Ch2_souflige

= = George Souflis Honors Physics Chapter two: 1D Motion

=Constant Speed= toc Lab: A Crash Course in Velocity (Part 1) Honors Physics


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

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


 * Hypotheses: ** I predict the CMV to move at a speed of 30 m/s. I traced with my finger for one second along a ruler and I think 30 cm/s is reasonable.

Position-Time Data y=30.025x R 2 =0.99991


 * Discussion questions **

Why is the slope of the position-time graph equivalent to average velocity? Slope is ∆y/∆x. Since x represents time and y represents position in a position-time graph, slope= (change in position)/(change in time). That, in turn, is equivalent to ∆d/∆t. By definition, that expression is equal to average velocity.

Why is it average velocity and not instantaneous velocity? What assumptions are we making? Instantaneous velocity only describes velocity at one point. Because of the several data points made during the run, average velocity is more appropriate. We assume that the CMV does not move at a completely constant velocity.

Why was it okay to set the y-intercept equal to zero? This is acceptable because we set the starting position at zero. In other words, the beginning of the run started when the CMV went from stationary into constant motion.

What is the meaning of the R2 value? This value evaluates the accuracy of the trend line made for the data points.

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? The slope of this graph would be less steep than our current graph, putting the graph “below” ours.


 * Conclusion **

The speed (or average velocity) of the CMV was, after graphical analysis, determined to be 30.025 cm/s. This is determined by the slope in the equation y=30.025x. Looking at the equation ∆d=vt, it is clear that the slope equals velocity in a constant speed graph. This result was surprisingly similar to my hypothesis of 30 cm/s. Several sources of error may have contributed to inaccuracies. Accidental shifting of the ruler and incorrect reading of the ruler are two issues that could be solved by using a measuring instrument such as a tape measure that is level with the spark tape. Also, having a more precise instrument would eliminate the inaccuracy associated with estimating the hundredth decimal of each measurement.

Class Notes 9.6.11: Constant Speed
Constant Speed = ∆d/∆t, no acceleration.

Notes: Graph Shapes (At Rest and Constant Speed)
No slope = at rest Positive/Negative Slope = Constant speed away/Constant speed towards

Lesson 1: Describing Motion with Words
Kinematics is the study of motion. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.

-Distance is the amount of ground an object has covered, and is measured by speed.

-Displacement describes how an object has changed position after motion. Velocity is the rate at which the object changes position.

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.

-The concept of direction awareness is much clearer. Distance, and therefore speed, do not take into account the direction of travel. Displacement, and therefore velocity, //do// take direction into account, because certain movements can cancel each other out if they are made in opposing directions (i.e 4m West + 4m East = 0m displaced).

-The meaning of constant speed vs. average speed is clearer. Constant speed refers to a linear, unchanging rate of travel. Average speed is used for an object that changes its speed throughout the travel.

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

-Everything is understood.

What (specifically) did you read that was not gone over during class today?

-Scalar and Vector quantities were not gone over in class. These tie in nicely with distance&speed and displacement&velocity. -Scalar quantities describe distances, which are measured by speed. They are ignorant of direction. -Vector quantities describe displacement, which is measured by velocity. They are direction-aware.

Notes: 9.9.11: Motion Diagrams and Ticker Tape Diagrams
V=∆d/∆t +This describes average speed as well, but do not take into account distance. __Types of Motion__ -at rest -constant speed -increasing speed }}}}}}}}}}}}}}}}}}}}}}}}}}}}Acceleration (changing speed) decreasing speed __(disregard underscores)__ || a=0 || **._._._._.** || --> a || **._.._..** || <-- a || **._._.__._.** || Ticker Tape gives a numerical value (can measure distance between marks), but do not indicate direction. - Signs are arbitrary. But are useful in describing direction.
 * __Types of Motion__ || __Motion Diagrams__ || __Ticker Tape__
 * at rest || v=0, a=0 || **.** ||
 * constant speed || --> --> -->
 * increasing speed || --> -> -->
 * decreasing speed || -> > -->

9.9.11

Lesson 2: Describing Motion with Diagrams
What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.

Vector diagrams were very easy to understand. They indicate velocity by using an arrow. The arrows length corresponds to a faster or slower velocity. Where the arrowhead points represents the direction the object is travelling. The lesson neglected to mention that acceleration arrows can be added under the velocity arrows to show acceleration.

Ticker Tape diagrams were easy to understand, especially after the CMV lab. Marks are made onto a tape at a constant interval as an object travels. After the run, the distance between each mark can be measured to determine changes in speed. Closer dots indicate a slow speed, while further spread dots indicate a faster speed.

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.

None.

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

None.

What (specifically) did you read that was not gone over during class today?

It was not specifically mentioned that vector diagrams represented vector quantities, and therefore track velocity&displacement. Ticker tape diagrams, being linear in nature, to scalar quantities and track speed&distance.

= Acceleration =

9.12.11

Class Activity: Graphing Acceleration
No Motion Graph

Constant Motion Graph Fast Motion Graph

Slow Graph


 * Discussion questions **
 * 1) How can you tell that there is no motion on a…
 * 2)  position vs. time graph - There will be no change in position over time.
 * 3)  velocity vs. time graph - velocity will equal zero.
 * 4)  acceleration vs. time graph - acceleration will be zero.


 * 1) How can you tell that your motion is steady on a…
 * 2)  position vs. time graph - linear slope, steady rise
 * 3)  velocity vs. time graph - constant positive value, no slope.
 * 4)  acceleration vs. time graph - similar to above, no slope w/ constant positive value.


 * 1) How can you tell that your motion is fast vs. slow on a…
 * 2)  position vs. time graph - linear slope, steep rise vs. linear slope gradual rise
 * 3)  velocity vs. time graph - constan t positive value, no slope vs. same thing
 * 4)  acceleration vs. time graph - constant positive value, no slope vs. same thing


 * 1) How can you tell that you changed direction on a…
 * 2) position vs. time graph - The slope became negative.
 * 3) velocity vs. time graph - The values became negative.
 * 4) acceleration vs. time graph - The values became negative.


 * 1) What are the advantages of representing motion using a…
 * 2) position vs. time graph - Shows clearly how far and object has traveled from the starting position.
 * 3) velocity vs. time graph - Better indicator of constant speed because shows no slope.
 * 4) acceleration vs. time graph - Clearly shows any change in velocity.


 * 1) What are the disadvantages of representing motion using a…
 * 2) position vs. time graph - The graph is misleading because it appears to show an increase in speed.
 * 3) velocity vs. time graph - Does not track distance.
 * 4) acceleration vs. time graph - Does not track distance or motion.


 * 1) Define the following:
 * 2)  No motion - no change in position. Object is at rest.
 * 3)  Constant speed - rate of change in position is constant over the entire run.

Notes: 9.13.11: The Big 5
v-t graphs tell the most info: distance, motion, and acceleration. +Negative values means towards the origin, positive away from origin +Slope of v-t graph = acceleration +Area = displacement

__Kinematics__ Acceleration (a) m/s^2 v= ∆d/∆t only for average or constant speeds...

v= (vi+vf)/2 only for average speed

a= (vf-vi)/∆t -> vf=vi+at

0.5(vi+vf) = ∆d/∆t

∆d = 0.5(vi+vf)t

∆d=vit + 0.5at^2

vf^2=vi^2 + 2a∆d

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

The definition of acceleration was already clear to me. I fully understand the concept that acceleration is the change in an object's velocity over time. Positive and negative values indicate direction. Constant acceleration vs. non-constant acceleration is also an easy idea to understand. A constantly accelerating object has an acceleration that changes at an unchanging rate. A non-constant accelerating object has different rates of acceleration at different times.

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.

Unfortunately I can't think of anything to write here.

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

The square relationship between total distance and the time of travel is confusing. The reading did not explain it in a way that I understood. I understand what it means, but I do not understand why it works or under what conditions it works.

What (specifically) did you read that was not gone over during class today?

Although we discussed how as a vector quantity acceleration regards direction, we did not discuss them in reference to vector diagrams. An object whose velocity is decreasing would have an acceleration opposite to the direction of motion - also making the acceleration a negative value.

9.14.11

Lab: Acceleration Graphs
w/ Kaila Solomon

Hypothesis: Like half of a parabola. Hypothesis: Change in velocity over time.
 * 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:**


 * Important Notes:**
 * Be sure to write your hypothesis BEFORE doing the experiment.
 * All data should be recorded in an organized spreadsheet in Excel.
 * You need to document your procedure. Decide with your lab partner how you intend to do this.

a) Interpret the equation of the line (slope, y-intercept) and the R2 value. The equation is a curve. For this reason, a polynomial trendline was used. As compared to the R2 value of a linear trendline, .898, the polynomial trendline an impressive .999 value. The same occurred in the "slowing down" graph.
 * Analysis:**

The curve, as a quadratic, has the form y=Ax2+Bx. In this case, x is time and y is position. Upon further examining, a related equation can be derived from the graph using this information (∆d=0.5at2+v i t). Logically, the slope of x2, or the A value, is half the acceleration and the B value is the initial velocity. In our case, a 20.796 cm/s/s acceleration (and a -33.03 cm/s/s acceleration for the decreasing speed graph) and an initial velocity of 2.7483 m/s (although it should be zero).

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.)

Acceleration of Cart (+two tangent lines) Instantaneous speed... ...at halfway = 18.38 m/s ...at end = 40.76 m/s

c) Find the average speed for the entire trip. 19.3625 cm/s

Acceleration Data


 * Discussion Questions:**

What would your graph look like if the incline had been steeper? The graph would’ve shown a faster acceleration, represented by a steeper curve.

What would your graph look like if the cart had been decreasing up the incline? The graph would look like the inverse (reflected over the line y=x) of the increasing speed graph.

Compare the instantaneous speed at the halfway point with the average speed of the entire trip.

These two speeds are incredibly similar. 18.38 m/s at halfway, and 19.3625 m/s as an average.

Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense?

The tangent line passes only through one point on the curve. A linear equation passing through just that one point would have a slope that represents the speed at that point. This is because slope is ∆y/∆x, which in this case is position/time.

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

Velocity of Cart over Time Graph

The results show that the graph is a curve. It was not half of a parabola as I had predicted, but similar in concept (y=Ax2+Bx). The graph can be used to find acceleration, or change in velocity over time, as I expected. More specifically, the A value of the equation of the curve gives ½ of the acceleration. Also, the equation shows the initial velocity of the object, the B value. In our curve, y=10.398x2+2.7483x, we found the acceleration to be 20.796 cm/s2, and the initial velocity to be 2.74 cm/s (although it should technically be 0). Sources of error include differences in the steepness of the incline, differences in the speed of the cart, and differences in measuring the distance between points. Multiple trials using the same cart and incline, measured by an exact device, would make for a constant result.
 * Conclusion: **

Lesson 3: Describing Motion with Position vs. Time Graphs
What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.

I already understand the meaning of slope in a p-t graph. Slope determines velocity, because when x is time and y is position, ∆y/∆x clearly indicates a velocity. This is true for constant speed graphs and acceleration graphs. A linear equation has a constant slope, indicating a constant speed. A curved line has a changing slope, indicating a changing speed.

I already understood how direction is depicted in the graphs. Rightward, upward slopes indicate a positive velocity. Leftward, downward slopes indicate a negative velocity.

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.

Unfortunately I can't think of anything to write here.

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

None.

What (specifically) did you read that was not gone over during class today?

None.

Lesson 4: Describing Motion with Velocity vs. Time Graphs
What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.

Most of what was discussed in the reading was already clearly understood. Velocity-time graphs are horizontal for a constant motion, positively sloped for increasing acceleration, and negatively sloped for decreasing acceleration. V-t graphs' slopes represent acceleration. Direction is shown by the region in which the graph resides: above the x-axis = positive direction, below the x-axis = negative direction.

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.

We only briefly covered how velocity-time graphs can be used to calculate displacement. I was unsure as to how to go about doing it. Now I understand that one finds the area of a specific portion of the graph to find the amount displaced during that time frame. This works for constant motion (rectangles) and acceleration (triangle or trapezoid).

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

None.

What (specifically) did you read that was not gone over during class today?

None.

Lab: A Crash Course in Velocity (Part II) w/ Kaila Solomon, Ben Weiss, Maddy Weinfeld.


 * Objectives ** :

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.

CMV1: 68.628 cm/s CMV2: 30.025 cm/s

(68.628)t+(30.025)t=600 t=6.08 seconds

CMV1's distance = (68.628)(6.08)=417.26 cm CMV2's distance = (30.025)(6.08)=182.55

The CMV's will collide after 6.08 seconds.

__Runs (Distance of each CMV when the two collided)__
 * Run # || CMV1 (distance in cm) || CMV2 (distance in cm) ||
 * 1 || 413.05 || 186.95 ||
 * 2 || 420.24 || 179.76 ||
 * 3 || 415.12 || 184.88 ||
 * 4 || 414.63 || 185.37 ||
 * 5 || 417.54 || 182.46 ||

__Results:__ After CMV1 traveled an average of 416.116 cm, the CMV's collided. Our hypothesis was accurate. CMV1 traveled 183.884 cm on average.

__Analysis:__ Percent Error and Difference The algebraic equations used to form our hypotheses proved incredibly accurate. We can accredit the small error to the fact that we used a different CMV than originally intended; the CMV we intended to use broke down. Also, it was difficult to accurately measure the exact point of collision because the two CMV's did not travel in perfectly straight paths, and we had to 'eyeball' the point of collision.

media type="file" key="New Project 3 - Mobile.m4v" width="300" height="300"


 * 1) 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.

Same CMV velocities were used.

(68.628)t-100=(30.025)t t=2.59 seconds

CMV1's distance = 177.75 cm CMV2's distance = 77.76 cm

CMV1 will catch up to CMV2 in 2.59 seconds, at position 177.75 cm. CMV2 will have traveled 77.76 cm.

__Runs (Distance of each CMV when CMV1 caught up to CMV2)__ __Results:__ After CMV1 traveled an average of 174.85 cm, it caught up to CMV2 (which had traveled 80.728 cm on average).
 * Run # || CMV1 (distance in cm) || CMV2 (distance in cm) ||
 * 1 || 173.25 || 82.25 ||
 * 2 || 178.53 || 76.97 ||
 * 3 || 174.68 || 80.82 ||
 * 4 || 170.99 || 84.91 ||
 * 5 || 176.81 || 78.69 ||

__Analysis:__

Percent Error and Difference These runs had a larger range of percent differences. Obviously we accredit some error to the fact that we were, again, using a different CMV than intended. However, other sources of error include our own discrepancy over where to call the measurement (where exactly CMV1 was to be considered "caught up") and the ramps we used to keep the CMVs on a straight track. CMV1 had to run over a pair of tracks, which did not run as smoothly as one ongoing track would run. Also, althought we assumed simultaneous starting time, there was of course a reaction time difference between the two starters after they heard "Go!".

media type="file" key="New Project 4 - Mobile 1.m4v" width="300" height="300"


 * Available Materials ** :

Constant Motion Vehicle, Tape measure and/or metersticks, Masking tape (about 30 cm/group), Stop watch, 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?

at 300 cm, exactly halfway. x-t graph for catching up. x-t graph for collision
 * 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.


 * 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?

V-t Graph No, there is no way to find where they meet up.


 * Conclusion **

After CMV1 traveled an average of 174.85 cm, it caught up to CMV2 (which had traveled 80.728 cm on average). The algebraic equations used to form our hypotheses proved incredibly accurate. We can accredit the small error to the fact that we used a different CMV than originally intended; the CMV we intended to use broke down. Also, it was difficult to accurately measure the exact point of collision because the two CMV's did not travel in perfectly straight paths, and we had to 'eyeball' the point of collision. In collision portion, CMV1 traveled an average of 174.85 cm before catching up to CMV2 (which had traveled 80.728 cm on average). These runs had a larger range of percent differences. Obviously we accredit some error to the fact that we were, again, using a different CMV than intended. However, other sources of error include our own discrepancy over where to call the measurement (where exactly CMV1 was to be considered "caught up") and the ramps we used to keep the CMVs on a straight track. CMV1 had to run over a pair of tracks, which did not run as smoothly as one ongoing track would run. Also, although we assumed simultaneous starting time, there was of course a reaction time difference between the two starters after they heard "Go!". To accommodate these errors, I think a single straight track with an automated, simultaneous release system would be ideal. Additionally, the use of video could be very useful. We could playback the runs and find the point of collision/catching up very exactly.

Egg Drop Project
Final Picture

Calculations for A:

d=8.5 m v initial = 0 t = 1.638 seconds a = ?

∆d = v(initial) + 0.5(a)(t^2) 8.5 = 0+0.5(a)(1.638^2) 6.34 m/s/s = a

Results:

The drop did not go as planned. While falling, the structure turned on its head. The box therefore had no board to brace it upon impact. The egg was completely scrambled after the drop.

Brief Analysis of why broke:

A main for the project's failure was the fact that it turned on its head. The lack of padding on the top made for a disastrous outcome. Also, the box weighed 158.02 g with the egg, a bulky size. We tried to sacrifice the efficiency of the box for more padding, a poor decision.

What would you do differently?

In retrospect, we should have constructed a cone rather than a box. A cone would have spread the force from impact over a longer period of time, instead of all at once. We were too narrow-minded in our approach.

10.3

Lesson 5 Notes
A free falling object is an object that is falling under the sole influence of gravity. Free-falling objects do not encounter air resistance and accelerate downwards at a rate of 9.8 m/s/s. 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. || || || ||

If the velocity and time for a free-falling object being dropped from a position of rest were tabulated, then one would note the following pattern. Observe that the velocity-time data above reveal that the object's velocity is changing by 9.8 m/s each consecutive second. A position versus time graph for a free-falling object is shown below.
 * **Time (s)** || **Velocity (m/s)** ||
 * 0 || 0 ||
 * 1 || - 9.8 ||
 * 2 || - 19.6 ||
 * 3 || - 29.4 ||
 * 4 || - 39.2 ||
 * 5 || - 49.0 ||

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 velocity versus time graph for a free-falling object is shown below.

Observe that the line on the graph is a straight, diagonal line. 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 dropped from a position of rest, the object will be traveling 9.8 m/s at the end of the first second, 19.6 m/s at the end of the second second, 29.4 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 vf = g * t

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. The actual explanation of why all objects accelerate at the same rate involves the concepts of force and mass. The acceleration of an object is directly proportional to force and inversely proportional to mass. Increasing force tends to increase acceleration while increasing mass tends to decrease acceleration. Thus, the greater force on more massive objects is offset by the inverse influence of greater mass. Subsequently, all objects free fall at the same rate of acceleration, regardless of their mass.

Lab: What is the acceleration of a falling body?
Objective: **What is the acceleration of a falling body?** Hypothesis: The acceleration of a falling body should be 9.8 m/s/s (980 cm/s/s). The position-time graph on the left shows a curve as expected. An object in freefall constantly accelerates 981 cm/s/s. Though quantitatively the graphs are not entirely accurate, the graphs qualitatively shows the basic shape that occurs for an object in freefall. In the velocity-time graph, which should be linear, the slope should equal the acceleration. 856.73 cm/s/s was our result based on our data. The data for the v-t graph came from finding the instantaneous speed at the mid-time of every 0.1 second interval. Instantaneous speed was found by doing final position (at every .1 interval) minus initial position over final time minus initial time. For example, (1.58-0)/(0.1-0) = 15.8 cm/s. Our R 2 values showed that our data points were consistent with the expected graph shapes (values of above .98).
 * Class Data ||  ||
 * || ** Period 2 ** ||
 * || 754.43 ||
 * || 856.73 ||
 * || 851.07 ||
 * || 891.38 ||
 * || 891.12 ||
 * || 798.13 ||
 * || 710.65 ||
 * || 755.87 ||
 * || **// 659.39 //** ||
 * || **// 1225.4 //** ||
 * Average || ** 839.417 ** ||
 * x-t and v-t graphs**

While we were a significant percentage off of the theoretical value for freefall acceleration (981 cm/s/s), we were very similar to the class average (838.417 cm/s/s)
 * % error and difference**

Not entirely. I expected the graph to be negative because the object is starting at zero and moving downward. However, I did expect it to be linear, which it was. Yes. I was not at all surprised to see a curved x-t graph. An accelerating object will always have a curved x-t graph. Compared to the class average (839.417 cm/s/s) we found only a 2.06% difference to ours (856.73 cm/s/s).
 * 1) Does the shape of your v-t graph agree with the expected graph? Why or why not?
 * 1) Does the shape of your x-t graph agree with the expected graph? Why or why not?
 * 1) How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.)
 * 1) Did the object accelerate uniformly? How do you know?

Yes. Objects in freefall are only acted on by gravity, which has a constant value of g=981 cm/s/s. The v-t graph has a linear slope. The slope of a v-t graph is acceleration.

There virtually should be no reason for acceleration due to gravity to be higher than 981 cm/s/s. A heavier object would not change the result. Throwing the object downward would not either because it would still be only acted on by gravity after leaving your hand. There are several reasons why it could be lower, however. For one, air resistance. If the objects surface area is large relative to its mass, it will fall slower. In our case, friction also plays a large role that can slow down the object's descent. The spark tape needs to run through our fingers and through the spark timer to fall, which can considerably slow down the freefall.
 * 1) What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?

Conclusion:

We found the accleration of our freefall object to be 856.73 cm/s/s, 12.67% error relative to the theoretical value of 981 cm/s/s. Still, we matched the class average of 839.417 cm/s/s fairly well, with only a 2.06% difference. Our results indicate that our object fell slower than the acceleration of gravity. There are several sources of error which contributed to this result. Friction played a large role in the object's descent. The spark tapes needs to run through our fingers and through the spark timer to fall, which can considerably slow down the descent. The measurements were also a source of error. It was difficult to keep the long spark tape from shifting around. In spite of all this, the results were decent. To fix these errors, I would use a motion detector instead of a ticker tape, because that would prevent friction and create a more accurate graph in turns of measurement.