Ch2PolewayM

toc  Chapter 2 = = =Class Notes=

Free Fall




= = =1D Kinematics Lesson 1= B) 1. Scalars and vectors was already fully explained in class, and while reading over the lesson, I already had a well understanding of them. Scalars have to do with size only and vectors have to deal with both size AND direction. 2. There was nothing I was confused about in this lesson 3. I understood everything fully. 4. Everything that I read about was gone over in class.

C) 1. I already fully understood the difference between distance and displacement from what we learned in class. I knew that displacement had to do with direction and that distance was the total area traveled that had nothing to do with direction. 2. I wasn't confused about any of this from class and I didn't need any clarification 3. I understood everything fully. 4. I learned that distance is a scalar quantity because it has to do with size only, while displacement is a vector quanitity because it has to do with size and direction. We learned about scalars and vectors, but we never said how distance and displace were those quantities.

D) 1. From class we learned the difference of speed and velocity and I already knew about the differences. Speed is the rate of change of a position, or in other words how fast something is traveling. In contrast, velocity is the rate of change in position in a certain direction. It also is how fast something is traveling, but it has to do with direction. If speed of something is 50 mph then the velocity would be 50 mph in a certain direction. 2. I wasn't confused about anything coming into reading this, so I didn't need any clarification. 3. I understood everything fully. 4. We never learned about instantaneous speed, which is the speed at any given time. The average speed is the average of ALL instantaneous speeds.

E) 1. I understood the definition of acceleration from class very well, how it is used, and the formula that goes along with it. Acceleration is a vector quantity like velocity because it has to do with direction. It is the rate at which an object's velocity changes. Whenever the velocity is changing, an object is accelerating, but if the velocity is constant there is no acceleration. If an object is traveling positively, then the acceleration will be going that way, but if the velocity is slowing down, then the acceleration would be going the opposite way. In addition, the acceleration formula was familiar to me and I fully understood it. To find acceleration you need to subtract the final velocity by the initital velocity and then divide it by the time travelled. With that you get the acceleration and have to use acceleration units with distance units over time units. 2. There was nothing I was confused about in class 3. Why are there two time units for acceleration. For example why is it m/s^2 other than just m/s? 4. We didn't really go over constant acceleration much in class, and I never realized that an object could be constantly accelerating at the same speed. It's quite a simple concept and I understood it right away. If the velocity is going up by the same amount every second, then object will be constantly accelerating. In addition with constant acceleration, I learned that the total distance traveled is directly proportional to the square of the time. That makes it very simple to find the distance every second.

=**1D Kinematics Lesson 2**= 1. I understood the vector diagrams and the ticker tape diagrams very well from our class discussion today. There wasn't much if anything I needed clarification on. Vector diagrams are diagrams that track speed by using arrows pointing that way. The larger the arrow, the faster the object is moving. If the arrows are all the same size and pointing the same way it means the object is going constant speed. If they continue to get bigger that means the onbject is speeding up. The ticker tape diagram is a tape attached to an object moving and it gets threaded through a device which makes dots. According to these dots, you can tell how fast the object is moving.

2. I understood the concept of ticker tape diagrams but I didn't know whether the object was moving fast or slow by looking at the tape. Now I know that it goes in .10 second intervals and the closer the dots are to each other, the slower the object is moving. The further the dots, the faster the object is moving.

3. I don't understand why the object would have its motion pointing downwards if it was thrown into the air. Why is the arrow pointing down if it is being throw UP in the air and obviously gaining speed?

4. Everything was fully explained in class.

=**Chapter 2 lab: Speed of a CMV**=

Purpose: Here we are going to graph the constant motion of a CMV onto a position versus time graph. After we want to calculate the speed of the CMV and analyze and explain the graph

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

Materials: spark timer + spark tape, meter stick, masking tape, CMV

Hypothesis 1: Not very precise because if the distance is larger than a meter stick you'll have no way of measuring that distance. Hypothesis 2: 7 mph Hypothesis 3: It shows the acceleration of an object

Length of laptop (cm): 33 cm

Data table: position vs time

Analysis: Our graph had a diagonal line going straight and not changing since the speed was constant. The slope of any position-time graph is the velocity of the object so we found the velocity simply by finding the slope of this graph, which was about 61.63 cm/s. Our results were very accurate because our R^2 value is .987. For perfect results, we want that value to be 1.00, which we were very close to.

1. The slope is equivalent to average velocity because slope is the change in y over the change in x. y is position and x is time. Average velocity is position over time, so the two are exactly equal. 2. It's average velocity because there's more than one point averaged together to create the slope. It cannot be instantaneous because if it were instantaneous then it would be one point creating the slope. Since there is more than one point, it is impossible for it to be instantaneous. 3. It was okay to set the y-intercept equal to zero because than the x-intercept will also be zero. Since the speed is constant, as the position rises so does the time. When the object is at rest, if one axis if zero, the other will be too. It will not ruin the graph and it gives a better perspective to what is going on. 4. This value determines how close your line is to a completely straight line. Once this value hits 1.00, or 100 percent, your line will be a perfectly straight line. As long as your decimal is close to 1.00, you'll get good results, and usually you ll be accurate when you range from .96 to .99. The most ideal value is to have .9999 5. If I were to add another graph of a CMV slower than mine it would lie below the line already plotted. This is because the position will always be lower then the first line when the time is always the same.

Conclusion: In our experiment, our blue CMV travelled at 61.63 cm/s, when converted to mph is only 1.38 mph. I found the velocity by taking the total distance traveled and divided it by the total time. In my hypothesis I believed that my CMV was traveling at about 7 mph, and I was very wrong. I also thought that the graph showed the acceleration of the CMV, but the CMV never really accelerated, as it stayed at a constant speed once it started traveling. It is possible that we cut the ticker tape off too early before the CMV was traveling at constant speed. Because of this, in the beginning of the graph we the CMV might be accelerating instead of traveling at constant speed. To minimize this error we could run the CMV for longer and have a longer ticker tape so there is no doubt about where the constant speed is. In addition, when measuring the cms between dots on the ticker tape we could have either skipped a dot by accident or measured wrong. By looking at our graph we would be able to fix that problem, as it would not be a straight line. Simply by measuring very carefully and looking over our data to make sure we didn't make a typing error, there should be no problems with our graph.

Position time graph: shows where you are located Slope=speed Velocity time graph: shows how fast you are moving Slope= accel area= displ. Acceleration time graph: shows how fast is your velocity changing

=**Activity: Graphical Representations of Equilibrium**=




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


 * 1) How can you tell that your motion is steady on a…
 * 2) position vs. time graph- slope- speed so with a straight line
 * 3) velocity vs. time graph- slope- acceleration so with a straight line
 * 4) acceleration vs. time graph- with a horizontal line at 0


 * 1) How can you tell that your motion is fast vs. slow on a…
 * 2) position vs. time graph- the slope would be steeper
 * 3) velocity vs. time graph- the slope would be steeper
 * 4) acceleration vs. time graph- N/A


 * 1) How can you tell that you changed direction on a…
 * 2) position vs. time graph- with a negative slope
 * 3) velocity vs. time graph- with a negative velocity
 * 4) acceleration vs. time graph- it will always be the same


 * 1) What are the advantages of representing motion using a…
 * 2) position vs. time graph- at any given time you will know the position of the object and you can find the speed simply by using the slope formula. Slope=speed
 * 3) velocity vs. time graph- you can determine the velocity at any time and you can find the area on a velocity vs time graph to find the displacement. You can find the acceleration by
 * 4) acceleration vs. time graph


 * 1) What are the disadvantages of representing motion using a…
 * 2) position vs. time graph- in these graphs you can't find the acceleration or the displacement
 * 3) velocity vs. time graph- In these graphs you can't find the position well
 * 4) acceleration vs. time graph- you cant find the velocity or position


 * 1) Define the following:
 * 2) No motion- the object is at rest or not moving
 * 3) Constant speed- the object is moving at the same speed for every interval of time

=**1D Kinematics Lesson 3**= 1. From class i already understood that the slope on a position versus time graph is equivalent to the velocity. The same goes the other way. The slope and velocity of the line tells everything about the graph. For example, if the slope is positive, than the velocity is positive. If the slope is negative than the velocity is negative. With the slope you can also tell how fast an object is moving. The smaller the slope, the smaller the velocity, and the bigger the slope, the bigger the velocity. In addition, if the slope is a straight line, than the velocity is constant. If the slope is changing though, then the velocity is accelerating. I also know that to find the slope/velocity of the graph you need to use the slope formula which is the change in y divided by the change in x. You pick two points on a graph and subtract the y values and divide that solution by the subtracted x values. 2. There was nothing I was confused about in class. 3. I understood everything fully 4. Everything was thoroughly explained in class. There was not one new thing I learned from reading this, as it was all review

=**1D Kinematics Lesson 4**= 1. I already knew that in a velocity versus time graph, the slope is equivalent to the acceleration of the object in motion. The line of the graph shows you what the acceleration is doing. If the acceleration is zero, than the graph will be a straight line. If the slope is positive than so is the acceleration and when the slope is negative so is the acceleration. If the velocity is traveling to the right or in the positive direction then it will be in the first quadrant of a graph, but if it is negative then it will be in the fourth quadrant. Once again, like in the position v time graph, to find the slope you do change in y divided by change in x. You pick two points on a graph and subtract the y values and divide that solution by the subtracted x values. 2. There was nothing I misunderstood in class 3. I understood everything fully 4. I learned how to find the area and displacement using a velocity versus time graph. We never learned this in class. There are three different instances, where you either can find the area of a triangle, square, or trapezoid using their respective formulas. Finding the area helps you determine the displacement of the object.

=Interpreting Position-Time Graphs=



=**Lab: Acceleration Graphs**= 9/16/11 Lab partner- Noah Pardes

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

Hypothesis: - It would have an upwards slope. It would get steeper as time goes on. - You would be able to find the position, velocity, and acceleration

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

Procedure: - First we got ticker tape, a ramp, a textbook to put the ramp on and a cart. We attached the ticker tape to the cart, turned on the timer, and let the cart ride down the ramp. This would show increasing speed. Then, we put the cart at the bottom of the ramp with the ticker tape attached and pushed it up the ramp. We were trying to find decreasing speed here. Even though it increases at the beginning, it eventually turns into decreasing speed. We then ripped the ticker tape off and measured with a ruler the amount of mm in between every dot.



Analysis: a) Interpret the equation of the line (slope, y-int) and the R^2 value The equation of the line going uphill is -23.49x^2 + 54.901x with an R^2 value of .99995. This has a negative slope because it is moving away from the spark timer and it has a curved line because it accelerates. The vehicle starts off fast, but then slows down near the end where it reaches the top of the ramp. The vehicle actually comes to a period of zero velocity. The R value when it was a linear trend line was more around .94, but when changed to a polynomial trend line, it had a much better result of .99995, showing how accurate we were. The equation of the line going downhill was 16.731x^2 + 8.9839x and an R^2 value of 1. This has a positive slope because it going towards the spark timer and it has a curved line because its accelerating. It starts off slower and speeds up near the end. The R value when changed to a polynomial trend was 1, which means our results were absolutely perfect. The y-intercept for both graphs was 0, showing that each vehicle started at rest a 0 cm/s.

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.) The instantaneous for the vehicle going uphill at the halfway point was 23.38 cm/s, where at the endpoint it was 1.75 cm/s. For the vehicle going downhill, its speed was 27.05 cm/s at the halfway point and 74.5 cm/s at the endpoint.

c) Find the average speed for the entire trip. The average speed for the vehicle going uphill is 26.45 cm/s, while the average speed for going downhill was 29.08 cm/s


 * Discussion Questions:**

1. What would your graph look like if the incline had been steeper? The graph going downhill would be steeper since the speed would increase faster. More distance traveled over less time, creating a steeper line. In contrast, for the uphill graph, the line would be flatter. This is because the vehicle would slow down quicker when going up a steeper incline.

2. What would your graph look like if the cart had been decreasing up the incline? The graph would start off steep with its highest velocity in the beginning and would flatten out over time until it finally reached 0 cm/s. As time passed, the distance traveled would keep getting shorter until no distance is traveled at all. Once it hits that point, it would fall back down the incline and increase speed and get a steeper slope again.

3. Compare the instantaneous speed at the halfway point with the average speed of the entire trip. The instantaneous speed at the halfway point of the uphill graph was very similar to the average speed of the entire uphill trip. There's only a slight difference of about 3 cm/s, showing how the two are very close related. In addition, the halfway point of the downhill graph was also very similar to the averaged speed of the downhill trip. This difference was only about 2 cm/s. These two values are closely related to each other, as the halfway point will usually be around the average speed.

4. Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense? The graphs are curved since they are changing speeds, and there are too many slopes to calculate to get a good answer. To find the slope, we draw a tangent, linear line which goes through one of the points on the graph. Then we find the slope of that line, which is the instantaneous speed. This is because slope equals velocity.

5. Draw a v-t graph of the motion of the cart. Be as quantitative as possible. Conclusion: In this experiment, we had two trials, one where a CMV went down a ramp and another where it was pushed up the ramp. We got two graphs from this, and I hypothesized that as speeds increased, so would the steepness. As speed decreased though, the graph would flatten out. I was correct because as speed increased so did the steepness, where as in the graph of the CMV going uphill, near the end it flattened out. This is because its speed was slowing down and it finally reached at rest. The graphs were curved because the speed was always changing, so our slope wasn't linear. The R^2 values of our graphs also showed us that our results were very accurate because they were very close to 1. For example, one was .99995 and one was actually 1. Then, we found the instantaneous speeds of the graphs at the halfway points and endpoints by drawing tangent lines. The halfway points were very similar to the average speeds so they were definitely related. In constrast, the endpoint speed of the graph going uphill was only 1.75 cm/s because it was almost at rest. For the graph going downhill it was very high at 74.50 cm/s. This was because the vehicle's speed was increasing at great speeds. This experiment showed us what the graphs would look like for increasing and decreasing speeds, and we used it to find instantaneous speeds. We then compared these speeds with the average speeds. Although it seemed successful, there could have been several errors in this experiment. For example, when doing the graphs we could have messed up the tangent line and plotted points that were completely incorrect. This would have messed up all of your speeds, and the only way to fix this is to double check your work. Furthermore, when doing the trials we could have messed up the calculations on the ticker tape by measuring invalid points at the beginning of the tape. To fix this we could measure a few points in so we the results we want.

=Cart Activity of Acceleration=

In the position-time graph the graphs are curved because of acceleration and changing speeds. Run 1 and run 6 are going away from the motion detector, where run 1 increases speed, as run 6 decreases speed. Run 1 goes down the ramp away from the motion detector. Run 6 is pushed up away from the motion detector and then comes back down the ramp towards the motion detector and increases speed. In run 3, the cart is pushed up the ramp towards the motion detectors, and it decreases speed. It then falls back down the ramp away from the motion detector and increases speed. Finally, run 4 goes down the ramp towards the motion detector, and it increases speed. The velocity time graphs are all straight diagonal lines. When they changes axes, that show a change in direction.

=**Lab: Crash Course**= Mike Poleway, Noah Pardes, Hella Tallas, Kouske Seki Purpose: In this experiment, we want to find where two constant motion vehicles, one fast and one slow, will crash when put six meters apart and move towards each other. In addition, we want to find the point where a faster CMV will reach a slower CMV if they are put one meter apart going in the same direction.

Objective: Both algebraically and graphically, solve the following 2 problems. Then set up each situation and run trials to confirm your calculations.
 * 1) 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.
 * 2) 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.

Materials: CMV, tape measure/meter stick, masking tape, stop watch, spark timer/tape

Sample Calculations We believe that the two CMVs will meet 362.57 cm away from the faster car

We believe that it will take the faster CMV to travel 189.78 cm before it reaches the slower CMV with a time of 5.43 seconds.

Procedure of Crash media type="file" key="Crash+Process.mov" width="300" height="300"

Procedure of Catch up media type="file" key="Catch+Up+Process.mov" width="300" height="300"

Crash trial and results: media type="file" key="Crash.mov" width="300" height="300" media type="file" key="Crash+3.mov" width="300" height="300" Crash Results

Catch up trial and results: media type="file" key="Catch up.mov" width="300" height="300" Catch Up Results

Analysis:





Discussion Questions:

1. Where would the cars meet if their speeds were exactly equal?

If the cars were exactly equal, in the crash test they would meet at 300 cm, which is half of the 600 cm. On the other hand, in the catch up test, the cars would never meet. The only reason the cars met in the initial experiment was because one was faster than the other. If they are the same speeds they will never meet.

2. Sketch position-time graphs to represent the catching up and crashing situations. Show the point where they are at the same place at the same time.



3. Sketch velocity-time graphs to represent the catching up situation. Is there any way to find the points when they are at the same place at the same time? No, you cannot find the points when they are at the same place at the same time because this graph doesn't show the position they are at.

Conclusion: This experiment definitely satisfied our results, as they were very accurate and spot on. The first experiment was putting a slow CMV at one end of 600 cm and putting a faster one on the other end. Then we would see where they crashed. Our theoretical value for the crash was 362.57 cm. One of our trials was extremely close at 360 cm and all our trials averaged out to an average experimental value of 370.25 cm. The percent error was only 2.12%, which is very good considering percents under 10 are accurate. Furthermore, our catch up results were also very accurate. In this experiment we put the slow CMV 100 cms away from the fast CMV, with both pointing in the same direction. When put on at the same time, we wanted to find where the two would meet. Our theoretical value for this experiment was 189.78 cm with a time of 5.43 seconds. We had three trials which were all very close to this point, and our closest was 187 cm. The average experimental value was 186.33, creating a percent error of only 1.82%. This is an amazing percent error, and it shows how accurate our results were. Although we had some accurate results, there were several flaws in our lab. For instance, the blue, fast CMV always drifted off in one direction, which could have easily hurt the results. In addition, to try and solve the drifting problem we put the CMV against a wall so it wouldn't drift. This could have slowed the vehicle down though, altering our results. Finally in most of our trials, the CMVs never actually crashed but actually passed each other. This made it harder for us to figure out the exact point where they met. Fixing the first problem is simple, as we could just find a CMV that doesn't drift off every time. For the other problem, when we found the results during the trials, we only videotaped one or two of them. If we videotaped all the trials, we could go back and watch them to make sure where the exact points they hit were. All in all, according to our percent errors our results were very accurate, yet we could have fixed some things to make them even better.

=**Egg Drop Project**= Mike Poleway and Matt Ordover

Our egg broke. Final structure=27.03 g Our final structure was made of several straws connected by tape that created a two layer box. Inside the box was shredded paper which was believed to cushion the fall. The egg was placed in the box, but it broke from the fall.



Analysis: The acceleration of our structure falling from the window was 8.67 m/s^2. Our acceleration was much higher than everyone elses, which is probably a main factor in our egg breaking. Since we had a small surface area and no parachute, our box travelled much faster than anyone else's structure.



Conclusion: At first, Matt and I created a design out of wood and rubber band. It was a very intricate design, where wood was glued together to create a shape and rubber bands held the egg in the middle. When dropped, the egg and the design were completely destroyed, so we knew we had to think of something entirely new. We shifted our focus to straws, which we believed would cushion the fall of the egg. We made a bottom of straws and then created a mini box for where the egg could sit. This didn't work, as the egg fell out but as the design stayed intact, we knew we had an area to work with. We created a box of straws with a lot of cushion and it looked very strong. It weighed 27.03 grams and when we dropped it from my roof, which was about 30 feet, the egg survived. This gave us false hopes, as we went into class the next day with high expectations. When I dropped it, the box hit the ground in 1.13 seconds. That was way to quick, and this made our acceleration a lot higher at 8.67 m/s^2. If I were to do this project again I would definitely use the same design because at a high distance it worked, but I would also add a parachute. With a parachute, the fall would be much slower, and the acceleration would be too. If there were a slower acceleration our egg would have survived, and a parachute would've helped us.

Freefall- any object only acted on by the force of gravity acceleration=-9.8 m/s^2

=Free Fall Lab=

What is acceleration due to gravity?
 * Purpose:**

-9.8 m/s^2 should be the acceleration due to gravity and the v-t graph should be a diagonal line moving up in a straight line away from the origin. The ticker tape diagram should have more dots close to each other near the end of the tape because the speed will be increasing. To find acceleration we can find the slope of the v-t graph, because the slope of that is acceleration.
 * Hypothesis:**

1. Get materials- ticker device, ticker tape, mass object (100 g), masking tape, timer, power stripe 2. Once you have your materials, you have to feed the ticker tape through the device and then attach it to the mass object with masking tape. 3. Then we walked over to the balcony and dropped it over the ledge. 4. After we had to record how many cm were in between each dot. 5. Graph results
 * Procedure:**


 * Results:**


 * Analysis:**

x-t Graph:

v-t Graph: The position time graph shows that over time the speed is increasing because of the curved line. If the position time graph was just a straight line, then we would know it was traveling at constant speed. In addition, the velocity time graph shows that the velocity was increasing as time went on. Our results were extremely accurate for the x-t graph because it had an r value of 0.9999. 1.00 is the best value you can get and we were very close to that. The slope also has a parabolic equation, showing why there's a curve for the graph. The v-t graph also had a linear equation, showing why it was a straight line. Our r value wasn't as accurate as the x-t graph but it was still very accurate. Our group's slope of the line was only 876.56, much less than the theoretical equation of 980 cm/s.This was caused because of the friction created when the ticker tape slid through the device, creating the object to move slower. Air resistance was also a factor, but it was a small factor


 * Analysis:**

Percent error: ((Theoretical-experimental)/(Theoretical)) x 100 ((980-876.56)/(980)) x 100= 10.5%

Our percent error was pretty high because of the forces other than gravity acting upon the free fall. Friction and air resistance caused this high percent error because free fall is only supposed to be gravity. If it was only gravity acting on this fall, then the percent error would be much different.

Percent difference: ((Avg. experimental-Indv. experimental)/(Avg. experimental)) x 100 ((834.03-876.56)/(834.03)) x 100= -5.10%
 * v-t Graph Class Data ||
 * 853.72 ||
 * 861.69 ||
 * 805 ||
 * 708.97 ||
 * 767 ||
 * 864 ||
 * 881.5 ||
 * 887.79 ||
 * 876.56 ||
 * Average: 834.03 ||

Our percent difference was only -5.10% compared to the class average, which is a very good result.


 * Discussion questions:**

1. Does the shape of your v-t graph agree with the expected graph? Why or why not? Yes our v-t graph agrees with the expected graph because as time increases, we were expecting the velocity to increase at the same time as well in a straight, diagonal line moving away from the x-axis.

2. Does the shape of your x-t graph agree with the expected graph? Why or why not? Yes, our x-t graphs agrees with the expected graph. We were expecting the line to curve up, because the speed was getting faster, and in an x-t graph as speed increases, the graph will curve upwards.

3. How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.) Compared to the rest of the class, our results differed 5.10 percent from the class average acceleration of 834.03 cm/s^2. Our acceleration was 876.56 cm/s^2, which was the third fastest in the class. This means our group was one of the closest to the ideal value of 980 cm/s^2. A difference of 5.10 % is a very good result though, as we were pretty accurate and close to the class average.

4. Did the object accelerate uniformly? How do you know? No the object didn't accelerate uniformly because the R^2 value was not 100%.The v-t graph R value would have to be 100%, but ours was only 97.7%. That is still a very accurate result, yet if you want it to be uniform, it has to be exactly 100.

5. What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be? There are several factors that made acceleration due to gravity lower than it should be. When other factors act upon the fall other than gravity, such as friction and air resistance, then the fall will be much slower. With this, our object wasn't actually in a free fall. In contrast, if we wanted the acceleration due to gravity to be higher than it should be, we could throw the object instead of dropping it.

I said that the v-t graph would be a diagonal line moving away from the origin and my hypothesis for that was correct. Since the acceleration was constant, the v-t graph maintained a straight line. On the other hand, the x-t graph had a parabolic shape because the speed was not constant. Since the object was increasing speed, that graph curved. Our mass fell with an acceleration of 876.56 cm/s^2, which was one of the fastest in the class. The ideal acceleration was 980 cm/s^2, and we were one of the closest groups to that acceleration. That gave us a percent error of 10.5%, which really isn't an ideal value for an experiment. All the groups in class had a percent error around there, so our percent difference wouldn't be as high. Ours only had a 5.10 percent difference from the class average acceleration of 834.03cm/s^2. Our results were very close to those of the class, but not as close to the ideal acceleration which we ultimately wanted to get to. There were many reasons for this high percent error. First, there was possibly friction caused by the spark tape going through the spark timer. In addition, there was some air resistance that could have caused the object to fall slower. Since our object was so small though, there was not a large surface area, so air resistance wasn't a big deal. Unfortunately, friction is impossible to get rid of, but you could make it a little better by making sure the ticker tape is being fed through the device cleanly. For the air resistance, this is once again difficult to get rid of when throwing it off the balcony. When measuring the dots on the tape, it is possible we could have made some errors, which me and my partner almost did a few times. It was very tedious for us because our spark timer was set on 1/60 of a second, as opposed to the normal .01 second. There were many more dots on our ticker tape, and this made it much longer for us to get our results. This gave us a better chance to make more mistakes because of so many dots. If we did this experiment again, we would put the spark timer on .1 seconds instead of 1/60. Another problem was that we couldn't find out the exact point where the object dropped on the ticker tape, because before we dropped the object, the spark timer was on. The object was in our hands at that time, and we could have recorded some useless data right there. To fix this, we could try dropping the object and putting on the spark timer at the same exact time. In addition, we could get a longer piece of ticker tape and cut off the beginning portion to right when it started falling.
 * Conclusion:**

=1D Kinematics Lesson 5=

Introduction to Free Fall
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. The dot diagram at the right depicts the acceleration of a free-falling object. The position of the object at regular time intervals - say, every 0.1 second - is shown. 
 * Free-falling objects do not encounter air resistance.
 * All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s (often approximated as 10 m/s/s for//back-of-the-envelope// calculations)

The Acceleration of Gravity
This numerical value for the acceleration of a free-falling object is such an important value that it is given a special name. It is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. The symbol g represents it. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude. g = 9.8 m/s/s, downward

( ~ 10 m/s/s, downward)

Representing Free Fall by Graphs
One means of describing the motion of objects is through the use of graphs - __[|position versus time]__ and __[|velocity vs. time]__ graphs.  A position versus time graph for a free-falling object is shown below.

 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. Slow to fast. 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.  A velocity versus time graph for a free-falling object is shown below.

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. The object is moving in the negative direction and speeding up. 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.

How Fast? and How Far?
The velocity of a free-falling object that has been dropped from a position of rest is dependent upon the time that it has fallen. The formula for determining the velocity of a falling object after a time of t seconds is

vf = g * t
where g is the acceleration of gravity. 

Example Calculations:
At t = 6 s

vf = (9.8 m/s2) * (6 s) = 58.8 m/s At t = 8 s

vf = (9.8 m/s2) * (8 s) = 78.4 m/s

The distance that a free-falling object has fallen from a position of rest is also dependent upon the time of fall. This distance can be computed by use of a formula; the distance fallen after a time of t seconds is given by the formula.

d = 0.5 * g * t2
where g is the acceleration of gravity (9.8 m/s/s on Earth).

Example Calculations:
At t = 1 s

d = (0.5) * (9.8 m/s2) * (1 s)2 = 4.9 m At t = 2 s

d = (0.5) * (9.8 m/s2) * (2 s)2 = 19.6 m At t = 5 s

d = (0.5) * (9.8 m/s2) * (5 s)2 = 123 m

(rounded from 122.5 m) The diagram below (not drawn to scale) shows the results of several distance calculations for a free-falling object dropped from a position of rest.

The Big Misconception
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?" 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. More massive objects will only fall faster if there is an appreciable amount of air resistance present. 



Topic Sentence: Objects fall in free fall when gravity is the only force acting upon them, so they won't be going under air resistance. Acceleration is due to gravity and it is always accelerating at 9.8m/s^2. In free fall, all objects travel at the same speed, no matter what the mass is. All x-t graphs for free fall are going to be curved because the object is accelerating and it will start slow and end with a downward velocity. For the v-t graphs, it starts at rest where v=0. It will be a straight diagonal with the slope equaling 9.8. The velocity and distance of the object dropped are dependent on the time the object takes to hit the ground and can be found through equations.