CH2_braunsteinc

= = toc Constant Speed =Lab: a crash course in velocity= Honors Physics
 * hypothesis:** the cmv will move 20 feet in 15 seconds, or 40.64 cm per second


 * Objective: ** What is the speed of a Constant Motion Vehicle (CMV)?

Constant Motion Vehicle, Tape measure and/or metersticks, spark timer and spark tape
 * Available Materials ** :


 * data table:**


 * analysis**: the cmv moved at a steady pace of 37.962 cm per second.

The slope is the change in x over the change in y, which in terms of our graph was the change in position over the change in time. Change in position over change in time is the formula used to calculate average velocity. Average velocity is the average of all the instantaneous velocities, and the instantaneous velocity is the velocity of an object at a given time. We are assuming that the CMV was going the same velocity at all times. The y intercept is set to zero because the CMV started at one position and moved from there. No time had passed when it was at the starting position so its velocity was zero. The R squared value is the accuracy of our points to the trendline, it shows us how well we measured the centimeters on the spark tape. The other graph’s trendline would be lower and less steep than ours because it moved less centimeters in the same time that ours moved more centimeters, therefore its velocity would be slower. The velocity oh the CMV is the slope, so it’s line wo uld be less steep than ours.
 * Discussion questions **
 * 1) Why is the slope of the position-time graph equivalent to average velocity?
 * 1) Why is it average velocity and not instantaneous velocity? What assumptions are we making?
 * 1) Why was it okay to set the y-intercept equal to zero?
 * 1) What is the meaning of the R2 value?
 * 1) If you were to add the graph of another CMV that moved more slowly on the same axes as your current graph, how would you expect it to lie relative to yours?


 * Conclusion **

Our hypothesis was that the car moved 20 feet in 15 seconds, which is 40.64 cm per second. Our results showed that the CMV moves 37.962 cm per second. Our hypothesis was fairly accurate because it was only 2.678 cm per second off, which is not a big distance. We had to round to the nearest hundredth place, which was difficult to round to due to the miniscule size of a hundredth of a centimeter. Also, the meter stick could have shifted during measuring, the measurements on the spark tape could have been taken when they were not at even intervals (meaning the CMV was not at constant speed yet). Additionally, we used a meterstick which is about half a centimeter off the surface we placed it at, so we had a different perspective when viewing the dots than we would have if the measuring tool had been flush to the surface. If we had to redo this lab, to minimize issues we could use more accurate measuring tools. We could also use something like a measuring tape that lies flat on a surface and can be taped down to prevent shifting while we are measuring. We could also measure the intervals between dots on the spark tape to assure that they are equidistant so that we know the CMV was at constant motion when we used the data.

=homework september 8, 2011:=

I read about distance and displacement which I understood well from class discussion on wednesday. Distance is the total amount traveled regardless of direction. Whereas displacement is the amount traveled with regards to direction, its like net amount traveled. One can walk in a 10 foot circle, and the distance they traveled is 10 feet, however their displacement is 0. I was a little confused about speed and velocity. It was confusing because I didn't understand why they were different because I had never been taught that they were different from each other, that they weren't interchangeable. Now I know that speed is a scalar quantity, but velocity is a vector quantity. Speed is the rate that something covers a //distance// in. Velocity is the rate that something changes its //position,// it measures the rate of //displacement//. I understand everything from the reading, its all pretty straightforward and clear. We didn't really discuss the terms //scalar// and //vector.// The reading gave a clear definition for each, scalar quantities are numerical and vector quantities are numerical and directional as well. For example, 20 feet is scalar, but 20 feet north is vector because a direction is included. In class we also didn't talk about instantaneous speed, which is the speed you're going at any given moment, whereas average speed is the average that you're going the whole time.
 * **What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.**
 * **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.**
 * **What (specifically) did you read that you still don’t understand? Please word these in the form of a question.**
 * **What (specifically) did you read that was not gone over during class today?**

=september 9th notes:=

average speed= total distance/ total time --> can also be found if you know two different speeds, just average them together

constant speed: not changing speed, the instantaneous speed is always the same for the time interval

equation for constant, instantaneous, average speed is always the same --> v= change in distance/ change in time

- at rest - constant speed - increasing speed - decreasing speed (acceleration is whenever you're changing speed) (do not use decelerate)
 * types of motion:**

at rest- v=0, a=0.
 * motion diagrams**: show direction of velocity

constant speed- v ......v ......v ->->> a=0

increasing speed- + v....... v............ v --->>> > +a decreasing speed- + v................. v....... v --->->---> <--- - a if moving in a different direction, change direction of the arrows (if going left, draw them to the right, if left, draw to left)
 * used to show direction*

[ x..............] at rest
 * ticker tape diagrams:**

[- - - - - - - ] constant speed

[x..x.....x..........x] increasing speed

[x.........x.......x....x..x]decreasing speed used when you need to take measurements
 * drawback: tick tapes don't show direction*

signs are arbitrary

=homework: september 9, 2011=

i understood ticker tape and vector diagrams from class. ticker tape diagrams are when you put a piece of ticker tape through a spark timer, on one side you attach the tape to an object which pulls the tape through the timer. the timer places a dot on the paper at even time intervals, either 10 times a second or 60 times a second. you then use the dots on the tape to figure out how far apart they are which you then use to figure out the speed of the object. vector diagrams are used to show motion in objects and direction. they use arrows to show the velocity of an object during motion. equal sized arrows that an object was at constant motion. increasing sized arrows show that the object was accelerating. they also show the direction of an object. everything was clear in class everything is clear the reading was pretty much going over what we had learned in class, so there wasn't anything new on the reading.
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 1) What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.
 * 1) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 1) What (specifically) did you read that was not gone over during class today?

=class activity 9-12-11= at constant fast motion:



at constant slow motion:

at rest: discussion questions:
 * 1) How can you tell that there is no motion on a…
 * 2) position vs. time graph- the position is always the same, theres a horizontal line
 * 3) velocity vs. time graph- there will be a horizontal line along the x-axis b/c the x-axis means your velocity is zero
 * 4) acceleration vs. time graph- horizontal line along the x axis


 * 1) How can you tell that your motion is steady on a…
 * 2) position vs. time graph- you have a straight line
 * 3) velocity vs. time graph- the line will be straight and horizontal (line isn’t on the x axis)
 * 4) acceleration vs. time graph- horizontal line along the x axis b/c constant speed has no acceleration


 * 1) How can you tell that your motion is fast vs. slow on a…
 * 2) position vs. time graph- slower is a less steep slope, faster is a steeper slope b/c you’re covering more distance quicker
 * 3) velocity vs. time graph- the higher the number the line is on, the faster its going.
 * 4) acceleration vs. time graph


 * 1) How can you tell that you changed direction on a…
 * 2) position vs. time graph- the line will begin to go down, the slope will be negative
 * 3) velocity vs. time graph- the number the line is on will be negative if you begin moving toward as opposed to away
 * 4) acceleration vs. time graph


 * 1) What are the advantages of representing motion using a…
 * 2) position vs. time graph- slope of a position-time graph is velocity
 * 3) velocity vs. time graph- tell you most information. You can tell acceleration from this graph. Slope of this graph is acceleration. Displacement is the area between the graph and the x-axis
 * 4) acceleration vs. time graph- change in velocity= area under graph.


 * 1) What are the disadvantages of representing motion using a…
 * 2) position vs. time graph- doesn't tell you an exact direction other than towards or away
 * 3) velocity vs. time graph- doesn't tell an exact direction
 * 4) acceleration vs. time graph- not enough information to tell you much of anything


 * 1) Define the following:
 * 2) No motion- not moving
 * 3) Constant speed- moving at the same speed consistently with no change

=kinematics= acceleration- rate that velocity in changing (a) m/s squared

v= change distance/ change in time only for average or constant speeds

v= v1+v2/ 2 ---only for average speed

a= final velocity- initial velocity/ change in time

or

final velocity= initial velocity +acceleration x time

1/2 (initial velocity + final velocity) = change in distance/ change in time --- (two velocity equations from above combined)

change in distance= 1/2 (initial velocity+ final velocity) x time

change in distance initial velocity x time + 1/2(acceleration x time squared)-- substitution

final velocity squared = initial velocity squared + 2(a x change in distance)--- substitution

=**homework 9-13-11:**= I understood what acceleration is. Acceleration is the rate an object changes its velocity, a vector quantity. Also, constant acceleration is when the rate at which an object changes its velocity is constant. For example, a car increases 2 m/p/s for 30 seconds, it is constant acceleration because the interval is always equal. When an object is speeding up, the acceleration is positive. When its slowing down, it has negative acceleration. everything was clear in class I didn't understand when it said "According to our RULE OF THUMB, when an object is slowing down, the acceleration is in the apposite direction as the velocity. Thus, this object has a negative acceleration . In Example D, the object is moving in the //negative// direction (i.e., has a //negative// velocity) and is speeding up. When an object is speeding up, the acceleration is in the same direction as the velocity. Thus, this object also has a negative acceleration ." I dont understand why the last one is negative acceleration. the last part, about negative acceleration wasn't gone over completely in class. If an object is moving in the negative direction but is speeding up, it has a negative acceleration-- was not gone over in class.
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 1) What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.
 * 1) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 1) What (specifically) did you read that was not gone over during class today?

=Graph Shape Worksheet:= starts going away quickly, then is at rest, then comes back slowly

=Acceleration Lab 9-14-11:=
 * Objectives:**
 * What does a position-time graph for increasing speeds look like?
 * What information can be found from the graph?


 * Hypothesis:** A position-time (x-t) graph for increasing speed looks like a curve upwards, starting less steep and becoming steeper as time goes on. This is because it is covering a greater distance in less time. From a position-time graph, we can tell the velocity, because we can figure out the slope of the line.


 * Procedure:**

materials: ramp, car, spark tape, spark timer, textbook For acceleration: We placed the ramp on a textbook to make an incline, fed a piece of spark tape through the spark timer and attached it to the car. We sent the car down the ramp and then measured the distance between the dots on the spark tape. After recording the distances between the dots we graphed them on excel. Spark tape dot distances being measured:

For uphill: We placed the car at the bottom of the ramp, so that it would have to go up hill. The car had spark tape attached to it, we then pushed it up the ramp and measured the distance between the dots, which we graphed on excel. Spark tape distances being measured again:

data table for down hill acceleration-
 * data and graphs:**

data table for up hill acceleration- graph-

a) Interpret the equation of the line (slope, y-intercept) and the R2 value. y=Ax 2 ﻿+ Bx y=distance & x= time change in distance= 1/2at 2 ﻿+(v ﻿i ﻿)t slope= 1/2 acceleration accelerating down hill= 15.54t 2 ﻿+11.795t R 2 ﻿=.99996 accelerating up hill= -25.32t 2 ﻿+90.698t R 2=.9999
 * Analysis:**

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.) instantaneous speed at halfway point= 45.7 cm/s (.4,0) (1, 27.44) (27.44-0)/(1-.4)=45.7

c) Find the average speed for the entire trip. average speed= total change/ total time accelerating up hill= (71.34-1.42)/ (1.8-.1) = 38.84

1. What would your graph look like if the incline had been steeper? It would be a steeper curve, because the points would travel a greater distance in a shorter amount of time 2. What would your graph look like if the cart had been decreasing up the incline? The cart, as shown above in the photo, would start out going fast, with a steeper slope. It would then curve the opposite way, and gradually have a less steep slope. 3. Compare the instantaneous speed at the halfway point with the average speed of the entire trip. instantaneous speed= 45.7 average speed=38.84
 * Discussion Questions:**

4. Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense? the tangent line only touches the one point whose speed we want to know. the tangent line is "parallel" to the point, so their slopes are the same. slope is the average velocity or speed in this case, therefore the slope of the tangent line is the instantaneous speed 5. Draw a v-t graph of the motion of the cart. Be as quantitative as possible.

Our results showed that on a position-time graph, increasing speed looks like a curve that gradually becomes steeper, indicating the speed increasing. Our hypothesis was accurate because we thought that the graph would curve, as opposed to being a straight line as is the case with constant speed graphs. By looking at the graph of our lab, it can be easily seen that it curves upward, slightly steeper at the end than at the beginning. Some sources of error may have been that we needed to estimate to the hundredths place of a centimeter, if we had had something more exact in measuring, there would be less room for error. The ticker tape may have shifted while we were measuring. Also, we may have begun measuring the distance between the dots too early or late. If we redid this lab, we could have used a more exact measuring device. Also, the ticker tape could have been more firmly, safely attached to the measuring device on the ramp.
 * conclusion**:

=homework 9-15-11:= lesson 3-
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) i understood fully from the class the concept of what constant motion looks like on a position time graph. Because the motion is constant, the graph will be a straight like that would always have the same slope. I also understood the concept of determining direction from a position-time graph. If it has a positive slope, its going away from the origin( or motion sensor like in class). If it has a negative slope, its going towards the origin (or motion sensor like 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 understood everything from class.
 * 5) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 6) everything is clear.
 * 7) What (specifically) did you read that was not gone over during class today?
 * 8) we went over everything in class

lesson 4:
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) The slope of a velocity-time graph is the acceleration. If the slope is zero, theres no acceleration. Also, the slope of a v-t graph is different from that of a position-time graph. on a p-t graph, if there was constant speed, the slope was a straight line with a slope. If theres constant motion on a v-t graph, theres a horizontal line with no slope.
 * 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 kind of confused about when the line was drawn in the negative y value region of the graph. now i know that it means that the object was going in the opposite direction, it was going in the negative direction
 * 5) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 6) i don't really understand the last section, with the leftward velocity and rightward acceleration
 * 7) What (specifically) did you read that was not gone over during class today?
 * 8) everything was gone over in class

=Lab: A Crash Course In Velocity= Date: 9-21-11 Lab Partners: Dani Rubenstein, Ali Cantor, Julia Sellman

objective: Both algebraically and graphically, solve the following 2 problems. Then set up each situation and run trials to confirm your calculations. the yellow car will travel 229.214 cm and the blue car will travel -372.321 centimeters by the time they crash into each other.

the blue car will travel 260.156 cm by the time it reaches the yellow car

**procedure:**

media type="file" key="lab cmv 2.mov" width="300" height="300" the two cars are placed facing each other, 6 meters apart. we marked the place where they collided.
 * car crashing:**

media type="file" key="new - Medium.m4v" width="330" height="330" we placed both cars facing the same direction, with the faster car one meter behind the slower one. we than observed to see when the faster car would meet up with the slower car.
 * cars catching up:**

**data:** crashing: catching up:

**accuracy of data:** **crashing** **percent error:** using the percent error equation, we determined how accurate our results were compared with the theoretical results we calculated above. our results were extremely accurate, they were only .57% off from the theoretical results. **catching up** **percent error:** here we used percent error again to find out how accurate our second results were. they too were extremely accurate, within .03% of the theoretical results **precision of data:** percent difference of blue: at the top of the page, the equation for percent difference is shown, it is the absolute value of the average of the experimental values minus the individual experimental value. then it is divided by the average of the experimental values, then multiplied by 100. for the blue car the percent differences for our five crashing results were: 1. .31% 2. .33% 3. .001% 4. .23% 5. .21% all our percent differences were extremely small showing that we used great precision when taking our measurements as they were all very close to the average. percent difference of the yellow car: 1. .51% 2. .55% 3. .004% 4. .38% 5. .34% again, our data for the yellow car crashing was very precise and all numbers were very close to the average. percent difference of catching up: percent difference: 1. 11.6% 2. 11.2% 3. 1.9% 4. .2% 5. 1.38% for this set of data, our percent differences were in a wider range than with the crashing cars part. however, our highest difference of 11.6% is still fairly close to the average of our data. ** Discussion questions ** crashing:
 * 1) Where would the cars meet if their speeds were exactly equal?
 * 2) If the speeds were exactly equal, the cars would meet at 300 cm.
 * 3) 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.

catching up: on these graphs, the cars at the same place at the same time when they intersect.


 * 1) Sketch velocity-time graphs to represent the catching up situation. Is there any way to find the points when they are at the same place at the same time? [[image:Screen_shot_2011-09-21_at_9.42.36_AM.png]]
 * 2) this is when the blue car catches up to the yellow car
 * 3) there is no way to find the points when they are at the same place at the same time

Our results were very accurate. Both were within less than 1% of the theoretical results, which is exceptionally accurate. For the first part of this lab, we calculated that the blue and yellow cars would crash once the blue car had gone 372.321 centimeters, for the second part we calculated that the blue car would catch up to the yellow after traveling 260.156 centimeters. The blue car kept veering off from the straight line we needed it go in, so we created a path for the cars using ramps. The use of ramps allowed our cars to go in straight lines, thus giving us better results. Our measurements for the crashing part of the lab were very precise. For both the blue and yellow car, the percent difference was less than 1%, showing that all our measurements were all around the same range of numbers. For the catching up part of the lab, our percent differences were higher because it was more difficult to measure where the cars were passing each other. The first source of error that we encountered was while doing calculations for part A, the distances added up to 601 instead of 600 because of the sig figs we used. This could have been avoided had we used more sig figs while doing our calculations. Another source of error could have been the estimating of where the cars collided/ caught up to each other because we were using humans to mark this there could have been delays in reaction time and being farther from the tape measure. We could have used some sort of camera or computer device to more accurately see where the cars caught up to each other. The batteries in the cars may have also been lower in energy than the first time we used them, causing the speed to be different than when we originally calculated it. This would have caused our percent errors to be higher. Additionally, the cars may have been started at slightly different times due to human reaction time, this would have affected how far each traveled seeing as one car moved faster than the other. If I were to change the lab to lessen the chances of error, I would use some sort of computer device to measure where the cars crash and pass each other. If I could use something like this for this lab, the measurements would be significantly more precise and accurate. Also, I would use a machine to start the cars so that they both are started at the same time, I would also assure that the cars traveled in straight paths and had fresh batteries. =
 * conclusion:**

= interpreting position-time graphs: d.) 1. speeding up 2. 200 m/s 3. 390 m/s 4. 240 cm/s

e.) 1. decreasing towards, increasing away 2. 31 m/s 3. instantaneous velocity= 0 4. =60 m/s 5. average velocity=0

f.) 1. 24 k/h 2. no speed 3. 12 km/h 4. 10.8 km/h 5. 0

g.) 1. 6 m/s 2. 0 3. 4 m/s 4. 2.7 m/s 5. 4 m/s 6. 3.6 m/s 7. 0

=

= =homework 10-3-11= There are 4 steps to follow, remembered easily using the terms T-RG-TS (Pronounced: TARGETS). Start by copying and pasting the entire page into an editable document.


 * 1) **//T//**//rivia//: Draw a line through or delete anything that seems trivial or frivolous (adjectives, similar examples, transition words)
 * 2) **//R//**//edundancies//: Draw a line through or delete any repetitive information or examples.
 * 3) **G**//eneralize//: Replace lists of specific items with general terms and phrases. (Ex: if text lists mirrors, lenses, prisms and thin films, you can substitute "optical materials" for all of those.)
 * 4) **//T//**//opic// **//S//**//entence:// Write a good topic sentence for the material (the subject and the author’s claim about it)

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: 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. if an object travels downward and speeds up, then its acceleration is downward. The Acceleration of Gravity A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth). It is known as the acceleration of gravity  - the symbol g . The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s/s. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude. ( ~ 10 m/s/s, downward)
 * Free-falling objects do not encounter air resistance.
 * All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s (often approximated as 10 m/s/s forback-of-the-envelope calculations)

<span style="background-color: #ffffff; color: #000000; font-family: Verdana; font-size: 9pt; text-decoration: none; vertical-align: baseline;">Observe that the line on the graph curves. 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 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. <span style="background-color: #ffffff; color: #000000; font-family: Verdana; font-size: 9pt; text-decoration: none; vertical-align: baseline;"> A velocity versus time graph for a free-falling object is shown below <span style="background-color: #ffffff; color: #000000; font-family: Verdana; font-size: 9pt; text-decoration: none; vertical-align: baseline;">. <span style="background-color: #ffffff; color: #000000; font-family: Verdana; font-size: 9pt; text-decoration: none; vertical-align: baseline;">Observe that the line on the graph is a straight, diagonal line. 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. Since the slope of any velocity versus time graph is the acceleration of the object the constant, negative slope indicates a constant, negative acceleration. <span style="background-color: #ffffff; color: #0099cc; font-family: Verdana; font-size: 11pt; text-decoration: none; vertical-align: baseline;">How Fast? and How Far? <span style="background-color: #ffffff; color: #000000; font-family: Verdana; font-size: 9pt; text-decoration: none; vertical-align: baseline;">Free-falling objects are in a state of <span style="background-color: #ffffff; color: #980000; font-family: Verdana; font-size: 9pt; vertical-align: baseline;">[|__acceleration__] <span style="background-color: #ffffff; color: #000000; font-family: Verdana; font-size: 9pt; text-decoration: none; vertical-align: baseline;">. Specifically, they are accelerating at a rate of 9.8 m/s/s. This is to say that the velocity of a free-falling object is changing by 9.8 m/s every second. Thus, the velocity of a free-falling object that has been dropped from a position of rest is dependent upon the time that it has fallen. The formula for determining the velocity of a falling object after a time of <span style="background-color: #ffffff; color: #ff0000; font-family: Verdana; font-size: 9pt; text-decoration: none; vertical-align: baseline;">t <span style="background-color: #ffffff; color: #000000; font-family: Verdana; font-size: 9pt; text-decoration: none; vertical-align: baseline;"> seconds is <span style="background-color: #ffffff; color: #ff0000; display: block; font-family: Verdana; font-size: 11pt; text-align: center; text-decoration: none; vertical-align: baseline;">vf = g * t <span style="background-color: #ffffff; color: #000000; font-family: Verdana; font-size: 9pt; text-decoration: none; vertical-align: baseline;">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 <span style="background-color: #ffffff; color: #ff0000; font-family: Verdana; font-size: 9pt; text-decoration: none; vertical-align: baseline;">t <span style="background-color: #ffffff; color: #000000; font-family: Verdana; font-size: 9pt; text-decoration: none; vertical-align: baseline;"> seconds is given by the formula. <span style="background-color: #ffffff; color: #ff0000; display: block; font-family: Verdana; font-size: 11pt; text-align: center; text-decoration: none; vertical-align: baseline;">d = 0.5 * g * t2 <span style="background-color: #ffffff; color: #0099cc; font-family: Verdana; font-size: 11pt; text-decoration: none; vertical-align: baseline;">The Big Misconception <span style="background-color: #ffffff; color: #000000; font-family: Verdana; font-size: 9pt; text-decoration: none; vertical-align: baseline;"> it was stated that the acceleration of a free-falling object (on earth) is 9.8 m/s/s. This value (known as the acceleration of gravity) is the same for all free-falling objects regardless of how long they have been falling, or whether they were initially dropped from rest or thrown up into the air. Free-fall is the motion of objects that move under the sole influence of gravity; free-falling objects do not encounter air resistance. More massive objects will only fall faster if there is an appreciable amount of air resistance present


 * topic sentence:** free falling objects accelerate at 9.8 m/s/s regardless of mass as long as there is no air resistance; the final velocity and distance are each dependent on time spent free falling.

=egg drop lab:= partner: Kaila Solomon

Sadly, our egg drop was unsuccessful, the egg cracked and was leaking. The egg fell out of the carrier before hitting the ground so there was nothing to take the impact of the fall.
 * description of results:**

Prior to the final drop, we tested our prototype several times, each time it did work. However, in order to make it lighter, we had the remove several sheets of paper from the parachute. The cover we used was also thinner to decrease weight, which led to the egg not being held in as well. The parachute also did not work as well because it was the same size as the box so there wasn't much air for it to resist.
 * analysis of why it did not work:**

we could have used a smaller, deeper box to put the egg in. We also could have made sure the parachute was larger than the object it was over. we could have made sure the cover to the box was more secured so that our egg would not fly out because we had paper in the box to cushion the egg upon impact. also, we could have made our egg carrier into the shape of a cone, which would have allowed the egg adequate space to slow down as opposed to stopping at one instant.
 * what we could do differently:**

first: we made a large box out of tinfoil and lined it with crumpled newspaper to break the eggs fall. then we made a smaller box (hidden by the cover) that had more newspaper in it to also cushion the eggs fall. then we placed a cover on the top to prevent the egg from falling out. we also had a multi-layer parachute to slow the fall.
 * prototypes:**

second: we used tin foil to make a box to put the egg in, then used several sheets of paper to make a parachute. we filled the box with shredded newspaper to break/cushion the fall of the egg

these are the calculations for the acceleration of our object: we calculated that our acceleration was 4.10 m/s/s. this is not the acceleration of gravity, 9.8 m/s/s/, because we used a parachute and our object had air resistance upon it.
 * calculations:**

=Freefalling Object Lab:= Objective: **What is the acceleration of a falling body?** **Hypothesis:** the slope of the line will be 981 because that’s the acceleration of a free falling object and the slope of a v-t graph is the acceleration.

The acceleration of a falling body is 9.8 m/s/s or 981 cm/s/s.


 * procedure:**

we took ticker tape, a spark timer, a weight, and masking tape and taped the weight to the ticker tape. then we dropped the weight off of the balcony in school while using the spark timer. this simulated free fall. then we took the ticker tape and taped it to the ground next to a tape measure and measured the distance between all the dots.
 * __ Analysis should include: __**
 * x-t and v-t graphs
 * discussion and interpretation of the equations of the graphs
 * % error and % difference
 * Sample calculations of velocity, % error, and % difference

for the v-t graph, the slope of the graph is the acceleration, so the acceleration of our object was 891.12 cm/s/s. this was not what we had expected because as we know, acceleration due to gravity in 981 cm/s/s, but ours was about 90 cm/s/s away from that. our data was very accurate according to our r^2 value, which was .99995, meaning we were also perfect with our results. for the position- time graph, the slope is the velocity of the object. because the velocity was increasing, the line is curved. the slope of this graph was 445.35 cm/s/s. our p-t graph was extremely accurate; our r^2 value was 1, meaning we were 100% accurate with our measurements.
 * results:**
 * analysis of graphs:**

for v-t graphs, the equation of the line is y=mx+b. we deduced that for this graph, the equation of the line could also be substituted as v= at+ __. to find out what the blank was we figured out that the equation was also like the equation v final = v initial + at. we could then substitute into the other equation and find that v final= at+ v initial. our final equation was y=891.12x +50.009

for the position- time graph, the equation of the line is y= Ax^2 + Bx. in our graph, the equation of the line was also y= 1/2at^2 + v initial (t). this means that the slope of the position time graph should be half that of the velocity time graph. our final equation for our p-t graph was y=445.35x^2+ 50.138x.

however, 445.35 x 2 does not equal 891.12. this happened because the p-t graph's x intercept is set to zero, but the v=t graph's is not.

we did the percent error for the theoretical acceleration/slope of the v-t graph. we subtracted the experimental value of 889.65 from the theoretical value of 981, then we divided the sum by the theoretical value, 981. we then multiplied by 100 and the product was 9.31% error, which is well in the realm of 20% that calculations should be in. for percent difference we took the class average of 839.417 and subtracted 889.65. we used the absolute value of the sum and divided it by 839.417, then we multiplied it by 100 for a final product of 5.98%.
 * sample calculations:**

this is how we calculated velocity for a given point in time. we subtracted the initial velocity from the final velocity, then divided it by the final time minus the initial time. we did this for our first point and we calculated that the velocity was 93.2 cm/s.

|| - we speculated that the graph would be negative, but it is not. Because when copying down the results we neglected the negatives. - the x-t was also predicted to be negative, but as with the v-t graph, we omitted the negatives. the shape however, is what we expected it to be. a curve, starting slower and gradually getting steeper and the velocity became higher. - our results were fairly similar to most of the class', with the exception of the two outliers. there was a 5.98% difference between our slope and the class' average slope. - yes because on the v-t graph there is one constant slope and one constant, linear line. - altitude and air pressure can affect the acceleration due to gravity.
 * Class Data ||  ||
 * || ** Period 2 ** ||
 * || 754.43  ||
 * || 856.73  ||
 * || 851.07  ||
 * || 891.38  ||
 * || 891.12  ||
 * || 798.13  ||
 * || 710.65  ||
 * || 755.87  ||
 * || **// 659.39 //**  ||
 * || **// 1225.4 //**  ||
 * Average || ** 839.417 **
 * Average || ** 839.417 **
 * Discussion Questions **
 * 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?

Our hypothesis of what the acceleration of a free falling body was incorrect as compared to our results, even though it is known that the acceleration of gravity is 9.8 m/s/s or 981 cm/s/s. Our results showed that the acceleration was 889.65 cm/s/s.Our results for this lab were fairly accurate however. There was only a 9.31% error from our acceleration and the theoretical acceleration. This is well within the realm of 20%. The r^2 value for our v-t graph (which gave us the acceleration) was .99995, which means that we were very accurate with our measurements. Our graphs turned out differently than we had assumed they would because the hypothesized graphs were both negative whereas when we took our measurements, we omitted the negatives because signs are arbitrary. As always, there was definitely room for error in the lab. Some sources of error may have been going through the spark tape provided friction, which gave it drag and slowed it down. Holding the spark timer horizontally could have also provided friction and drag. The tape and measuring tape could have shifted while measuring.
 * conclusion:**