Ch+2+tonuzire

=Chapter 2=
 * HONORS PHYSICS**

toc

= Lab: A Crash Course in Velocity (Part 1) =

Remzi Tonuzi and Julie Van Malden

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

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

**Hypothesis** Based on all of the information we have acquired this far, we estimate that the yellow car will move at average velocity of 15 m/s while the blue, which is seemingly faster, will move at an average velocity of 20 m/s.

**Procedure-**
 * Gather all required materials
 * String the spark tape through the spark timer.
 * Attach the end of spark tape to the CMV
 * Make sure the spark timer is on 10 Hz, and then turn on both the car and the spark timer.
 * When all of the spark tape has run through the timer, turn off both.
 * The tape will now have dots on it that represent the distance the car moved per every 1/10s.
 * Measure the distance between the dots and create a scatter graph interpreting your results.

**Discussion questions**
 * 1) Why is the slope of the position-time graph equivalent to average velocity?
 * 2) The slope of the position-time graph is equivalent to the average velocity because the slope is equal to rise over run, which is equal to change in y over the change in x, which is the change of the position over the change in time, which is the definition of average velocity.
 * 3) Why is it average velocity and not instantaneous velocity? What assumptions are we making?
 * 4) We are assuming that velocity remains constant or relatively close to constant throughout the experiment, so there would be no unexpected change in speed.
 * 5) Why was it okay to set the y-intercept equal to zero?
 * 6) It is okay to set the y-intercept equal to zero because when no time has passed (0 s) the car hasn’t began to move so its it has moved 0 cm/m.
 * 7) What is the meaning of the R2 value?
 * 8) R2 value is whether the plotted data’s line of best fit is closer to a line or a curve in the form of a percentage.
 * 9) 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?
 * 10) Being that the CMV moved slower it would go less cm/s so the slope of the line would be slighter, lying underneath the original line.

**Conclusion**

In our CMV lab we tested for the CMV of the fast yellow car. We found that the CMV had an average speed of 31.95 m/s, ruling our hypothesis to be inaccurate. We hypothesized that the CMV would have an average velocity of about 15 m/s, but our result proved this hypothesis to be incorrect. Sources like the age of batteries, obstacles in the way of the experimental path, and unleveled testing surface may have contributed to inaccuracies in our data. These sources could have lead to a change in the result and possible error in the data. To minimize the chance of error we could ensure the batteries in the CMV are functioning well by changing them and replacing them with new ones, make sure the testing path is level and clear of any obstacles that could interfere.


 * Time (s) || Position (cm) ||
 * 0.0 || 0.00 ||
 * 0.1 || 3.19 ||
 * 0.2 || 6.35 ||
 * 0.3 || 9.55 ||
 * 0.4 || 12.78 ||
 * 0.5 || 15.89 ||
 * 0.6 || 19.08 ||
 * 0.7 || 22.39 ||
 * 0.8 || 25.55 ||
 * 0.9 || 28.81 ||
 * 1.0 || 32.01 ||
 * Fast Yellow ||  ||
 * Fast Yellow ||  ||

1D Kinematics, Lesson 1 Method 2a

 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) I know that scalars are quantities that are fully described by magnitude (a numerical value) alone. Also, vectors are defined by magnitude and direction. Distance is a scalar quantity. If an object moves 5 meters forward and 5 meters back to the original position, the distance is 10 meters. The displacement, which is focused on direction and the amount of distance from the starting point. If an object is moved from 0 meters to 10 meters the right, the distance is 10 m and the displacement was 10 m to the right/ east. So, distance is focused on a covered distance and displacement gives attention to direction.
 * 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 had forgot until I read about displacement and its difference between distance. Also the definition of vector was a little blurry until I read the web page. I realized that displacement is the distance from the starting position to a new position in a different direction. Also if it moves back in the same direction (5m east then 5m west) the displacement is 0m.
 * 5) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 6) I understood everything in the passage.
 * 7) What (specifically) did you read that was not gone over during class today?
 * 8) Pretty much everything was covered in class except I think the second example.

__Types of Motion__: There are only four types of motion. __Motion Diagrams__ Signs are arbitrary (random): Up (+), Down (-), Right (+), Left (-) Think of it as a Coordinate Plane (If speed is going diagonally, turn the plane accordingly... e.g. diagonally upward towards right would be considered right on the coordinate plane.)
 * At rest _ v=0, a=0
 * Constant _ V --> V --> V --> V (no change in velocity)
 * a = 0 (no increase in speed, acceleration is 0)
 * Increasing V --> V --> V -> V
 * a -> [Acceleration Point in SAME direction as velocity]
 * Decreasing V ---> V ---> V -> V
 * <-- a _ [Acceleration Point in OPPOSITE direction from velocity]

If the signs for velocity and acceleration are both (-) or (+), it is still increasing speed.



1D Kinematics, Lesson 2 Method 2a
1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. a) Ticker Tape is a common way to analyze the motion of an object. It is attached to a moving object and is dragged through a “ticking” machine. Every tick represents .10 seconds and it shows the history of the object’s motion by leaving a dot at every tick. The dots also indicate the object's change in position over a certain amount of time. If the space becomes larger, then it is increasing its speed (accelerating) and if the dots are closer together then it is slowing down (decelerating). Ticker tape indicates a constant velocity or acceleration as well.



b) Vector diagrams show the direction and number value of a vector quantity by a vector arrow. It can describe the velocity of a moving object during its motion. The size of the arrow determines whether the speed is constant or if the object is accelerating. Smaller to larger arrows means acceleration and same size arrows mean constant speed. Vector diagrams can be used to represent acceleration, force, and momentum. Vectors can be used to represent direction and size of quantity. 2) What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.

a) The reading was just a review of what was learned in class.

3) What (specifically) did you read that you still don’t understand? Please word these in the form of a question. a) I understand everything.

4) What (specifically) did you read that was not gone over during class today? a) Everything was gone over in class that was read on this page. Graphical Representations of Equilibrium

1. How can you tell that there is no motion on a… 1. position vs. time graph 2. velocity vs. time graph 3. acceleration vs. time graph On the at rest graph, we can tell that there is no motion because when we look at the graph, the line has very few fluctuations that go above or below zero. The line is nearly horizontal with a slight positive slope.

2. How can you tell that your motion is steady on a… 1. Position vs. time graph - The line rose at a gradual rate and than evened out and continued for a short time at the same pace

1. Velocity vs. time graph - There is a very small inline that leads to a pretty steady line slightly above 0.

1. Acceleration vs. time graph - For the most part the line remains exactly on 0. It does fluctuate slightly above or below 0 on a few occasions.

3. How can you tell that your motion is fast vs. slow on a… 1. position vs. time graph- The faster the motion is, the shorter it takes to go a greater distance or have a larger change in position. The slower a motion is, the longer it takes for something to change or move positions. 2. velocity vs. time graph- The faster motion will have a faster/higher velocity then the slower motion, so the faster motions line should be slightly above that of the slow motion line. 3. acceleration vs. time graph- Because during the experiment the fast and slow walk were constant, the acceleration remains close to 0 throughout.

4. How can you tell that you changed direction on a… 1. position vs. time graph- When the line, or position, starts at a high number and then gets lower as time goes on, the motion that is being preformed is moving towards/closer the sensor, but when the position starts at a low number and gets higher, the motion being performed is moving farther away from the sensor. 2. velocity vs. time graph- The velocity for the line going away from the sensor was at a slightly higher rate than the line of the test coming toward the sensor. 3. acceleration vs. time graph- Because there is no acceleration during either of these motions, you couldn’t tell that there was a change in direction because the acceleration would remain around 0 in both ways.

5. What are the advantages of representing motion using a…
 * 1) position vs. time graph –
 * 2) The movement of an object staying still will not change its position. This means that the graph has a straight line representing no change in its position. It can also show if there is a large change in position indicating that an object is moving faster than another. Also it can show the displacement depending on how far an object moves forward and back. It will show how much distance is covered at a certain speed.
 * 3) It can show that an object is moving towards a certain point and back by the negative and positive intervals of the graph. Negative would be moving closer to the original position and positive would be moving away from the original position.
 * 4) It shows whether an object is increasing its speed, decreasing its speed, or staying at a constant rate for a certain amount of time over the course of a distance.
 * 5) velocity vs. time graph
 * 6) acceleration vs. time graph

6. What are the disadvantages of representing motion using a… >
 * 1) position vs. time graph –
 * 2) Occasionally, one of the plotted points will be on the zero or another number, which represents either change in the moving position or no change at all from the starting point.
 * 3) A possible disadvantage for representing our data using a velocity graph would be that there are some peaks and valleys in the beginning and end of the data. It is also difficult to see the data because of its closeness unless you adjust the scale.
 * 4) A possible disadvantage for representing our data using an acceleration graph would just be because the acceleration for this specific lab was 0 it's very hard to see any truly outstanding differences between the data.
 * 1) velocity vs. time graph
 * 2) acceleration vs. time graph

7. Define the following:
 * 1) No motion – When an object is at rest and is not changing its position. Speed, velocity, and acceleration equal 0.
 * 2) Constant speed – When an object is going at the same speed for the entire time that it is moving from one position to another.

At Rest -

Constant Speed - Slow and Fast

Constant Speed - Towards and Away

=** LAB ACCELERATION GRAPHS **=


 * Remzi Tonuzi, Joe Miller, Julie Van Malden **


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

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

a) Interpret the equation of the line (slope, y-intercept) and the R2 value. b) Find the instantaneous speed at halfway point and at the end. (You may find this easier to do on a printed copy of the graph. Just remember to take a snapshot of it and upload to wiki when you are done.) i. Slope of tangent line= (8-2)/(.77-.37)= 6/.4= 15 cm/s i. Slope of tangent line= (8-0)/(.79-.55)= 8/.24 = about 33.3
 * Analysis: **
 * 1) When we constructed our line of best fit we tried two different lines before we came to the correct one (a polynomial). When we compare the two lines we find that the polynomial is closer to 1.0 (0.99979) than that of the linear, which was only about 0.934… This means that the polynomial line contains almost 100% of our points compared to only about 93.4. Our polynomial line has a slope of 13.67 and a y intercept of 0.7452.
 * 1) These are the tangent lines for the halfway point speed and the end point speed. Using these I used the coordinate points, (.37,2) and (.77, 8) from the tangent line for the slope of the instantaneous speed at the halfway point. I also used the coordinates (.79,8) and (.55,0) from the tangent of the end point line to find the slope/ instantaneous speed at the end.[[image:Graph_1.png]]
 * 2) The instantaneous speed is about 15 cm/s
 * 1) The instantaneous speed at the end is about 33.3 cm/s

c) Find the average speed for the entire trip.
 * 1) Average speed = total distance/ total time
 * 2) Average speed= 14.48 cm/ 1 s
 * 3) The average speed is 14.48 cm/s

- An upward incline with a positive slope - The graph will show us how fast an object travels a certain distance. The graph will display the distance at which our cart travels in 1.0 second.
 * Hypothesis: **

- Step 1- Get materials - Step 2- Lean Ramp on textbook and put cart at the top of the ramp - Step 3- Feed ticker tape through timer and attach to cart - Step 4- Turn on ticker time and release car - Turn off ticker timer and interpret results
 * Procedure: **

i. Avg. Speed: 14.48 i. 33.3 cm/s - If a line is tangent to a curve it touch at only 1point. If you know the slope of the tangent line you know the slope of the point that you chose on the curve,a which is included on the tangent line.
 * Discussion Questions: **
 * 1) What would your graph look like if the incline had been steeper?
 * 1) What would your graph look like if the cart had been decreasing up the incline?
 * 1) Compare the instantaneous speed at the halfway point with the average speed of the entire trip.
 * 2) Avg. Speed = ∆d/∆t = 14.48 cm/1 sec
 * 1) Inst. Speed = 3.75cm/.5 sec
 * 1) Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense?
 * 2) Draw a v-t graph of the motion of the cart. Be as quantitative as possible.
 * 3) [[image:Graph_3.png]]

**Conclusion:**


 * After taking starting the lab we measured our results for 1.0 second and came up with pretty decent results. When we placed a polynomial line of best fit we found that our equation for the line was: (y=13.67X^2+0.7452x) and our R^2 value was extremely close to being 1 it came out as (0.99979). For this laboratory we created two individual hypotheses. The first was: An upward incline with a positive slope and the second was: The graph will show us how fast an object travels a certain distance. It will display the distance at which our cart travels in 1.0 second. Both of our hypotheses seemed to be accurate with the data we received. You can see the labels on the graph represent position and time that proves our second hypothesis and you can see an upward curve with a positive slope, which proves our first hypothesis. Some sources of error that could have affected our results are the fact that the ticker timer starts and a person may not release the car at a perfect time, which creates a few extraneous dots. Another error that could have occurred was in the measuring of the distance between one dot and the next on the ticker tape, which could have led to slightly off results. In order to minimize these errors there are a few things that can be done. The first thing that I would recommend doing is starting on the ticker tape a few dots in to assure that it was recording the time when the cart was moving and not just the wait time in between the launch and timer start process. The second way we could minimize error is to use other methods of measuring the space in between the ticker tape dots. We could have used different measure devices like a flat ruler or a tape measure. Another option is to have two people measure it and see if the results they produced were comparable. **

Remzi Tonuzi

2(a) summary – Lesson 1(e)
1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. a) Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing velocity. If an object’s velocity is increasing (rate at which it changes position increases), the acceleration is increasing. Constant Acceleration is when the acceleration is increasing by the same amount each time, e.g. 3 m/s each second. Constant acceleration is not to be confused with constant velocity. If an object is changing its velocity – whether it is by a constant amount or varying amount – then it is accelerating. Constant Velocity = 0 Acceleration. 2) What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding. a) I was confused on how to used the 3) What (specifically) did you read that you still don’t understand? Please word these in the form of a question. a) How do you apply the average acceleration formula? 4) What (specifically) did you read that was not gone over during class today? a) The direction of an acceleration of a vector.

Lab:

**1-D Kinematics - Lesson 3 and Lesson 4**
Format 2a 1.What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. a) In a position – time graph, if the slope is constant (constant (+) velocity), then the line will be straight and diagonal towards the right. If the velocity is increasing in the p-t graph, then the line will be curved showing increased velocity. "As the slope goes, so goes the velocity." The velocity show what characteristics the slope has, and the slope does the same for the velocity. If the velocity is constant, then the slope is constant (a straight line). If the velocity is changing, then the slope is changing ( a curved line). If the velocity is positive, then the slope is positive (moving up and right). This can be applied to any motion. In a p-t graph, the steeper the line/slope the higher the velocity. If it is less steep, then it is slower. Negative Acceleration – moving in a negative direction and speeding up, or positive direction and slowing down. Positive Acceleration – Moving in a negative direction and slowing down, or positive direction and speeding up. Small slope = small velocity; negative slope = negative velocity, etc. The slope can change how the line looks. If it is a constant slope and then abruptly stops, it will be a diagonal line and then a straight line, representing no change. b) Finding Slope – y2 – y1/ x2 – x1 c) Constant velocity means no acceleration. The shapes of the velocity vs. time graphs for these two types of motion - constant velocity motion and accelerated motion disclose an important principle: the slope of the line on a velocity-time graph reveals useful information about the acceleration of the object. If the acceleration is zero, then the slope is zero (e.g., a horizontal line). If the acceleration is positive, then the slope is positive (e.g., an upward sloping line). If the acceleration is negative, then the slope is negative. This principle is used in virtually any motion. 2) What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding. a) Curved lines have changing slope; they may start with a very small slope and begin curving sharply (either upwards or downwards) towards a large slope. In either case, the curved line of changing slope is a sign of accelerated motion (i.e., changing velocity). I didn’t understand it fully until I read again what was gone over in class. 2) What (specifically) did you read that you still don’t understand? Please word these in the form of a question. a) There isn’t a topic from the reading and class that I don’t understand. 3) What (specifically) did you read that was not gone over during class today? a) Everything was gone over.

=LAB: Crash Course in Velocity (Part II)= Remzi, Madison, Ben, Julie **Objectives**:

Both algebraically and graphically, solve the following 2 problems. Then set up each situation and run trials to confirm your calculations. 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.
 * 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.
 * [[image:hpmsteele/Screen_shot_2011-09-23_at_1.42.33_PM.png caption="Screen_shot_2011-09-23_at_1.42.33_PM.png"]]
 * [[image:hpmsteele/Screen_shot_2011-09-23_at_1.41.16_PM.png caption="Screen_shot_2011-09-23_at_1.41.16_PM.png"]]

1 and 2 -



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

**Interpreting Your Results:** Be sure to provide concrete evidence (data and calculations) for the success of your experiment.

**Discussion questions:** >>
 * 1) Where would the cars meet if their speeds were exactly equal?
 * __For problem crashing situation__: At exactly 300 cm (middle). If the CMVs had equal speeds, they would be covering the same amount of distance over the same amount of time. Therefore, if they started at opposite ends, and the distance between them was 600 cm, they would meet at 300 cm.
 * __For problem catching up situation:__ They would never meet because they start 1 m apart and have the same speeds. There would be no way for one to catch up to 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.
 * Position-time graph crashing
 * Position-time graph catching up
 * [[image:CMV_Lab_Part_II_(3).jpeg width="640" height="839"]]
 * 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?
 * Velocity-time graph crashing and Catching Up
 * [[image:CMV_Lab_Part_II_(3)_1.jpeg width="800" height="1160"]]
 * No. It is not possible to do this by using a velocity-time graph. This is because velocity-time graphs do not show the position of a moving object. The slope of a line on a velocity-time graph represents acceleration, not position.


 * Analysis:**
 * Error analysis for crashing situation:
 * [[image:hpmsteele/photo_(2).JPG width="576" height="1003" caption="photo_(2).JPG"]]
 * Error analysis for catching up situation:
 * [[image:hpmsteele/111.JPG width="576" height="1003" caption="111.JPG"]]
 * [[image:hpmsteele/ss.JPG width="578" height="1005" caption="ss.JPG"]]


 * Conclusion:**

For the crashing situation our hypothesis proved correct. Using algebra, we hypothesized that the CMVs would crash at 239.80 cm. Our experimental results were 195.56 cm (Trial 1) and 220.50 cm (Trial 2). To ensure that our data is accurate and precise, we calculated the percent difference and percent error. Percent difference compares each individual trial to the average of all trials. Percent error shows how close our experimental values are to our theoretical values. The percent difference for both Trial 1 (195.56 cm) and Trial 2 (220.50 cm) was 5.99%, which is relatively low and very consistent. The percent error for Trial 1 (195.56 cm) was 18.44 %, which isn’t that low. However the percent error for Trial 2 (220.50 cm) was 8.05%, which is very low. If I am remembering correctly, we messed up Trial 1 because the ramp we used was crooked. This may account for the difference in percent error between Trial 1 and Trial 2.

For the catching up situation our hypothesis proved incorrect. Using algebra, we hypothesized that the blue CMV would catch up with the fast yellow CMV (100 cm away from the blue CMV) at 299.19 cm. Our experimental results were 165.5 cm (Trial 1), 175.25 cm (Trial 2), and 167.25 cm (Trial 3). We did further analysis situation by calculating the percent difference and percent error. The percent error for Trial 1 (165.50 cm) 44.68%, the percent error for Trial 2 (175.25 cm) was 41.43%, and the percent error for Trial 3 (167.25 cm) was 44.02%. The percent error for all three trials was high, yet extremely consistent. This led me to think that something was wrong with one of my CMVs, rather than my calculations (which I checked meticulously!). The percent difference for Trial 1 (165.50 cm) was 2.26%, the percent difference for Trial 2 (175.25 cm) was 3.50%, and the percent difference for Trial 3 (167.25 cm) was 1.23%. The percent difference for all three trials is low and extremely consistent, which confirmed my suspicion of something being wrong with the CMVs. Since the percent difference is so low and consistent, the experiment must have been conducted correctly.

Upon watching my procedure video a few times, I noticed that the yellow CMV was very slow. I knew that each metal track I used in my experiment (to keep the CMVs from veering off) was 120 cm long. While watching the video, I counted 4 seconds from the time the yellow CMV started moving to the time the blue CMV caught up to it. In that time, the yellow CMV only traveled about 70 cm. I know this because the yellow CMV started 100 cm away from the blue CMV, and ended at about 170 cm (170-100=70). I then calculated the speed of the yellow CMV (v = 170 cm – 100 cm / 4 seconds) and got 14.5 cm/s. This did not correlate with the speed of the yellow CMV we used for the other trial (v = 31.95 cm/s). This makes total sense because I had to do the catching up trial a day later. I found my yellow CMV in Mr. Vanucci’s room, which could explain why it ran slower a day later (other students used it, therefore the batteries lost charge).

The source of error included the batteries in the CMVs as well as the friction between the track we created to keep the CMVs on a straight path. Other classes used the CMV used in my trial. The reason the battery was a source of error is because the battery was used by the other classes while I used them after they did. I did my trial after school a few days later. Brand new batteries would have corrected this problem. The friction between the track and the CMV was because the tracks weren't perfectly aligned, thus creating friction. Also, the CMVs have a tendency to veer off in a different direction. The CMVs hit the sides of the track also causing friction. To correct this problem, I would make sure that the CMV stays straight to reduce the amount of friction.

Egg Drop Lab
Remzi Tonuzi and Lindsay Marella September 27th, 2011

Two sheets of computer paper taped together to make a cone, two sheets of shredded notebook paper on the inside.
 * Design:**

2 pieces of computer paper 1 piece of notebook paper Scotch tape
 * Materials:**

Prototype 1: This was our heaviest design, because at first we wanted to see how much shredded paper was needed to cushion the egg. We found that 2 sheets were plenty. Prototype 2: On our second test we took a single piece of paper and ripped it by hand which resulted in larger pieces and less absorption. The egg was unprotected, which resulted in a dry crack. Prototype 3: This was our final device, and we played it a safe, by using one sheet of heavily shredded notebook paper. It worked perfectly; the egg stayed intact, and (I believe) was the lightest in the class of the survivors. The only amendment that can be made to our design is possibly using less paper in the cone part to make it even lighter.
 * Results:**
 * || Prototype 1 || Prototype 2 || Prototype 3 ||
 * Egg weight (g) || 54.12 || 68.43 || 55.89 ||
 * Device || 34.99 || 13.80 || 18.49 ||
 * Device + egg || 89.11 || 82.23 || 74.30 ||


 * Acceleration:**

d- 8.5m Vi- 0 (egg was dropped) t- 1.57 (average of 1.97 and 1.16) a= ?

d=Vit + 1/2at 8.5m = (0m/s)(1.57s) + (1/2)(a)(1.57s^2) 8.5 = 1.23a a = 6.90m/s^2 This is less than g=9.80m/s^2 by 2.90m/s^2.

=1D Kinematics - Lesson 5 (Method 1)=

Free fall is when an object that is falling is only acted upon by the force of gravity. Ticker Tape diagrams can show the acceleration of an object during free-fall. The acceleration due to gravity is -9.8 m/s^2 on Earth. If an object is thrown up at a certain velocity, the negative velocity will slow it down (decelerating), but if the object is falling, whether with an initial velocity of 0 or some value, it is accelerating/ speeding up. Another way to represent the acceleration due to gravity is g. g= -9.8m/s/s (downward). g~10.0m/s/s. The formula for finding acceleration is A = ∆v/t.

Free fall can be represented by p-t and v-t graphs. -> V-T graph shows that it starts from rest and accelerates by -9.8m/s every second. Vf = g * t finds the ending speed of the object in free fall.

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

To find the distance traveled in free fall, the equation d= 1/2 g (or a) * t^2 When an object is dropped: Mass doesn't affect the rate at which the object falls. It has no influence on acceleration in free fall. *AIR RESISTANCE IS IGNORED* = = =Class Notes:=

9/7/11:
Speed and Types of Motion Notes: Average speed – take the total distance and total time

Constant speed – Going the same speed the entire time.

Instantaneous speed – It is the speed that you are going now. It is the change in speed.

Equation for all the above:

V = ∆d/∆t

m/s

Practice Problems




Free Fall Lab
Remzi Tonuzi and Julie Van Malden

**Objective**: What is the acceleration of a falling body?


 * Hypothesis**- The acceleration due to gravity should be around 9.8m/s2. The V-T graph should start at the origin and go down in a straight line with a negative slope of -9.8m/s2. You can tell the acceleration due to gravity by looking and finding the slope of the line from the data set.

Data Table -

Position Vs. Time Graph -



In this graph, y=Ax^2 + Bx is equal to d=v1t +1/2 at^2. This graph exhibits that A is half of the acceleration and B is the initial velocity. In this case, half of the acceleration is 4.59m or 459 cm, while the initial velocity is 44.78 cm/s or .4478 m/s. d=v1t +1/2 at^2 y=Ax^2 + Bx A= 1/2 acceleration B= initial velocity

Velocity Vs. Time Graph -



The line in this graph shows that y=mx+b, or vf= vi + at. Because the slope = velocity/time, it has the same value as acceleration. It also shows us that our group's acceleration is 907.4 cm/s (9.074 m/s). Also, it shows that they y intercept is -39.9.

% Error



% Difference




 * Discussion Questions-**
 * 1) Does the shape of your v-t graph agree with the expected graph? Why or why not?
 * 2) The shape of our graph doesn't agree with the V-T graph expected. This is because we made the velocity a positive while it was expected to have a negative slope.
 * 3) Our graph had a positive slope rather than a negative one.
 * 4) Does the shape of your x-t graph agree with the expected graph? Why or why not?
 * 5) Our graphs don't match since our falling positions were made positive rather than a negative position. This is because we decided to make our starting point zero and the distances that followed it to be positive, instead of measuring how high we were, using that as the starting point, and recording all the points that lie below it to be negative.


 * 1) How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.)
 * 2) Our result was closer to the actual speed rather than that of the class's.
 * 3) We had a % difference of 12.6 from the class and a % difference of 7.45 from the acceleration of the actual free fall.


 * 1) Did the object accelerate uniformly? How do you know?
 * 2) No. This is because the friction created between the ticker tape timer and the ticker tape going through it.
 * 3) This slowed down the acceleration of gravity in free fall thus slowing down the rate at which our object fell to the ground.


 * 1) What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?
 * 2) Factors like friction would cause the acceleration due to gravity to be lower then it should be. Factors like incorrect measure of the tickertape could cause
 * 3) the acceleration due to gravity to appear to be faster then it actually is.


 * Conclusion-**
 * The hypothesis that we generated wasn't correct due to the fact that many different variables were involved. G = 9.8 m/s/s (Acceleration = 9.8 m/s/s) as we**
 * expected except the friction between the ticker tape and the spark timer slowed down the acceleration of gravity. The V-T graph we made was incorrect and didn't**
 * come out as we expected since we had assumed that the line would have a negative slope. Because we measured the values as positive rather than negatives,**
 * it caused** **our slope to be positive, so down had a velocity and position that were both positive. Our percent error was about 7.45%. This was caused by the friction**
 * and the fact that our first dot was not marked precisely as the free fall object dropped. Also error could occur by inaccurate measuring. We measured in cm so**
 * our measurements weren't as accurate. If I could change this lab to address error, I would try to reduce the amount of friction of the ticker tape between the**
 * spark timer. I would try to get it through without any resistance so that acceleration due to gravity isn't affected as much. Using a better measurement system/tool**
 * would also benefit my results.**