slope determines the average velocity, and individual plots represent the instantaneous velocity. Since we are using the slope of the line to find velocity, and not individual plots on the graph, we are finding the average velocity. We are making the assumption that the CMV is going at a constant speed.
    3. Why was it okay to set the y-intercept equal to zero?What is the meaning of the R2 value?
      • It is okay to set the y intercept equal to zero in this situation because the y axis measures position, and as our CMVs do not travel backwards, it is impossible for them to have a negative position. The CMVs start at a position of (0,0) because the time and position are both zero.
    4. What is the meaning of the R2 value?
      • The R2 value tells you how accurate your graph is. It does this by setting a trendline through all the points in the graph. Then, the R2 value tells you how close the trendline is to the points you plotted in your scatter gram. The closer to 1.0 it is, the more accurate your results.
    5. If you were to add the graph of another CMV that moved more slowly on the same axes as your current graph, how would you expect it to lie relative to yours?
      • I would expect the graph to have a smaller slope. This means that my original graph would have a steeper slope than the 2nd, new, graph.



Conclusion:



For this lab, there were three objectives: How precisely can you measure distances with a meter stick? How fast does a CMV move? What information can you get from a position time graph? My hypothesis for the first question was that you could measure up to the nearest millimeter. It turns out that you can guess the distance between two millimeter marks, so you can get an even more precise measurement than I had predicted. I had also hypothesized that a CMV could travel 75 cm/s. But, my data shows that the speed of the CMV was actually 44.12 cm/s. Therefore, my hypothesis was wrong. I had also hypothesized that you could get two pieces of information from a position time graph: instantaneous velocity and average velocity. The data my partner and I collected supports this hypothesis.



There are many sources of error that could have contributed to inaccuracies. For example, the meter stick could have moved while my partner and I measured the distance between the dots. This problem could have been remedied by taping down the meter stick. Also, it was hard to get an accurate reading of the distances because the meter stick is made of wood, and the centimeter and millimeter marks do not sit directly on the page, next to the dots that my partner and I were measuring. Instead, they sit a couple of centimeters up, so depending on what angle you choose to read the measurement at, you could have different distances. Using a tape measure could have eliminated this source of error. It is important to learn how to take the proper measurements of a distance because in real life, a couple of centimeters could make a huge difference. For example, if an architect designs a building, but accidentally makes the door frames a couple of cm too small, then the doors can't fit. This would be a huge problem.



Overall, I feel that this lab was a success, and my partner and I made the best of the materials that were available to get the most accurate data.




Class Notes: Graph Shapes (At Rest and Constant Speed)

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Lesson 2: Describing Motion With Diagrams


A,B,C



  1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
    • There were two things that I read about that I already understood from our class discussion. They are:
      • Ticker tape diagrams
        • Ticker tape diagrams are used to describe motion. A long tape is threaded in a specialized machine, and dots are created on the tape. A dot is created X times a second, so the further apart the dots, the faster the tape is going. The closer the dots, the slower it is going. If the dots get more crowded on the paper towards the end, it is showing acceleration. However, this device only measures scalar quantities, as direction of movement is not recorded.
      • Vector Diagrams
        • Vector diagrams are diagrams with arrows and labels. There are two kinds of arrows. Ones that are labeled with "a" show acceleration (whether is is positive or negative) and arrows that are marked with "v" show velocity. If the object moving has positive velocity, the arrows get larger. Negative velocity means the arrows get smaller. If an object is traveling at a steady rate, the arrows are the same size.
  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 have any questions from class that the reading helped to clarify.
  3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
    • I do not have any questions.
  4. What (specifically) did you read that was not gone over during class today?
    • We went over everything in the reading in class.

Class Notes: Speed, Diagrams, Ticker Tape Diagrams


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Motion diagrams show direction of velocity and aceleration

Constant speed:
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Increasing Speed:
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Decreasing Speed:

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The Big Five Physics Equations (For Motion)



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Lesson 3: Describing Motion with Position vs. Time Graphs

  1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
    • I already knew from our classroom discussion that if two position time graphs of constant velocity were to be compared, the one that depicts the faster velocity would have the larger slope and the steeper graph.
    • Before reading this, I also knew that slope is equal to rise over run.This is a concept that I have explored both in math and in physics.The formula for slope, m= (Y1-Y2 / X1-X2) is something that I was familiar with before I read this lesson.
  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.
    • A concept I had trouble with during class was whether position time graphs had a negative y axis.I was confused because velocity-time graphs could have negative y-values.This reading helped me see that it is impossible for an object to have a negative position.
  3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
    • I do not have any questions, as the questions I had from class were answered in the readings.
  4. What (specifically) did you read that was not gone over during class today?
    • I believe everything in the reading was covered in class.


Lesson 4: Describing Motion with Velocity vs. Time Graphs

  1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
    • I already knew from our classroom discussion that if the velocity-time graph was a straight horizontal line with a slope of 0, then the object that was represented would be either moving at a constant speed or at rest.
    • From class, I learned that the area of the region between the line of and the axes in a velocity time graph show the displacement.The information in the reading only reaffirmed my knowledge.
  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 have any questions that the reading helped to clarify.
  3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
    • One question I have is: What if there is a negative velocity?Would displacement still be shown by the area of the region bounded by the line and axes?
  4. What (specifically) did you read that was not gone over during class today?
    • I believe everything in the reading was covered in class.

Class Notes: Increasing and decreasing Speed Graphs


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Activity: Graphical Representation of Equilibrium

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Cart and Ramp Class Activity

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Position time graph practice 9/15


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  1. Describe the motion of the car (qualitatively) during each segment.
    • AB at rest
    • BC constant towards
    • CD at rest
    • DE constant away
    • EF at rest
    • FG constant towards
    • GH constant away
  2. Describe the change in position of the car during each segment.
    • AB none
    • BC 10 to 0 (-10)
    • CD none
    • DE 0 to -16 (-16)
    • EF none
    • FG -16 to 0 (16)
    • GH 0 to 14 (14)
  3. Calculate the velocity of the car during each segment.
    • Distance / time
    • AB
    • BC -10/2 = -5 m/s
    • CD
    • DE -16/ .5 =-32 m/s
    • EF
    • FG 16/1 = 16 m/s
    • GH 14/2 = 7 m/s
  4. What is its average speed for the entire 12-s?
    • Total distance/ Total time (add displacements without signs)
    • 56 m/11 s = 5.1 m/s
  5. What is its average velocity for the entire 12-s?
    • Displacement/ time
    • Final position – initial position = displacement
    • 4/11 = .44 m/s
  6. What is the acceleration of the car during each segment?
    • 0 for each segment because they are constant speeds

Acceleration Graphs Lab: September 16

Partner: Nicole Tomasofsky

Objectives:
What does a position time graph for increasing speed look like?What information can be found from the graph?Hypothesis:
  • What does a position-time graph for increasing speeds look like? It is a curved line, with a positive slope. The slope is getting steeper.
  • What information can be found from the graph? You can find the average speed and instantaneous speed.
Available Materials:
Spark tape, spark timer, track, dynamics cart, ruler/meterstick/measuring tape
Procedure:
  1. Gather materials. This includes: Spark tape, spark timer, track, dynamics cart, ruler/meterstick/measuring tape.
  2. Set up the ramp by placing the end of the track on a textbook.
  3. Set up the spark timer by plugging it in and placing it at the top of the track.
  4. Run a piece of spark tape through the spark timer and tape it to the cart.
  5. Set the spark timer to 10 Hz
  6. Set the cart at the top of the track and let it roll down the ramp.
  7. Save this piece of spark tape.
  8. Now, set the cart at the bottom with the timer.
  9. Push the cart up, so the tape runs through the timer.
  10. Save the tape and make sure there are at least 10 spark marks on each piece of tape.
  11. Using a meter stick, measure the distance between the dots on both pieces of tape.
  12. Record this data in an excel spreadsheet and analyze.
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  • Halfway point = 0.60 s
    • 2 points on line: (0,6) (0.5, 28)
    • slope = 44
    • instantaneous speed = 44 cm/s
  • End = 1.20 s
    • 2 points on line: (0,31) (.3,36)
    • slope = 50/3
    • instantaneous speed = 50/3 cm/s

c) Find the average speed for the entire trip.
Decreasing speed: distance traveled/ time elapsed = average speed
= 66.40cm/2.0s
=33.20 cm/s
Increasing speed: distance traveled/ time elapsed = average speed
=49.98 cm/1.20 s
=41.65 cm/s

Discussion Questions:
  1. What would your graph look like if the incline had been steeper?
    1. The graph would have a larger slope and a steeper graph. The curve of the graph would be more predominant, since the acceleration would be greater.
  2. What would your graph look like if the cart had been decreasing up the incline?
    1. The initial curve would be steep and noticeable (since the cart is traveling faster in the beginning), but the line would flatten out. Also, the line would go away from the origin.
  3. Compare the instantaneous speed at the halfway point with the average speed of the entire trip.
    1. Increasing speed graph:
      1. half way point has a velocity of 25 cm/s.
      2. average velocity is 33.20 cm/s.
      3. Average velocity is faster
    2. Decreasing speed graph:
      1. half way point has a velocity of 44 cm/s.
      2. average velocity is 41.65 cm/s.
      3. Halfway point velocity is faster.
  4. Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense?Draw a v-t graph of the motion of the cart. Be as quantitative as possible.
    1. You cannot find the slope (or change) one point--you need two points. A tangent line offers a slope, and since it touches only the point on the graph it is tangent to, it is representing the slope of the line. The slope of the one line shows the speed at that specific moment, which is instantaneous speed.
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For my hypothesis, I stated that you could find the instantaneous speed and average speed from a position time graph. From this lab, I learned that you can also find determine initial velocity, and average acceleration. I also hypothesized that the position time graph for increasing speed would look like a curved line with a slope that is getting steeper. The graph is also moving away from the origin. This hypothesis proved to be correct.
Tehere are multiple possible sources of error for this experiment. For example, since we had two strips of tape, maybe we got them mixed up. This could have been easily remedied by labeling the strips of tape. Also, another error that could have occurred was that our measures of the distance between the dots were inaccurate. Maybe the tape moved around while we were measuring. This could have been fixed by taping the stick to the meter stick.

Lab: A Crash Course in Velocity


Stephanie Wang
Partner: Nicole Tomasofsky
Purpose: The purpose of this lab is to determine the distance two CMVs must travel in order to crash. Also, we are trying to see how far a slower CMV must travel before a faster CMV catches up to it.

Procedure/Materials:

Crashing:


Catching up:


Data and Calculations:

Speed of CMV 1 (blue): 44.159 cm/s
Speed of CMV 2 (yellow): 12.897 cm/s


Part A) Find another group with a different CMV speed. Find the position where both CMV’s will meet if they start at least 600 cm apart, move towards each other, and start simultaneously.
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They will crash at 464.38 cm from the blue car and 135.62 cm from the yellow car if the two cars start at 600 cm apart.

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Part 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.

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The yellow CMV (slower) will travel 41.25 cm, and the blue CMV (faster) will travel 141.25 cm before the faster CMV catches up with the slower CMV. (This is assuming the blue CMV starts 1 meter, or 100 cm, behind the yellow CMV)
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Analysis
Percent error=(Theoretical – experimental)/theoretical x 100
Percent difference= (avg. experimental value-individual experimental value)/avg. experimental x 100

Part A)
%Error = ((464.38-435)/464.38)*100 = 6.3%
Other percent errors:
  • 5.7%
  • 7.4%
  • 7.0%
  • 7.2%

%Difference = ((433.2-435)/433.2) * 100 = 0.42%
Other percent differences:
  • 11%
  • 7.4%
  • 2.8%
  • 5.1%

Part B)
%Error = ((141.3-122)/141.3)*100 = 13.66%
Other percent errors:
  • 4.5%
  • 11.5%
  • 10.8%
  • 12.2%

%Difference = ((126.2-135)/126.2) * 100 = 6.97%
Other percent differences:
  • 3.33%
  • .95%
  • 1.58%
  • 1.74%


Discussion questions
  1. Where would the cars meet if their speeds were exactly equal?
    1. In Part A, the two cars would meet in exactly the middle. In Part B, the two cars would never meet, and would maintain the same difference between them throughout.
  2. Sketch position-time graphs to represent the catching up and crashing situations. Show the point where they are at the same time.
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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?
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No, there is not any way to find the points when they are at the same place at the same time.

Conclusion
There were two parts of this lab, a crash test, and a passing test. For both parts, my partner and I ran 5 trials. We did this to get the most accurate results possible. Using the v= d/t formula, my partner and I figured that for part A, the two cars would crash at 464.38 cm from the blue car. For part B, we calculated, with the same formula, that the slower car (the yellow one) would pass the faster car (the blue one) at 141.3 m. The five trials we ran, with the two cars crashing, yielded these results: 435 cm, 438 cm, 430 cm, 432 cm, 431 cm. The five trials for the catch-up part of the lab yielded these results: 122 cm, 135 cm,125 cm, 126 cm, 124 cm.

For both parts of the lab, our results were close to our theoretical results, with less than 10% error on most runs. Our lowest percent error was 4.5%, and our highest was 12.2%. Our percent difference was between .95% and 11%, so we could probably work on getting more precise results.

However, I feel like this is pretty good, considering all the things that could have gone wrong. A possible source of error was the cars. Perhaps we recorded the velocity of the cars wrong, and got erroneous calculations as a result of that. This could be easily remedied by measuring the velocity of the cars with a spark measure before starting the lab. Also, the batteries might have been running low in our CMVs, making them go slower and slower, instead of at constant speed. This could have been fixed by checking the batteries before starting the lab, and making sure they still had power. Furthermore, since the cars are moving, it was sometimes difficult for us to see where exactly the cars crashed or passed. This could have contributed to our large percent error and percent difference. To fix this, we should have videotaped the cars and the tape measure, so we could look back and see where the cars met and passed. All things considered, this lab was a success, especially when considering all the possible sources of error.

Egg Drop Project

Partner: Hella Talas

Project:
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Description:
Our project was made of about twenty bendy straws, a sheet of printer paper, sewing string, and a rubber band. The straws are glued together in what seems like a haphazard fashion. In reality, the straws stick out in every direction because we could not predict which side the structure would land on, and we wanted the egg to be protected on every side. Inside the mass of straws, there is a rectangular pocket, big enough to hold an average egg. This is, of course, where we put our egg. The rubber band is there so the egg doesn't fall out, mid drop. The straws sticking out of the bottom cushion the landing for the egg, as they absorb some of the impact. The paper parachute also created drag, and helped slow our project down so there would be less of an impact on the contraption.

Mass:
81.97 g (with egg)
24.85 g (without egg)

Results:
The egg survived the drop fully intact, with no breakage or leakage of any kind.

Analysis:
Acceleration:
distance =8.5 m
average time of fall =2.45 s

d= vt + .5at^2
8.5 = .5a (2.45)^2
a= 2.83 m/s^2

Comparison to 9.8 m/s^2
d= vt + .5at^2
8.5= .5(9.8)(t^2)
8.5=4.9 t^2
t= 1.32 s

My project took longer to drop than an object that is only affected by gravity (as opposed to my project, which was affected by wind, air resistance, etc.)

Sources of error/ what I would change:
A possible source of error was the timing. It was hard to get an exact time for when the project landed, because my reaction time could have gotten in the way. To get a more accurate time, I should have taken times from 5 or more people who also timed my drop, instead of the three people I did take times from. Since the times varied by around 1 second, more data would definitely be helpful.
I am pleased with the final project. It was effective, protected the egg, and was lightweight. If I were to redo this project, I would make my design lighter by making my project more minimalist. I would use less straws, and make they all shorter. I would also use a more light-weight substance to hold the straws together, as hot glue is pretty heavy. Since my egg did not break or fracture, I do not need to provide better protection for the egg.


Lesson 5: Free Fall and the Acceleration of Gravity


Introduction to Free Fall
Topic Sentence: Free-falling objects are influenced only by gravity, have an acceleration of -9.8 m/s/s, and speed up (velocity increases).

A free falling object is an object that is falling under the sole influence of gravity. 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.
The object travels downward and speeds up. Therefor, its acceleration is downward.

The acceleration of gravity
Topic sentence: The acceleration of gravity is 9.8 m/s/s downward, although there can be slight variations due to altitude.
9.8 m/s/s, downward (on Earth) is known as the acceleration of gravity. Physicists have a special symbol to denote it - the symbol g. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude.
To accelerate at 9.8 m/s/s means to change the velocity by 9.8 m/s each second.
external image U1L5b1.gif
Representing Free Fall by Graphs
Topic sentence: Free falling objects have a curved x-t graph that starts slow and finished with a large velocty and a v-t graph that starts from rest and has a slope of -9.8 m/s/s.

A position versus time graph for a free-falling object is shown below.
external image U1L5c1.gif

The curved line on this position versus time graph signifies an accelerated motion. The position-time graph reveals that the object starts with a small velocity (slow) and finishes with a large velocity (fast). The slope of any position vs. time graph is the velocity of the object. The small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity. Finally, the negative slope of the line indicates a negative, downward, velocity.

A velocity versus time graph for a free-falling object is shown below.
external image U1L5c2.gif
The line on the graph is a straight, diagonal line. This signifies an accelerated motion. 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. This means the object has a negative acceleration. The slope of any velocity versus time graph is the acceleration of the object, so there is a constant, negative acceleration.

How Fast? and How Far?
Topic Sentence:The formula used to determine the velocity of an object at any time after the object is dropped is Vfinal = g x t
The velocity of a free-falling object is changing by 9.8 m/s every second. If dropped from a position of rest, the object will be traveling 9.8 m/s (approximately 10 m/s) at the end of the first second, 19.6 m/s (approximately 20 m/s) at the end of the second second, etc. The formula for determining the velocity of a falling object after a time of t seconds is

vf = g * t
The value for g on Earth is 9.8 m/s/s (gravity). The above equation can be used to calculate the velocity of the object after any given amount of time when dropped from rest.

The Big Misconception
Topic sentence: All objects in free fall have the same acceleration, regardless of mass.

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. A more massive objects do not accelerate at a greater rate than a less massive object in free fall because 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. 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.


Free Fall Class Notes:
  • When an object is dropped, initial velocity is 0
  • Only gravity acts as a force
  • Vfinal = gravity x time
  • Gravity=9.8 m/s/s
  • When an object is thrown, v=0 at max height.
  • Acceleration is never 0 because there is always gravity.
  • v-t graph for free fall objects
  • Screen_shot_2011-10-13_at_4.44.23_PM.png
  • Acceleration always points down
  • x-t graph for item in free fall
  • Screen_shot_2011-10-13_at_4.48.18_PM.png
  • Object will take same to go down as up if displacement is 0
  • Free fall speeds are identical for all objects
  • g can change depending on environment
  • d= .5gt^2
  • Velocity and acceleration are negative

Problem with thrown ball:

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  • It takes the same amount of time to go from the person's hand to the max height as it does from the max height to the top of the cliff. So, T3 = (.82 x 2) = 1.64 s

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Level 2 Problems:
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1st part: ball falls down the well and hits botton
2nd part: sound travels up

Ball (free fall)
initial velocity = 0
accelerating
a= -9.8 m/s/s
t = 2.5 - time of sound
-d = vit + .5at^2
-d = .5 (9.8)(2.5 - time of sound)^2 = -340 x time of sound

2.5 = time of sound + time of ball

Sound (constant speed)
v= d/t
340 = d/ time of sound
340 (time of sound) = d

d = d
.5 (9.8)(2.5 - time of sound)^2 = -340 x time of sound = -340 (time of sound)
t = .089s
d=20.26 m

The well is 30.26 m deep.

Second question:
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Part 2 of second question:

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Free Fall Lab: 10/4

Partner: Nicole Tomasofsky
Purpose/objective:
What is acceleration due to gravity? What will the v-t graph look like for an object in free fall? How will you find "g" from this graph?

Hypothesis:
The acceleration due to gravity is 9.8 m/s/s downward. The velocity time graph will be a straight diagonal line that starts at zero and goes downward with a slope of -9.8. Since the slope of a v-t graph is the acceleration, and the acceleration of gravity is "g," the slope of the v-t graph should be equal to -9.8.

Procedure:
Take a ticker tape and run a length of thread through it. Attach a weight to the tape and drop it off the second story balcony. Measure the velocity and acceleration by counting the ticker marks.

Materials:
  • Spark timer
  • Spark tape
  • Masking tape
  • Mass (weight)
  • Tape measure

Position Time Table and Graph:


Screen_shot_2011-10-08_at_9.54.52_PM.png

Analysis for Position Time Graph:
Our graph is polynomial instead of linear. This is because the mass accelerated, or gained speed as it fell. All increasing speed x-t graphs are polynomial. Our x^2 value is 357.48. The theoretical x^2 value is around 500 or 600, so our x^2 value is off by quite a bit. Our r^2 value is .99998. This means that our graph is .002% from being perfect. This is because a perfect graph has an r^2 value of 1. The r^2 value shows how close the trendline matches up with the plotted points. So, our results are pretty accurate. Also, if the equation of the line was put in a y=Ax^2 + Bx + C format, 2 times A would equal the slope of the v-t graph, or the acceleration. Our A value times 2 is 714.96, which is pretty close to the slope of our v-t graph, which was 708.97.


Velocity Time Graph and Table

Screen_shot_2011-10-08_at_9.53.31_PM.png
Analysis for Velocity Time Graph:
The slope of the v-t graphs shows acceleration. Since the slope of this graph is 708.97, the acceleration of our mass is 708.97 cm/s/s. Since our object was technically supposed to be in free fall, and objects in free fall are all supposed to have the same acceleration of 981 cm/s/s, our acceleration was slower than the expected acceleration. The y intercept is not zero because the initial velocity is not zero. The initial velocity is not zero because the object did not start at rest. The r^2 value of our graph is .99507. This means that the trendline fits our plotted dots 99.507% of the time, which shows that our results are extremely accurate.

Class Data:

Screen_shot_2011-10-08_at_10.09.32_PM.png

Sample Calculations:

Mid time = (final time + initial time) / 2
Example:
mid time = (0 s + .1 s) / 2
mid time = .05 shttp://hpwangste.wikispaces.com/Ch2_WangS

Velocity = change in position/ change in time
Example:
Velocity = (31.42 cm - 0.0 cm) / (.1 s - 0.0 s)
Velocity =314.2 cm/s

Percent error = (theoretical - experimental) / theoretical x 100
Percent error = (981 - 708.97) / 981 x 100
Percent error = 27.7 %

Percent difference = (average experimental value - individual experimental value) / average experimental value x 100
Percent difference = (834.03 - 708.97) / 834.03 x 100
Percent difference = 14.99 %

Discussion Questions:
  1. Does the shape of your v-t graph agree with the expected graph? Why or why not?
No, the shape of our v-t graph was different from the one we predicted. Although we were right in that the v-t graph is a straight linear line, we did not account for the negative displacement or the negative acceleration in our graph. The displacement is negative because the final position of the mass was lower than the initial position (the floor is lower than the balcony), and a small positive number minus a bigger positive number always equals a negative number. This, divided by the elapsed time, which is positive, would result in a negative number (this is the formula for velocity). So, if we took in consideration the negative displacement and negative acceleration, our v-t graph would have a negative slope, but otherwise look the same.

  1. Does the shape of your x-t graph agree with the expected graph? Why or why not?
The shape of our x-t graph agrees with the predicted graph. All qualitative increasing speed x-t graphs look similar. They are all curved lines, with positive slopes that increase over time. This is the graph that we predicted, and this is the graph of our results. The x-t graph looks like this because the speed is increasing.

  1. How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.)
Our percent difference was 14.99 %. This means that our results were 14.99% percent different from the average result of the class. Our acceleration was 708.97 cm/s/s. The average class acceleration was 834.03 cm/s/s. The class average was closer to the expected acceleration, 981 cm/s/s than our result. This means that our classmate's results, as a whole, were more accurate than ours.

  1. Did the object accelerate uniformly? How do you know?
The object does accelerate uniformly. You can tell because the v-t graph is a linear line. Since the acceleration is the slope, and the slope of linear lines is constant throughout, the acceleration is the same for the entire graph.

  1. What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?
If the mass was thrown down over the balcony, it would have a higher acceleration than gravity. But in most cases, the acceleration of the mass should be lower than that of gravity because the spark timer caused friction, which slowed the mass down.

Conclusion
Our hypothesis was partially correct. My partner and I predicted that the acceleration due to gravity would be 9.8 cm/s/s downward. This is a fact that we already knew from class and from homework readings. However, according to our lab, acceleration due to gravity would actually be 7.09 cm/s/s. But, this is wrong and can be explained by experimental errors. We also hypothesized that the v-t graph would look like a linear line with a slope of -9.8. Since we did not account for the negative displacement in our experiment, used cm instead of m, and had experimental errors, the slope of our v-t graph turned out to be 708.97. We were correct in stating that "g" could be found from the v-t graph's slope, though, as the slope of the v-t graph equals acceleration.
There were many experimental errors that could have occurred that altered our results. First, our object was not really in free fall, since the spark timer created friction. So, there was no way that the acceleration of our mass could have been -9.8 m/s/s. Also, one of us could have been standing on the tape or holding it back as it fed through the spark timer, causing errors in our results. Another error that could have occurred was if we used a defective spark timer, or set it to 60 Hz accidentally. Furthermore, we could have mistaken some of our measurements of the dots. To address all of these issues in a future experiment, a spark timer should not be used. Instead, the object should be dropped next to an extended tape measure, with a camera taking a picture of it every 10 seconds. This ensures that the object is truly in free fall, with no friction to change the results. This also means that it is impossible to get wrong results because of broken spark timers.
However, I feel that our results were not too inaccurate, since our percent error was only 27.7 %, so I feel that there is no need to redo this lab.