Ch2_TomasofskyN

Chapter 2

toc ** Dimensional Kinematics **
Lesson 1; Physics Classroom HW 9/7/11

1. 2.  3.  4.
 * Displacement:
 * How far an object has gone from its original spot after traveling the total distance
 * Vector term
 * Needs direction
 * Distance
 * The total length an object traveled
 * Scalar term
 * Does not need a direction
 * Speed + Velocity
 * When finding average speed is the displacement at all applicable
 * Average speed only deals with total of distances and time
 * Velocity concept makes sense that it is basically speed with direction
 * But average velocity is different because deals with the change in position on top
 * The change in position is simply the displacement and the time is total time not change in ti
 * Is the change in velocity the complete change or the change from time to time because, if the object is not free falling, will the change in velocity between time intervals be the same
 * Objects slowing down and moving in opposite directions, why do they have to speed up in a way (ex go from a velocity of -8 to -2) to stop?
 * Acceleration
 * Can not be negative
 * Deals with how an object changes velocity
 * Constant Acceleration
 * Unlike a free,falling object with gains acceleration as goes on
 * Change in velocity same amount each time


 * Scalars and Vectors
 * Scalar: quantity that does not need a direction (I.e. speed)
 * Can really be any quantity
 * Vector: quantity that DOES need a direction (I.e. velocity)

** Class Notes 9/8 **

 * Constant Speed **


 * Terms: (September 6th) **

At Rest (v=0) * Average Speed: All speeds together without change Cosntant Speed: Unchanging * Instantaneous Speed: At any given moment

^^^Same Equation: v =change in distance/ change in time

Increasing Speed * Decreasing Speed *

Accelearation: Change in velocity including speeding up and slowing down

-Can be shown by motion diagram -Draw Arrows [qualitative representation of motion]
 * = The four types of motion
 * At rest (v=0, a=0)
 * Constant Speed ---> ---> ---> (v = unchanging, a = 0)
 * Increasing Speed --> > --> (v= increasing, a = points in same direction)
 * Decreasing Speed --> > --> (v = decreasing, a = points in opposite direction)

** Describing Motion With Diagrams **
Lesson 2; Physics Classroom HW 9/8

1.
 * Vector Diagrams
 * Represent motion
 * Use arrows to show velocity (also then show direction)
 * Size of arrow varies, for example if something is accelerating
 * Arrows will increase as they go on then
 * Ticker Tape Diagrams
 * Trail of dots represents motion of the object, the history of it
 * Will clearly then show if it is accelerating

2. Learning how to read the ticker tape was something that was unclear but not anymore after the reading. As shown in the diagrams below: the closer the ticks are to each other the slower the object is moving.

The diagram below further explains this that if an object is accelerating, or gaining speed the dots will slowly become farther apart.

3. There was nothing from the reading that was confusing or that I have further questions on. 4. Short reading that emphasized what we learned in class. No new info.

** The Speed of a CMV **
Lab 9/9 Nicole Tomasofky, Stephanie Wang

Objectives
 * How precisely can you measure distances with a meter stick?
 * How fast does my CMV move?
 * What information can you get from a position-time graph?

Equipment


 * Spark timer + spark tape
 * Meter Stick
 * Masking tape
 * CMV

Hypothesis
 * To the 10th. Measurement is precise but there is definitely room for user error.
 * About 8 miles/hour because a person's average walking speed is approximately 6 miles/hour. The cart will go quicker but not much than an average walking speed
 * The instantaneous/average/constant speed of a moving object.

Procedure
 * Get equipment
 * Tape the spark take (shiny side up) to your CMV
 * Turn on the Spark timer and the CMV
 * Spark Timer will pull spark tape through describing the motion of the CMV
 * Measure the distance between dots on ticker tape (starting from zero each time)
 * First couple dots are not applicable because it is when the CMV is starting

Measurements (Using Meter stick)
 * Laptop: 32.8 cm
 * Class's answers varied by about 1 cm. No one in the class was precise.
 * Always include known values and then guess one more (in between milimeters)
 * When it is exactly on the line need to put 0
 * Always use 4 sig figs with meter stick (that goes up to milimeters)
 * 10 hz = to dots/sec

Graph: Data Points:

** Analysis **

 * y=44.159x (equation of trendline


 * y = change in distance
 * 44.159 = velocity
 * x= change in time


 * r^2 values: tells how close your data was to the actual trend line
 * r^2 = .99797 (i.e. 99.8%)
 * Very close data, anything below 97% is considered not good data

Objective Questions Answered
 * Using a meter stick, you can get as precise as 2 decimal places. The last decimal place is a guess because it falls in between two milimeter marks which is not marked on meter stick.
 * My CMV moves on an average of 44.156 cm/s
 * A position-time graph shows the distance, or position, at any given time -- these are the points. But it also shows the average time -- this is the trendline

** Discussion Questions **
Slope is change in my graph would be, change in position over the change in time, which can also be described as the change in distance over the change in time. The equation for velocity is just this, change in velocity over the change in time.
 * 1. Why is the slope of the position-time graph equivalent to average velocity? **

** 2. Why is it average velocity and not instantaneous velocity? What assumptions are we making? ** This is average velocity and not instantaneous because it is the velocity over an entire second, whereas instantaneous velocity would only be the velocity at a specific instance. However, we are making the assumption that our CMV starts and ends at the same speed and does not accelerate at all.

It is okay to set the y-int to zero because we started at zero. Our cars were stopped when we started which is zero. The only way we would need a y-intercept lower than zero is if we are slowing down which would mean a negative acceleration.
 * 3. Why was it okay toset the y-intercept equal to zero? **

The r value tells us how closer our points are to the trendline that. The closer value the r value to 1 the better percentage we have and the closer our points are. It means we probably made less mistakes.
 * 4. What is the meaning of the R2 value? **

A graph that is slower than our CMV would have a slope that is stepper than our graph. Since slope is equivalent to the average velocity, a slower CMV would have a slower velocity. It traveled less distance in the same amount of time, so the number would be bigger resulting in a stepper slope.
 * 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? **

** Conclusion **
After collecting our results from measuring the spark tape dot distances and graphing, we found that our CMV moves at an average of 44.159 cm/s. Our hypothesis about the meter stick was off because to be precise you need to use your judgement for the last questionable place. Our CMV moves a lot slower than our hypothesis moving at about .9878 miles per hour after converted. But our hypothesis about the position-time graph is correct because our graph shows instantaneous speed, the dots, and average speed and constant speed which are the same since the vehicle did not accelerate, the trendline. Our results, according to our r squared value is about 99.797% correct. Error could have resulted from where we started to measure, because if we hadn't started from the dot that had already reached constant speed, all of our measurements would be a little off. Another reason why our measurements could be off is because the ruler may not have been perfectly straight on the ticker tape. We tried to keep the ticker tape as straight as possible, held between two laptops, but it might not have been as straight as it could have been. If we did the lab again we could run the test again for starters to get more data. We could start from the last dot and measure our way towards the beginning to be sure it starts at a constant speed. Also we could have measured more than once to make sure the ruler had not moved in between measurements.

** Constant Motion Activity **
At Rest:

Fast (Away from Sensor)

Slow (Away from Sensor)

Slow (Towards Sensor)

Slow + Fast (Both away) (Together for Comparison)


 * Discussion Questions: **
 * 1) How can you tell that there is no motion on a…
 * 2) position vs. time graph
 * -Horizontal line (Not necessarily on axis though)
 * 1) velocity vs. time graph
 * -on axis
 * 1) acceleration vs. time graph

(Based on the slow away graph, because results are best)
 * 1) How can you tell that your motion is steady on a…
 * 1) position vs. time graph
 * -steady curve up
 * 1) velocity vs. time graph
 * -almost a straight line at the axis
 * 1) acceleration vs. time graph
 * -unsteady line that goes up and down


 * 1) How can you tell that your motion is fast vs. slow on a…
 * 2) position vs. time graph
 * -Faster has a larger slope making the curve steeper (positive) Slower speed has less of a steep curve (negative)
 * 1) velocity vs. time graph
 * -Pretty similar lines
 * 1) acceleration vs. time graph
 * -pretty similar lines


 * 1) How can you tell that you changed direction on a…
 * 2) position vs. time graph
 * -Towards is a negative slope. Away has a positive slope
 * 1) velocity vs. time graph
 * Pretty similar; if anything slightly follows the same pattern listed above
 * 1) acceleration vs. time graph
 * No noticeable difference


 * 1) What are the advantages of representing motion using a…
 * 2) position vs. time graph
 * -Can track how far you’ve gone in how many seconds and compare
 * 1) velocity vs. time graph
 * Can see which direction you are traveling whether it be away or towards the sensor
 * 1) acceleration vs. time graph
 * It is hard to tell since we did not focus on acceleration


 * 1) What are the disadvantages of representing motion using a…
 * 2) position vs. time graph
 * Tracks motion of even your arm movements
 * 1) velocity vs. time graph
 * Does not show directly which direction you are going, need to understand how the graph works. Also if you are going to either side or up or down it would not track that. It can only track forward and backward
 * 1) acceleration vs. time graph
 * It is hard to tell since we did not focus on acceleration


 * 1) Define the following:
 * 2) No motion
 * -There is no movement at all
 * -Graph is a complete horizontal line
 * 1) Constant speed
 * -Speed that does not change, it does not accelerate negatively or positively. It is almost impossible to maintain a constant speed when walking because walking involves stopping and changing your weight distribution

Class Notes: (At Rest vs. Constant Speed)

** Acceleration **
Lesson 3; Physics Classroom HW 9/13


 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * Acceleration is a change in velocity, meaning it is when an object speeds up or slows down. An object in constant motion, like the CMV's we observed has no acceleration, but when we walk we actually do because we stop and switch our weight.
 * Average acceleration is an average of the accelerating times. If an object is not accelerating the same amount each time (constant acceleration), if would be beneficial to find its average acceleration
 * [[image:Screen_shot_2011-09-13_at_4.01.12_PM.png]]
 * 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.
 * Acceleration needs to have a unit of time attached because it deals with the change over a time period
 * Positive acceleration --> speeding up --> same direction as its velocity
 * Negative acceleration --> slowing down --> opposite direction as its velocity
 * 1) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * Everything was well explained in the class and reading
 * 1) What (specifically) did you read that was not gone over during class today?
 * Free-falling motion: as an object free falls it has a positive acceleration (Not constant) as it falls. It can be found by squaring each distance


 * The Big Five of Kinematics **

= =

** Describing Motion With Position-Time Graph **
HW: 9/14


 * 1) ** What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. **


 * Constant Velocity= straight line
 * Rightward = positive slope
 * Changing Velocity (Accelerating) = Curve
 * [[image:Screen_shot_2011-09-14_at_9.19.51_AM.png]]
 * Slope relates directly to acceleration and velocity!
 * 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. **


 * Rightward velocity is the same as positive velocity it just means that the line will be traveling to the right.
 * 1) ** What (specifically) did you read that you still don’t understand? Please word these in the form of a question. **


 * Is there a way to show which direction an object is moving in a position time graph? besides rightwards or leftwards (Positive/negative)?
 * 1) ** What (specifically) did you read that was not gone over during class today? **


 * Everything in the reading was pretty much went over in class.

** Describing Motion with Velocity-Time Graphs **
Lesson 4 Hw: 9/14


 * 1) ** What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. **


 * Constant velocity= slope is zero (horizontal line)
 * Slope tells you acceleration 
 * Quadrant tells you positive/negative velocity
 * [[image:Screen_shot_2011-09-14_at_9.37.45_AM.png]]
 * 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. **


 * Speeding up and slowing down on a graph
 * Look at the number increasing or not (Don't worry about sign)
 * [[image:Screen_shot_2011-09-14_at_9.39.40_AM.png]]
 * 1) ** What (specifically) did you read that you still don’t understand? Please word these in the form of a question. **
 * I understand everything
 * 1) ** What (specifically) did you read that was not gone over during class today? **
 * How to properly plot a graph when you have numbers to coincide with your data
 * Plot only the velocity

Cart Graphs Class Activity
Position Time Graphs+ Velocity-Time Graphs Acceleration Graphs

Decrease Acceleration (Towards Sensor: up the angle)

Increase Acceleration (Away from Sensor: down the angle)

Class Notes (Increasing + Decreasing Speeds)

** Acceleration Graphs Lab **
Lab 9/16 Nicole Tomasofsky + Stephanie Wang

Objevtives
 * What does a position-time graph for increasing speeds look like?
 * What information can be found from the graph?

Hypothesis
 * We know that on a position-time graph constant speed is a straight line with a slope that is not zero. I think the graph will be a curved line because the position will not be steady with the time.
 * You can find the average speed and instantaneous speed it shows distance and time, both factors that attribute to speed.

Procedure
 * Set up ramp (with only one book lined up exactly to the end)
 * Set up ticker tape by taping it to the back of the cart
 * Measure motion
 * Once with cart going down the ramp (Increasing speed)
 * Once with pushing up the ramp (Increasing Speed at first but then Decreasing)

Data: Graph:

Shape: Curve
 * Our r squared values were not very high (96.28% and 97.07%) with a linear line
 * After testing multiple trend lines, we found that the polynomial line gave us the highest r squared values (99.88% and 99.99%)
 * Because our line is curved we have an x squared in the equation
 * y= ax^2 + bx
 * y= change in position
 * x= change in time
 * Last number is initial velocity
 * This number is not zero because our ticker tape might have started seconds before we started the cart meaning the movement would not have started exactly at a tenth of a second.
 * Same formula as change in d=1/2at^2 + v(first)t

** Analysis: **

 * a)Interpret the equation of the line (slope, y-intercept) and the R2 value. **
 * Decreasing Speed:
 * Linear Line = y =42.219x + 43131 / r^2 = .96279
 * Polynomial Line= y = -24.762x^2 + 72.133x - 1.1345/ r^2= .99888
 * IncreasingSpeed:
 * Linear Line= y = 33.167 x - 6.9016/ r^2 = .97068
 * Polynomial Line= y= 10.663x^2 + 11.842x - .1485/ r^2= .99991
 * Our results were not at a constant speed, therefore the line is not represented best with a linear line
 * Our r^2 values show that our data is not very accurate compared to the linear line
 * Instead we used a polynomial line after playing around with different types of lines
 * Our r^2 values show that our data is much more accurate compared to this line
 * The slope represents instantaneous speed in both cases


 * 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.) **

*Good results, end point should have a faster speed since it is increasing
 * Increasing Speed:
 * (1,22.60) (1.5,38)
 * (38-22.60)/(1.5-1) = 30.8 cm/s 
 * (2, 66.40) (2.1,70)
 * (70-66.40)/(2.1-2) = 36 cm/s 

*Good results, end point should have slower speed since it is decreasing c)Find the average speed for the entire trip
 * Decreasing Speed:
 * (1,54)(.60, 34.20)
 * (54-32.20)/(1-.60) = 49.50 cm/s 
 * (1.20, 49.98) (1.5.54)
 * (54-49.98)/(1.5-1.20) = 13.4 cm/s 
 * Average Speed = distance traveled/time
 * Average Speed [Increasing] = 66.40cm/2.00s = 33.20 cm/s
 * Average Speed [Decreasing] = 49.98cm/1.20s = 41.65cm/s

** Discussion Questions: **

 * 1) ** What would your graph look like if the incline had been steeper? **


 * If the incline was steeper, the cart would have had a larger increasing and decreasing speed.
 * The graph would have had a larger slope which means that the curve would have curved not as wide.
 * 1) ** What would your graph look like if the cart had been decreasing up the incline? **


 * In a decreasing speed incline (see decreasing speed above) it is traveling away from the axis. The slope decreases as it goes up the incline.
 * 1) ** Compare the instantaneous speed at the halfway point with the average speed of the entire trip. **


 * The instantaneous speed for increasing speed at the halfway point was lower than the average speed for the entire trip. It was approx. 10.85 cm/s slower.
 * The instantaneous speed for decreasing speed at the halfway point was faster, approx. 7.85 cm/s faster.
 * 1) ** Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense? **


 * **I**nstantaneous speed is the slope of the tangent of the line because the tangent of the line goes through only one point of the slope, which ensures that it is at that instant. Slope of a line in a p-t graph tells you speed. Because you only plot the points the position of the points, meaning the distance, it does not tell you speed right away.


 * 1) ** Draw a v-t graph of the motion of the cart. Be as quantitative as possible. **



** Conclusion **
After observing our cart with both an increasing and decreasing speed, we found that our hypothesis was correct. We predicted the way in which the graphs would look like, being a curved line instead of a straight line. We also accurately predicted that it was possible to find instantaneous speed and the average speed from the information on a position-time graph. For the increasing speed trial, when the cart went down the incline, we found that the instantaneous speed at the midpoint was 5.2 cm/s less than the instantaneous speed at the end of the graph. This not only validated that our motion was increasing speed, but it also made our average speed more precise. The average of the two instantaneous speed (36+30.8)/2 = 33.1 which is only .1 away from the average speed we found doing distance over time. Moreover, for the decreasing speed trial, when the cart went up the incline, we found that the instantaneous speed at the midpoint was 36.1 cm/s faster than the instantaneous speed at the end of the graph. This shows that our motion was decreasing. However, when averaging out the two instantaneous speeds (49.50 + 13.4) /2, the answer is 31.45 cm/s, very different from the answer of average speed when calculating distance over time which was 41.65 cm/s. Our data was more accurate in the increasing speed according to our r^2 value, showing that it was 99.99% accurate. However, our r^2 value in the decreasing speed was less accurate at 99.89%. This goes along side with the math we did, the two average speeds we found were a lot closer in the increasing speed than it was in the decreasing speed. There was definitely error when drawing the tangents on our graphs because it was all done by hand. To check and make it more efficient, we could use a computer program to put the same line with the same slope that we found to see if it crosses through point we intended it too. In addition, we could have plotted more points on the drawn-in tangent and averaged out slopes in that way. Also, One reason why our decreasing speed trial was off was because we had less data points for our decreasing speed. If we had let the cart travel higher up the incline we could have gotten better results, that would give us more data points in general but specifically more data points of a decreasing speed which only happens at the end. All in all though, our lab was pretty successful.

Quantitative Graph Interpretation
(HW) 9/19





Crash Course Lab
9/23/11 Nicole Tomasofsky + Stephanie Wang

Objectives
 * Compare expected results of a problem to the actual results of a situation.

Hypothesis (Calculations of when cars would crash/ meet up)

Cars will crash @ 464.38 cm



Cars will Catch up @ 1.413 m (141.3 cm)

Lab Materials:
 * CMV
 * Tape Measure
 * Masking Tape
 * Stop Watch
 * Spark Timer
 * Spark Tape

Procedure:

Crashing media type="file" key="Movie on 2011-09-23 at 11.31.mov" width="300" height="300"

Catching Up media type="file" key="Movie on 2011-09-23 at 11.36

Data: Average Crash [Experimental value]: 433.2 cm Average Catch Up [Experimental Value: 126.2 cm

Analysis:


Crashing CMV (Percent Error): 6.7% Catch Up CMV (Percent Error) : 10.7% Any percentage below 10% is considered good
 * Our crashing CMV results were good with only a 6.7% percent error
 * The percent differences of our crashing CMV were very small, all being less than 1 percent
 * Our catching up CMV results were not as good - had a 10.7% error
 * The percent differences of the catching up CMV were much larger, specifically Trial 1 where our results greatly varied

Discussion Questions:

 * 1) Where would the cars meet if their speeds were exactly equal?
 * If the cars were to start at opposite sides, they would crash after traveling the exact same difference (in this lab specifically 300 cm. If they are traveling at the same speed, they would travel the same distance in the same amount of time.
 * If the cars were to start 1 m apart, they would continue to travel 1 m, apart, one never catching up to other. If they were so start at the exact same time at the same speed they would travel the same distance, one always staying 1 m apart from the other.
 * 1) 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.



-Although they have traveled different times, but at same distance so they have crashed -Blue car moved faster, larger slope
 * At same position at 10 seconds



-Different distances, but at same position
 * At same position at 4 seconds

Due to the fact that my lab group only had 2 people, it was difficult to record time, start both cars, record on video, and mark crash/catch up time. Therefore, we only have time for one crash/catchup (trial 1). That is why I compared these times with the average of all the trials.***

The velocity time graphs would only show the speed of each car. Since the speed is constant, and on a v-t graph constant speed is a horizontal line, there would be no different in the crash or catchup velocity time graphs. Therefore is no way to find when they are at the same place at the same time.
 * 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? **

**Conclusion**
After completing the lab, the results my lab partner and I predicted were off compared to results we got after completing the lab.For crash test, our results were pretty good with only a 6.7% error, but for our catch up lab our results had a 10.7% percent error. In this type of lab, any percentage below 10 is considered good. We had errors for many reasons. First of all, the carts were not started at the exact same time since my partner and I did this. Once the carts started they did not always move in a straight line messing up results, by how much time it took the CMV to travel a distance. In addition, they would crash or catch up in an instant, an instant that is easily missed or misinterpreted by the human eye. If we were to do this lab again, it would be beneficial to have more people, one person recording where the carts crashed or caught up, one person recording the time, and one person each starting the carts. Also the carts did not sway to the side our results would have had much less error.

Egg Drop Project
Nicole Tomasofsky + Maddie Margulies 9/28/11



Results:
 * Mass of egg: 60.47 g
 * Mass of egg + final container: 94.47 g
 * Times Collected: (1.31 s, 1.29 s, 1.53 s)
 * Average Time = 1.38 s

Analysis:
 * d = 8.5 m
 * V1 = 0 m/s
 * t = 1.38 s

D=v1 * t + 1/2 *a*t^2 8.5= 0(1.38) + 1/2a(1.38)^2 a=8.93

a= 8.93 m/s2 g (acceleration of gravity) = 9.8m/s2 Our final project slowed down the acceleration a little, compared to the natural acceleration of gravity, but we did not slow it down that much. Mostly the successful projects had a much smaller acceleration compared to the acceleration of gravity. They achieved this through using parachutes to change the air resistance.

Class Accelerations (m/s2)
 * 2.95
 * 6.43
 * 1.75
 * 3.36
 * 8.93
 * 7.36
 * 9.33
 * 3.64

Our egg did not survive the fall, therefore there are a lot of things we could improve on in this project. My partner and I tried to focus on lowering the impact, which is why we padded the center part with paper, making sure the egg never touched the ground, etc. But, just the impact of the egg, even though it didn't move, hitting the paper/straws was enough to crash it. If we had made our project more aerodynamic, meaning it would have less air resistance, we would slow down the acceleration and have a greater chance of it not cracking. In addition, if we focused on changing the impact of the egg hitting the ground, we should have given the egg somewhere to go, meaning instead of simply padding the egg, we should have given the egg somewhere to go. If it had room once hit to slowly fit into a spot without just going straight down and hitting the straws, it would have been more successful.

The Acceleration of Gravity
Lesson 5; Physics Classroom Hw 10/3 *Please Note: The cross through method did not save right on the wiki. Therfore, assume things that are not bolded or underlind are crossed out because they are not important or key words.

It was learned in the [|previous part of this lesson] that a free-falling object is an object that is falling under the **sole influence of gravity**. A free-falling object has an acceleration of **9.8 m/s/s**, downward (on Earth). This numerical value for the acceleration of a free-falling object is such an important value that it is given a special name. It is known as the ** acceleration of gravity ** - the acceleration for any object moving under the sole influence of gravity. A matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a **special symbol to denote it - the symbol g .** The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s/s. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude. We will occasionally use the **approximated value of 10 m/s/s** in The Physics Classroom Tutorial in order to reduce the complexity of the many mathematical tasks that we will perform with this number. By so doing, we will be able to better focus on the conceptual nature of physics without too much of a sacrifice in numerical accuracy.

Recall from an [|earlier lesson] that **acceleration** is the rate at which an object changes its velocity. **[ Change in velocity over change in time]** It is the ratio of velocity change to time between any two points in an object's path. To accelerate at 9.8 m/s/s means to **change the velocity by 9.8 m/s** each second.

^Free Falling Motion - Free falling objects are only affected by the acceleration of gravity (g = 9.8 m/s/s) therefore has a changing acceleration each second it is dropping


 * How Fast and How Far **

Free-falling objects are in a state of ** [|acceleration] .** 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. 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, 29.4 m/s** (approximately 30 m/s) at the end of the third second, etc. Thus, the velocity of a free-falling object that has been **dropped from a position of rest is dependent upon the time that it has fallen**. The formula for **determining the velocity of a falling object after a time of t seconds is**

(velocity final = acceleration * time falling) *Velocity dropped at any time when dropped from rest*
 * v **** f **** = g * t **
 * The distance ** that a free-falling object has fallen from a position of rest **is also dependent upon the time of fall**. __This distance can be computed by use of a formula; the distance fallen after a time of ** t ** seconds is given by the formula__

(Distance = .5 * acceleration * time falling 2
 * d = 0.5 * g * t **** 2 **

-Distance, time, and velocity are directly reliant on each other in free fall. Since it travels at the same acceleration if you know one or the other you can find all missing variables.


 * Representing Free Fall by Graphs **

Early in [|Lesson 1] it was mentioned that there are a variety of means of describing the motion of objects. One such means of **describing the motion of objects is through the use of graphs - [|position versus time] ** and ** [|velocity vs. time] graphs.** In this part of Lesson 5, the motion of a free-falling motion will be represented using these two basic types of graphs. A position versus time graph for a free-falling object is shown below. Observe that the line on the graph curves. [|As learned earlier], **a curved line on a position versus time graph signifies an accelerated motion.** Since a free-falling object is undergoing an acceleration (g = 9.8 m/s/s), it would be expected that its position-time graph would be curved. A further look at the position-time graph reveals that the object starts with a **small velocity (slow) and finishes with a large velocity (fast).** Since the slope of any position vs. time graph is the velocity of the object ( [|as learned in Lesson 3] ), **the small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity**. Finally, the negative slope of the line indicates a negative (i.e., downward) velocity.

[Start: Small inivial velocity/small slope End: Large velocity/large slope]

Observe that the line on the graph is a **straight, diagonal line**. As learned earlier, a diagonal line on a velocity versus time graph signifies an **accelerated motion**. Since a free-falling object is undergoing an acceleration (g = 9,8 m/s/s, downward), it would be expected that its velocity-time graph would be diagonal. A further look at the velocity-time graph reveals that the object **starts with a zero velocity** (as read from the graph) and finishes with a **large, negative velocity**; that is, the object is moving in the negative direction and speeding up. An object that is moving in the negative direction and speeding up is said to have a negative acceleration (if necessary, review the [|vector nature of acceleration] ). Since the slope of any velocity versus time graph is the acceleration of the object ( [|as learned in Lesson 4] ), **the constant, negative slope indicates a constant, negative acceleration.** This analysis of the slope on the graph is consistent with the motion of a free-falling object - an object moving with a constant acceleration of 9.8 m/s/s in the downward direction.



[Start: Zero Velocity, End: Big slope, Negative velocity]

One can tell if an object is free-falling from observing the position-time graphs and velocity-time graphs. Position time graphs have a curved line, showing accelerating, starting with a smaller slope line and getting larger, showing change in velocity. Whereas velocity-time graphs show a straight line showing acceleration, starting at zero velocity and then ending with a larger negative slope, showing the negative velocity.

[|Earlier in this lesson], 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 **. Yet the questions are often asked "doesn't a more massive object accelerate at a greater rate than a less massive object?" "Wouldn't an elephant free-fall faster than a mouse?" This question is a reasonable inquiry that is probably based in part upon personal observations made of falling objects in the physical world. After all, nearly everyone has observed the difference in the rate of fall of a single piece of paper (or similar object) and a textbook. The two objects clearly travel to the ground at different rates - with the more massive book falling faster. The answer to the question (__doesn't a more massive object accelerate at a greater rate than a less massive object?)__ **is absolutely not**! That is, absolutely not if we are considering the specific type of falling motion known as free-fall. Free-fall is the motion of objects that move under the sole influence of gravity; **free-falling objects do not encounter air resistance.** __More massive objects will only fall faster if there is an appreciable amount of air resistance present.__
 * The Big Misconception **

[Only amount of air resistance affects mass]

Mass has no affect on objects in free-falling motion. It also does not matter whether you throw an object up or it drops from rest, because the constant g remains the same, and in free fall it is only that constant that affects the object.

Free-Fall Lab
10/4/11 Nicole Tomasofsky. Stephanie Wang

Objective: What is acceleration due to gravity?

Hypothesis:
 * Acceleration due to gravity is 9.8 m/s/s. A velocity time graph would be a diagonal straight line starting at the origin going down, into the negative direction. The slope of the line would show the acceleration, and should be -9.8 m/s/s

Analysis :

 * Mass = 100g **



*We deleted the last data point (position at .7 seconds) because it was an outlier on our graph and messed up our results. We probably measured the last point wrong.

y= 353.67x^2 + 267.39x r^2= .99997 Shape
 * We choose a polynomial trendline since our data is is not in a straight line and is curved
 * x^2 value was much lower than expected (should be 500-600)
 * Our value shows that our data is 99.99% accurate compared to the trendline. This result is very good.
 * Shows increasing speed away. Meaning from where we dropped the mass, it is increasing away as it travels from the drop spot, which is what actually happened.



y= 708.97x + 267.34 r^2= .99507 Y-Intercept Shape
 * Indicates that slope of line (acceleration) is 708.97 cm/s/s
 * Linera line because it is straight line
 * Indicates that our results are 99.51% accurate
 * Relatively strong numbers
 * The y-int is not set at zero at this graph because first point is far from zero
 * One possible explanation is that we did not use the first dot on our ticker tape paper because it was in a cluster. Or, we did drop the mass right at the 1/10 of a second.
 * Straight line showing increasing speed. The ideal slope would have been 981 (the acceleration of gravity). Therefore, the ideal line would have been steeper than our actual line. However, the steepness of our line is close to the ideal steepness.

Sample Calculations:



Percent Error: 27.73% Percent Difference: 14.9%
 * For this lab, is appropriate to get between 10-20% error. Although our percent error was higher than expected in relation, our percent error is not terribly high
 * The class had an average velocity of 834.03 cm/s. With the wide number of results we had a 14.9% difference is not a bad number considering how easy it was to achieve the percent errors.

** Discussion Questions **

 * 1) ** Does the shape of your v-t graph agree with the expected graph? Why or why not? **

The v-t graph does not agree with what we expected. In our hypothesis, we expected to drop to start at zero and go negatively down. Our graph is different because in our calculations we did not account for the displacement of the the mass or the actual acceleration. In actuality, the mass would have a negative displacement because displacement is final position-initial position, with its final position being zero and its initial position being how hight the balcony was. Also, our acceleration would be negative because it is being dropped. Therefore, our v-t graph is not incorrect it just does not account for the negative displacement and the negative acceleration.


 * 1) ** Does the shape of your x-t graph agree with the expected graph? Why or why not? **

Our x-t graph is the same as the the expected graph. This makes sense because as the time increased the position it the mass traveled also decreased which is what the x-t graph should have shown.


 * 1) ** How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.) **

Compared to the class, we had a 14.9% difference. Our acceleration, being in the 700s was lower than most of the classes which was in the 800s. However a 14.9% difference on this type of lab is not necessarily bad. There was a lot of possible differences in errors and everyone in the class was off from the theoretical acceleration.


 * 1) ** Did the object accelerate uniformly? How do you know? **

You can tell that our object had an increasing velocity, because on the x-t graph, the slope of the line which is velocity, is not constant. A curved line would mean that it is not constant. However, our acceleration was uniform. On a v-t graph, the slope which is acceleration is the same throughout the graph. The v-t graph is a straight line showing that the acceleration is uniform, accelerating at the same rate each second.

Our number was lower than the acceleration of gravity. This could be for many reasons including that the spark tape was slowed down through friction when going through the spark timer. The tape could not have a constant flow through the timer because of friction, or a hand holding it back. A number could be higher than the acceleration of gravity it it was thrown down by accident or if it was measured wrong.
 * 1) ** What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be? **

** Conclusion **
In conclusion, after completing the free-fall lab, we found that our hypothesis was somewhat correct. We knew from prior knowledge that the acceleration of gravity was -9.8 m/s/s, however our lab did not prove this correct. Our object actually had an acceleration of 7.01 m/s/s. In addition our v-t graph was not expected with what a v-t graph of an object in free-fall shows. We were off in our results for many reasons. Our accelerations were less that the acceleration of gravity because of friction of the tape going through the spark timer. Our 27.73% error comes from this issue as well as calculation issues with not starting at the actual first point, or having our first point not being exactly at the 1/10th of a second. Our v-t graph varied because we did not account for the negative displacement it actually had or the negative acceleration. If we were to do the lab again, it would be best to ensure that we are holding the spark tape timer vertically and not horizontally to try to eliminate friction as much as we can. We could also do this experiment multiple times and compare results to try to ensure that our first was truly at the first 1/10 of a second. In order to make our v-t graph more accurate we should account for the negative displacement and the negative acceleration of our object. All in all though our 14.9% difference with the rest of the class proved that even though our results were off they were not drastically off from the rest of the classes. Even though our lab was not perfect, it did help explain acceleration of an object in free fall.

Free Fall Class Notes
10/5