Ch2_BarinskyL

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Chapter 2.

Lesson 1
9/7/11
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) From the class discussion, I had already understood what distance and displacement were and the difference between them. Distance is a scalar quantity that describes how much length an object has covered in total. Displacement is a vector quantity that describes how far an object has travelled in relation to its original position. If a ball rolls 3 ft, hits a wall, and rolls back 2 ft in the same direction, its distance rolled is 5 ft and its displacement is 1 ft.
 * 3) I also understood how to find the average speed of an object. The distance travelled is divided by the time it took to travel. If someone drives 100 miles in 4 hours, their average speed is 100mi/4h = 20 mph.
 * 4) 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.
 * 5) Before the reading, I was confused about the difference between speed and velocity. I now understand that speed is a scalar quantity that describes how fast an object is moving, while velocity is a vector quantity that describes the rate at which an object changes its position. Speed is like the rate at which an object covers distance with no indication of its starting and ending positions. Velocity relates to the direction that the object is traveling. If an object moves 5 ft north in 1 minute, its speed is 5 ft/min and its velocity is 5 ft/min, north.
 * 6) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 7) I do not have any questions after reading the material.
 * 8) What (specifically) did you read that was not gone over during class today?
 * 9) The difference between instantaneous speed and average speed was not mentioned in class, but after reading the definitions, they are easy to understand. Instantaneous speed is the speed of an object at any given time, while average speed is an average of all instantaneous speeds. If I am walking at 2 mph for 3 seconds and then at 5 mph for another 2 seconds, my instantaneous speed at 2 seconds is 2mph and my average speed for the entire 5 seconds is 3.2 mph.

Notes: Constant Speed
9/8/11

Average speed: average Constant speed: unchanging Instantaneous speed: speed at a certain time For all : aV = Δd /Δt

4 types of motion: at rest, constant, increasing, decreasing

Acceleration: change in velocity (speeding up or slowing down)

Motion diagram show velocity and acceleration

At rest V=0, a=0

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

increasing v.... v...... v -> > > a - -> acceleration points in the same direction as velocity

decreasing v.......... v...... v > -> --> ...a <-- acceleration points in opposite direction from velocity

Lesson 2
9/8/11
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) From the class discussion, I had already understood how ticker tape is used. It is attached to a moving object and marks are made on the tape every x seconds. These marks can be observed afterwards to analyze the movement of the object. A larger gap between marks indicates a faster speed than a smaller gap. Therefore, if the gap sizes are increasing along the tape, the object was accelerating. If the gap sizes are decreasing along the tape, the object had a negative acceleration. If the gap sizes remain constant, the object had a constant speed.[[image:Screen_shot_2011-09-08_at_9.07.53_PM.png width="410" height="178"]]
 * 3) I also understood how vector diagrams are used. If the sizes of the arrows in a diagram are all the same, the velocity remains constant. If the arrow sizes increase frame by frame, the velocity is increasing (accelerating).
 * 4) 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.
 * 5) I was not confused about anything during class that was made clear after the reading.
 * 6) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 7) What I was a little unclear with from class and still am a little confused with is certain vector diagrams like the one below.
 * 8) [[image:Screen_shot_2011-09-08_at_9.14.41_PM.png width="162" height="174"]]
 * 9) Why are the arrows pointing downwards, and why are they all the same size?
 * 10) What (specifically) did you read that was not gone over during class today?
 * 11) Everything that I read was mentioned in class.

Lab: Speed of CMV
Lab Partner: Garrett Almeida 9/9/11

__Objectives__:1) How precisely can you measure distances with a meter stick?2) How fast does your CMV move?3) What information can you get from a position-time graph? __Hypotheses__:1) I believe we can measure distances in millimeters with a meter stick.2) The CMV travels 50 cm/s.3) We can get the CMV's speed and velocity.-The speed will be the slope of the position-time graph. Since we will have the position, we can find the velocity with the below equation. __Materials__:Spark timer & spark tape, meter stick, masking tape, CMV __Procedure__:1. Attach one end of the spark tape to the CMV using tape and insert the remainder of the spark tape into the spark timer.2. Turn on the spark timer, then turn on the CMV.3. Let the CMV run until the length of the spark tape has passed through the spark timer.4. Isolate eleven points on the spark tape. With a meter stick, measure the distance from the first point to the second point, first to the third, first to the fourth, etc.5. Enter the measurements in a data table showing time and position. This allows you to make a chart and find the average speed of the CMV. __Data__:



__Analysis__: As evident from the collected data, there are no outliers or skewed data points that suggest the failure of this experiment. The data is believable and makes sense. All of the points fall within close proximity to the line of best fit. This line was set to be linear because it appears as though the points fall in a straight line, and the R 2 value of 0.988 suggests its accuracy.

__Discussion Questions__:
 * 1) Why is the slope of the position-time graph equivalent to average velocity?
 * Slope is defined by the change in y over the change in x. In this case, the slope is change in cm over the change in seconds. Average velocity is the change in position over the change in time. Therefore, the slope is the average velocity.
 * 1) Why is it average velocity and not instantaneous velocity? What assumptions are we making?
 * If we used instantaneous velocity, we would only have the velocity of the CMV at a certain time utilizing only two data points. Using all of the data points allows us to get an average velocity. We are assuming that the CMV is indeed going at a constant speed for the entire second that we collected data from.
 * 1) Why was it okay to set the y-intercept equal to zero?
 * We want the line of best fit to pass through (0,0) because we know for sure that at 0 seconds the CMV had a velocity of 0.
 * 1) What is the meaning of the R2 value?
 * The R2 value tells us how accurate the line of best fit is according to our data. The closer the value is to 1, the more accurate. If the value is 99, for example, the line is 99% accurate.
 * 1) If you were to add the graph of another CMV that moved more slowly on the same axes as your current graph, how would you expect it to lie relative to yours?
 * The graph of a slower CMV would have a smaller slope because it would have a lower velocity. It would appear on the graph less steep and extend under the line of the fast CMV.

__Conclusion__: Our CMV ended up traveling at 59.821 cm/s, almost 10 cm/s more than I had predicted. We could have measured the distance in mm with the meter stick, but the more accurate way was to measure in cm. I was correct when I had predicted to be able to get the speed, position, and velocity from the graph. A source of error in this experiment could have came from when we measured the distances between the points on the spark tape; the meter stick could have moved without us knowing, and it is harder to judge what the correct value on a measuring tool is when it is raised off of the surface. This could be fixed by taping down tape measure next to the spark tape for measurement. Another source of error could have came from deciding which points on the spark tape to use. The CMV could have not yet reached its full speed at the time its position was recorded and used as a data point. This could be fixed by using a longer piece of spark tape, letting the CMV run for a few seconds, and using the last points on the tape for the data.

Activity: Graphical Representations of Equilibrium
9/12/11 __At rest__: __Constant speed, slow__: (#3 away, #4 towards)

__Constant speed, fast__: (#5 away, #6 towards)
 * 1) How can you tell that there is no motion on a…
 * 2) position vs. time graph
 * 3) The position remains constant.
 * 4) velocity vs. time graph
 * 5) The velocity remains at 0.
 * 6) acceleration vs. time graph
 * 7) The acceleration remains at 0.
 * 8) How can you tell that your motion is steady on a…
 * 9) position vs. time graph
 * 10) The position changes at a steady rate (constant slope).
 * 11) velocity vs. time graph
 * 12) The velocity remains constant.
 * 13) acceleration vs. time graph
 * 14) The acceleration remains at 0.
 * 15) How can you tell that your motion is fast vs. slow on a…
 * 16) position vs. time graph
 * 17) The line of the faster speed will reach positions before the line of the slower speed (faster has larger slope).
 * 18) velocity vs. time graph
 * 19) The faster motion will show a larger absolute value of velocity.
 * 20) acceleration vs. time graph
 * 21) The absolute value of the acceleration will be larger on the graph of faster motion.
 * 22) How can you tell that you changed direction on a…
 * 23) position vs. time graph
 * 24) The line will form a corner. The position will suddenly decrease if it was increasing before.
 * 25) velocity vs. time graph
 * 26) The velocity will cross over the x axis and change signs.
 * 27) acceleration vs. time graph
 * 28) The acceleration will cross over the x axis and change signs.
 * 29) What are the advantages of representing motion using a…
 * 30) position vs. time graph
 * 31) You can see where the object is at what time and determine a speed.
 * 32) velocity vs. time graph
 * 33) You can see when the object changes direction and how fast it is going in that direction.
 * 34) acceleration vs. time graph
 * 35) You can see how rapidly the speed of the object is changing and what direction it is traveling.
 * 36) What are the disadvantages of representing motion using a…
 * 37) position vs. time graph
 * 38) Does not clearly show the velocity, and it may be hard to determine when the object is accelerating.
 * 39) velocity vs. time graph
 * 40) Does not show where the object is and its acceleration.
 * 41) acceleration vs. time graph
 * 42) Only compares the velocity of the object to its previous instantaneous velocity.
 * 43) Define the following:
 * 44) No motion
 * 45) The position remains constant over time. Velocity and acceleration are zero.
 * 46) Constant speed
 * 47) The rate of change between position and time remains constant. Acceleration is zero.

Notes: The Big 5
9/13 Example: With a cruising speed of 128 m/s, the French supersonic passenger jet Concorde is the fastest commercial airplane. Suppose the landing speed of the Concorde is 20.0 percent of the cruising speed, 25.6 m/s. If the plane accelerates at -5.80 m/s2, how far does it travel between the time it lands and the time it comes to a complete stop?

Lesson 1e
9/13/11
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) I understand that acceleration is a vector quantity. It is the rate of change of velocity. If an object is changing its velocity, it is accelerating. Acceleration can be negative.
 * 3) [[image:Screen_shot_2011-09-20_at_8.05.08_PM.png]]
 * 4) I also understand that if an object is slowing down, its acceleration is in the opposite direction of its motion. This is a general rule of thumb.
 * 5) 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.
 * 6) Constant acceleration is when an object's velocity is changing at a constant rate. This is different than an object with constant velocity, for this object is not accelerating. I just found this a little unclear at first but it makes sense to me now.
 * 7) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 8) I do not have any questions.
 * 9) What (specifically) did you read that was not gone over during class today?
 * 10) We did not talk about the motion of a free-falling object. These objects usually accelerate as they fall. If an object is accelerating by 10 m/s, it has constant acceleration. The total distance traveled is directly proportional to the square of the time.

Notes: Shapes of Graphs
9/14/11

At Rest and Constant Speed:

Increasing and Decreasing Speed:

Activity: Motion Graphs with Cart
9/14/11 #6: cart speeding up down ramp, away from sensor #15: cart speeding up down ramp, towards sensor #7: cart pushed up ramp towards sensor #11: cart pushed up ramp away from sensor

Lesson 3
9/14/11
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) I understood position-time graphs. If there is no acceleration and the object is moving at a constant rate, the line is straight (constant slope) (first graph). If there is acceleration and the object is speeding up, for example, the line is a curve (second graph.
 * 3) [[image:U1L3a2.gif width="213" height="121"]][[image:U1L3a4.gif width="222" height="118"]]
 * 4) I also understood how slope is important on a position-time graph. The slope is the rate at which y (position) changes in relation to x (time). A larger absolute value slope means that the object is moving faster. If the slope is positive, the object is moving away/right. If it is negative, the object is moving closer/left. From the slope, you can determine the object's speed. The slope is equal to the object's velocity at that point in time.
 * 5) 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.
 * 6) I was not confused about anything from class.
 * 7) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 8) I do not have any questions.
 * 9) What (specifically) did you read that was not gone over during class today?
 * 10) Everything in this section was gone over in class.

Lesson 4
9/14/11
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) I understood the important of the shape of a velocity-time graph. Unlike a position-time graph, constant speed is shown by a horizontal line. This is because although the position is changing, the velocity is remaining constant and is not changing. There is no acceleration in this case. If there is acceleration, the velocity is changing. This may be shown as a slanted line or a curve. the slope of which is the acceleration at a certain point. The first graph shows positive velocity with zero acceleration, and the second graph shows positive velocity with positive acceleration.
 * 3) [[image:U1L4a4.gif width="105" height="93"]][[image:U1L4a5.gif width="105" height="93"]]
 * 4) I also understood the important of slope of a velocity-time graph. If there is zero slope, there is no acceleration. If the line is slanted or curved, its slope is the object's acceleration at the point that the slope is calculated.
 * 5) 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.
 * 6) I understood everything from class.
 * 7) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 8) I have no questions.
 * 9) What (specifically) did you read that was not gone over during class today?
 * 10) To determine the slope of a velocity-time graph (acceleration), you first pick two points on the line and determine their coordinates. You then calculate the rise (difference in y-coordinates) and the run (difference in x-coordinates). You divide the difference in y-coordinates by the difference in x-coordinates.
 * 11) To determine displacement of an object, you find the area on the velocity-time graph.
 * 12) [[image:U1L4e2.gif width="277" height="106"]]

Lab: Acceleration Graphs
Lab Partner: Garrett Almeida 916/11 __Objectives:__ 1) What does a position-time graph for increasing speeds look like2) What information can be found from the graph? __Hypotheses__: 1) The line is a J-curve resembling that of an x^2 graph. The position increases with time at a faster rate at each point. -The slope of the line will be increasing as time goes on because the cart is accelerating and increasing its velocity. 2) We can find the instantaneous speed at each point and the average speed for the entire graph. -Speed is distance / time, both of which will be shown on the graph.

__Materials__: Spark tape, spark timer, track, dynamics cart, measuring tape

__Procedure__:

media type="file" key="lab acceleration grpahs procedure.mov" width="270" height="270" for increasing speed: media type="file" key="lab acceleration grpahs procedure 2 - increasing.mov" width="270" height="270" for decreasing speed: media type="file" key="lab acceleration grpahs procedure 3 - decreasing.mov" width="270" height="270"

__Data__:

__Analysis__**:** When adding the lines of best fit, we determined that the best choice would be polynomial lines. When we selected linear lines, the R 2 values were not high enough and the lines did not appear to fit well. The R 2 values when we selected linear lines for the increasing speed and decreasing speed graphs were 0.938 and 0.830, respectively. When we switched them to polynomial, they were both 0.999. Since the R 2 value is a measure of accuracy of the line of best fit, we decided that the polynomial lines with R 2 values extremely close to 1 would be more accurate. We agreed that the y-intercepts of both lines should be set to 0 because in this situation p=0 when t=0.



We can plug the A and B values from the equations into another equation (see image for increasing speed above) to find the acceleration and initial velocity. The acceleration for the increasing speed graph was found to be 32.378 cm/s, while the initial velocity was found to be 8.0125. The velocity at t=0 should be 0 since the cart is at rest at this time. This value is slightly off, possibly because the spark timer was not started at the perfect time and our interpretation of t=0 was actually in between two sparks. The average speed of the cart can be found by determining the average slope of the line. The average speed for the increasing speed graph is 24.21 cm/s, and the average speed for the decreasing speed graph is 20.46 cm/s. The instantaneous speed of the cart at a certain point can be found by determining the slope of the line at that point. The slope of a curved line at point X is the slope of the tangent line that goes through point X. The slope was found by printing out the graph, drawing a tangent line, and selecting points on the tangent to calculate the slope. For the increasing speed graph, the speed was 24.07 cm/s at the halfway point and 40.55 cm/s at the end. For the decreasing speed graph, the speed was 19.87 cm/s at the halfway point and 0.15 cm/s at the end.



__Discussion__:
 * 1) What would your graph look like if the incline had been steeper?
 * 2) The value of |A| would have been higher. The graph would be stretched vertically.
 * 3) What would your graph look like if the cart had been decreasing up the incline?
 * 4) It would resemble that of the left half of a -x 2 graph, and it would be shallower.
 * 5) Compare the instantaneous speed at the halfway point with the average speed of the entire trip.
 * 6) For both the increasing and decreasing speed graphs, the instantaneous speeds at the halfway point and the average speeds of the entire trip were very close to each other, differing by less than 1 cm/s.
 * 7) Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense?
 * 8) The tangent line at a point on a curve goes through only that point of the curve. Since the slope=speed, the slope of a curve at a specific point is equal to the slope of the line that goes through only that point (the tangent line).
 * 9) Draw a v-t graph of the motion of the cart. Be as quantitative as possible.
 * 10) This graph was made by plotting 3 known points on the v-t graph: (0,0), (0.5, 20.47), and (1,40.55). The second two were found when calculating the velocity at the halfway point and at the end. I set the trendline to be linear because we know that the acceleration is constant and the velocity graph in this case would be a straight line, increasing at a steady rate.
 * 11) [[image:Screen_shot_2011-09-22_at_7.59.31_PM.png]]

__Conclusion__: Our hypotheses were correct. The graph for increasing speed did end up having a J-curve and looking like the right half of a x 2 graph. The position increased with time at a faster at each point. We also found that we could find the instantaneous speed at each point and the average speed for the entire graph. To do this, all we needed was the position at certain t values to find the change in position and change in time. A source of error could be that the spark timer is not perfectly in sync with the cart, so, for example, the point could have been made on the spark tape .05 s before we actually assumed. This error could be fixed by figuring out exactly how hard to push the cart up the incline, for example, to make sure it takes exactly 1 second to reach a stop, and having a spark timer that started automatically at the forward motion of the cart. Another source of error could be the difference in interpretations of measurements between points. I measured the distances on the spark tape of the decreasing speed trial, while Garrett measured those for the increasing speed trial. What I interpret as 4.18, for example, Garrett may have seen as 4.15. The chances of this error influencing the data could be decreased by having the same person measure the distances for both trials.

Interpreting Position-Time Graphs
9/19/11

__//A. The graph below shows the motion of a car.//__
 * 1) Describe the motion of the car (qualitatively) during each segment. (__all answers shown below)__
 * 2) Describe the change in position of the car during each segment.
 * 3) Calculate the velocity of the car during each segment.
 * 4) What is its //average speed// for the entire 11-s?
 * 5) What is its //average velocity// for the entire 11-s?
 * 6) What is the acceleration of the car during each segment?
 * 7) Sketch a v-t graph that corresponds to this x-t graph.

v=d/t || Accelera-tion (m/s2) || 4) Average speed V = d / t = total distance / total time = 56 / 11 = 5.1 m/s
 * Segment || Description || Time (s) || Initial Position (m) || Final Position (m) || Displacement (m) || Velocity (m/s)
 * AB || At rest || 2 || 10 || 10 || 10 - 10=0 || 0/2 = 0 || 0 ||
 * BC || Constant || 2 || 10 || 0 || 0 - 10=-10 || -10/2 = -5 || 0 ||
 * CD || At rest || 1.5 || 0 || 0 || 0 - 0=0 || 0/1.5 = 0 || 0 ||
 * DE || Constant || 0.5 || 0 || -16 || -16 - 0=-16 || -16/0.5 = -32 || 0 ||
 * EF || At rest || 2 || 0 || 0 || -16 – -16 = 0 || 0/2=0 || 0 ||
 * FG || Constant || 1 || -16 || 0 || 0 - -16 = 16 || 16/1=16 || 0 ||
 * GH || constant || 2 || 0 || 14 || 14 – 0 = 14 || 14/2 = 7 || 0 ||

5) Average velocity V = d / t = change in position / total time = 14-10 / 11 = 0.36 m/s

7) V-t graph

__//B. The graph below shows the motion of a helicopter. (Label each endpoint with a letter, beginning with A at 0 s, B at 1 s, C at 2 s, D at 4 s, E at 5 s, F at 7 s, G at 8 s, H at 10 s.)//__ v=d/t || Accelera-tion (m/s2) ||
 * 1) Describe the motion of the car (qualitatively) during each segment. (__all answers shown below)__
 * 2) Describe the change in position of the car during each segment.
 * 3) Calculate the velocity of the car during each segment.
 * 4) What is its //average speed// for the entire 10-s?
 * 5) What is its //average velocity// for the entire 10-s?
 * 6) What is the acceleration of the car during each segment?
 * 7) Sketch a v-t graph that corresponds to this x-t graph.
 * Segment || Description || Time (s) || Initial Position (m) || Final Position (m) || Displacement (m) || Velocity (m/s)
 * AB || Constant || 1 || 0 || .4 || .4 || 0.4 || 0 ||
 * BC || Constant || 1 || .4 || 3.2 || 2.8 || 2.8 || 0 ||
 * CD || Constant || 2 || 3.2 || 0 || -3.2 || -1.6 || 0 ||
 * DE || Constant || 1 || 0 || 3.2 || 3.2 || 3.2 || 0 ||
 * EF || Constant || 2 || 3.2 || 12.8 || 8.6 || 4.3 || 0 ||
 * FG || Constant || 1 || 12.8 || 13.2 || .4 || 0.4 || 0 ||
 * GH || Constant || 2 || 13.2 || 7.2 || -6.0 || -3.0 || 0 ||

4) Average speed V = d / t = 24.6 / 10 = 2.46 m/s

5) Average velocity V = d / t = 7.2-0 / 10 = 0.72 m/s

7) V-t graph

__//C. The graph below shows the motion of an airplane. (Label each endpoint with a letter, beginning with A at (0,0) and ending with E at (100,0).)//__ v=d/t || Accelera-tion (m/s2) ||
 * 1) Describe the motion of the car (qualitatively) during each segment.
 * 2) Describe the change in position of the car during each segment.
 * 3) Calculate the velocity of the car during each segment.
 * 4) What is its //average speed// for the entire 100-s?
 * 5) What is its //average velocity// for the entire 100-s?
 * 6) What is the acceleration of the car during each segment?
 * 7) Sketch a v-t graph of the motion shown in the x-t graph.
 * Segment || Description || Time (s) || Initial Position (m) || Final Position (m) || Displacement (m) || Velocity (m/s)
 * AB || Constant || 40 || 0 || 400 || 400 || 10 || 0 ||
 * BC || At rest || 30 || 400 || 400 || 0 || 0 || 0 ||
 * CD || Constant || 20 || 400 || 200 || -200 || -10 || 0 ||
 * DE || Constant || 10 || 200 || 0 || -200 || -20 || 0 ||

4) Average speed V = d / t = (400 + 200 + 200) / 100 = 8 m/s

5) Average velocity V = d / t = 0 / 100 = 0 m/s

7) V-t graph

__//D. The graph below shows an x-t graph.//__
 * 1) Describe the motion of the object (qualitatively). __Moving away__
 * 2) Calculate the instantaneous velocity of the car at 0.2 seconds. __190 cm/s__
 * 3) Calculate the instantaneous velocity of the car at 0.4 seconds. __380 cm/s__
 * 4) What is its //average speed// for the entire 0.5-s? __240 cm/s__
 * 5) Sketch a v-t graph of the motion shown in the x-t graph. __(shown in right of picture)__



E. The graph below shows an x-t graph.
 * 1) Describe the motion of the object (qualitatively). __Moving away, slowing down, then moving towards speeding up__
 * 2) Calculate the instantaneous velocity of the car at 1 seconds. __31 cm/s__
 * 3) Calculate the instantaneous velocity of the car at 2.5 seconds. __0 cm/s__
 * 4) Calculate the instantaneous velocity of the car at 5 seconds. __-46 cm/s__
 * 5) What is its //average velocity//for the entire 0.5-s? __0 cm/s__
 * 6) Sketch a v-t graph of the motion shown in the x-t graph. __(shown in right of picture)__

^^equation used: Velocity = Δposition /Δtime = (p2 - p1) / (t2 - t1)

Notes: Catch-up Problems
9/21/11 A car is behind a truck going 30 m/s on the highway. The car’s driver looks for an opportunity to pass, guessing that his car can accelerate at 2 m/s2. He gauges that he has to cover the 15-m length of the truck, plus 10 m clear room at the rear of the truck and 10-m more at the front of it. In the oncoming lane, he sees a car approaching, probably traveling at 20 m/s. He estimates that the car is about 450-m away. Should he attempt the pass?



Since we calculate that the the truck will travel 177 m away and the other car with travel 118 m towards the main car in the time that it would take the main car to pass the truck, we can add these two numbers to get 295 m as the minimum amount of space there needs to be between the other car and the truck at the start. Since we are given that there is 450 m between them, and 295 < 450, we figure that the main car does have room to pass the truck.

Lab: Crash Course in Velocity
Lab Partners: Garrett Almeida, Matt Ordover, Maddie Margulies 9/23/11 __Objectives__: __Hypothesis/Calculations__:

Based on the above calculations, the carts will crash at 137.08 cm in the first situation and at 141.97 cm in the second situation.

__Materials__: Blue CMV, yellow CMV, tape measure

__Procedure__: Part 1:media type="file" key="lab 3 procedure part 1.mov" width="270" height="270"Part 2:media type="file" key="lab 3 procedure part 2.mov" width="270" height="270"

__Data__: Points at which the two carts met (cm):
 * Trial || Part 1 || Part 2 ||
 * 1 || 160 || 147 ||
 * 2 || 150 || 148 ||
 * 3 || 198 || 147 ||
 * 4 || 161 || 146 ||
 * 5 || 152 || 148 ||

__Analysis__: . . . ..

We ran five trials for each part of the experiment. Doing less may lead to inaccurate and confusing results. Percent error is a measure of accuracy determined by comparing the theoretical value (what we should have gotten) to the experimental value (what we did get). Percent difference is a measure of precision determined by comparing the average experimental value with each individual experimental value. It tells us how precise we were in our measurements from trial to trial. As a general rule, a percent error that is less 10 usually suggests the success of an experiment. For the first part of the experiment, the second trial proved to be the most accurate, yielding a 9.43% error. Aside from maybe the fifth trial, the rest proved to be rather inaccurate, especially the third one. All of the trials in the second part of the experiment proved to be very accurate, the highest % error being 4.25. I believe that the second part of the experiment yielded better results because it required the carts to travel for a smaller amount of time (theoretically 2.37 s) than the first part did (theoretically 7.74 s). This meant that the sources of error (discussed in Conclusion) had far less time to influence the data. Perhaps as time went on, the data became more and more inaccurate. As an example, the below calculations show how the % error and % difference were determined for the first trial of the first part of the experiment.

__Discussion__:
 * 1) Where would the cars meet if their speeds were exactly equal?
 * 2) In the first situation, the cars would meet at middle at 300 cm. The velocities of both cars and the time they would travel for would be equal, so the only variable that would be left in the velocity equation is distance.
 * 3) In the second situation, the cars would never meet. The second car would be moving away from the first car at the same speed as the first car is traveling, so the distance in between the two cars would never change.
 * 4) 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.
 * 5) [[image:Screen_shot_2011-09-25_at_12.56.54_PM.png width="362" height="106"]]
 * 6) The crashing point is where the two lines meet because that is when they are at the same position at the same time.
 * 7) 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?
 * 8) [[image:Screen_shot_2011-09-25_at_12.57.13_PM.png width="373" height="124"]]
 * 9) The crashing points of the carts cannot be found from v-t graphs without knowing some information about the carts' starting positions.

__Conclusion__: From the data, we can conclude that the CMVs meet at around 156 cm in the first situation and at around 147 cm in the second situation. Our calculations at the beginning of this experiment proved to be pretty accurate for the second part of the experiment, which yielded percent errors ranging from 2.84 to 4.25, but not for the first part, which yielded percent errors from 9.43 to 44.44. The percent differences ranged from 1.95 to 20.58 for the first part and from 0.14 to 0.82 for the second part. One main source of error is the path of blue CMV. The CMV would not move in a straight line, therefore taking more time to reach the position of the yellow CMV, adding to the distance of "crashing" point. Also, since the CMVs would not actually hit, we had to estimate the point at which they would crash by watching where they pass each other. This experiment could be redone more accurately if we put the CMVs on a track to ensure that they both move in a straight line and hit at some point. Another source of error is that our blue CMV that we had previously calculated the speed of was not working, so we had to use another blue one that may have travelled at a different speed than the one we used in our calculations. This could have been fixed by calculating the speed of the new blue CMV and using that value instead of that of the old one.

Egg Drop
9/28/11 __My Device:__ My design included a multi-layer cushioning of straws at the bottom (with spacing to minimize mass) followed by a spring-simulating bundle of straws to absorb the rest of the shock. This bundle contains straws positioned in a circle each with a small slit on the side facing the inside of the bundle. They were taped together at the top and bottom, but slits were made in the tape to allow for flexibility. When pressure is put on this feature of the device, the straws jut outwards. This allows the velocity of the egg cradled in straws above it to decrease over time before coming to a complete stop, minimizing the effect of the impact. There is also a long parachute attached to the top to minimize the acceleration of the device. The parachute has a small hole cut into the top so that some air can pass through and the device does not turn over midair. The other straws not mentioned serve to either hold everything together or prevent breakage upon falling over after impact. The benefit of this device being long and having weight at the bottom is that it falls straight and does not sway while falling.

__Results__: The egg survived the drop. It took 1.55 s to hit the ground. The device was 35.78 g and the egg was 58.56 g, totaling 94.34 g. __Analysis__:Calculation of acceleration: d = .5t + .5at 2 - 8.5 = (.5 x 1.55) + .5a(1.55 2 )a = -6.43 m/s 2 Not taking air resistance into account, the acceleration on a free-falling object due to gravity is -9.8 m/s 2. My parachute helped lower the absolute value of the acceleration of my device to -6.43 m/s 2. This plus the cushioning and shock-absorbing features was what allowed for the egg to survive the impact. If I had to do this project again, I would probably try to figure out a way to make more effective use of the spring-like bundle of straws. I think that it did help, but if it were perhaps not taped in place to the side supports, it would be able to compact freely from the rest of the device. Since it cannot do that with my current device, its shock-absorbing ability is limited.

Lesson 5
10/3/11

A free falling object is an object that is falling under the sole influence of gravity. Two important motion characteristics that are true of free-falling objects:
 * 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

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. The fact that the distance that the object travels every interval of time is increasing is a sign that the ball is speeding up as it falls downward. If an object travels downward and speeds up, then its acceleration is downward.

Acceleration of gravity (g): the acceleration for any object moving under the sole influence of gravity. G is different in different gravitational environments (different planets).



A position-time graph for a free-falling object is shown below:

Observe that the line on the graph curves. A curved line on a position-time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9.8 m/s/s), it would be expected that its position-time graph would be curved. The object starts with a small velocity (slow) and finishes with a large velocity (fast). Since the slope of any position-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-time graph for a free-falling object is shown below: Observe that the line on the graph is a straight, diagonal line. A diagonal line on a velocity-time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9,8 m/s/s, downward), it would be expected that its velocity-time graph would be diagonal. 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 has a negative acceleration. Since the slope of any velocity-time graph is the acceleration of the object, 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.

Free-falling objects are in a state of acceleration. 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. vf = g * t The above equation can be used to calculate the velocity of the object after any given amount of time when dropped from rest.

Example Calculations: At t = 6 s vf = (9.8 m/s2) * (6 s) = 58.8 m/s The distance that a free-falling object has fallen from a position of rest is also dependent upon the time of fall.

d = 0.5 * g * t2 The acceleration of a free-falling object (on earth) is the same for all free-falling objects regardless of how long they have been falling, or whether they were initially dropped from rest or thrown up into the air. Mass does not affect acceleration. Two objects //can// travel to the ground at different rates (rock falls faster than flat piece of paper), but more massive objects will only fall faster if there is an appreciable amount of air resistance present. Since another force is introduced, the object is no long freely falling. If the piece of paper was crumbled up into a ball, it would fall at a rate much closer to that of the rock.

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.

Notes: Freefall
10/3/11



Example Problem:

A stone is dropped and passes the length of a 2.2 meter window in 0.28 seconds. At what height was the stone dropped from above the window? Answer: 2.15 m

Lab: Freefall
Lab Partner: Garrett Almeida 10/4/11

__Objective__: 1) What is acceleration due to gravity? 2) What will the v-t graph look like? 3) How can you get the acceleration from the v-t graph?

__Hypotheses__: 1) -9.8 m/s2 2) Starting at (0,0), the line will be straight and extend to the bottom right. 3) The acceleration is the slope of the v-t graph (change in velocity over change in time).

__Materials__: Ticker tape, spark timer, mass, measuring tape

__Data__:





__Analysis__:

When adding a trendline to the x-t graph, choosing "linear" resulted in a R 2 value of 0.88. When "polynomial" was chosen, the R 2 value became 0.99, ensuring that this was the most accurate option. The graph appears as a curve because the mass's velocity is constantly increasing, resulting in a larger jump from initial position to final position each second. The data points on the v-t graph appear to fall on a straight line. When the trendline was made linear, the R 2 value was 0.99, confirming the visual interpretation. The graph is linear because the velocity is increasing each second at a constant rate because of the constant acceleration due to gravity (constant slope). We did not set the y-intercept to 0 because that might have made the line inaccurate (discussed in Conclusion).

Calculation of velocity at 0.15 s: This is the velocity at 0.15s because we have data for 0.1s and 0.2s. Using the data from 0.1s as the initial values and from 0.2s as the final values gives us a velocity for the mid-time between these two points, which is 0.15s.

The slope of our v-t graph is 853.72, as shown in the trendline's equation. Since the slope of a v-t graph is the acceleration (slope of v-t = Δy / Δx = Δv / Δt = acceleration), this value is the acceleration of the mass in cm/s 2. The difference in our slope compared to the class average is 2.36%. The slopes from each group ranged from 708.97 to 887.79. Theoretically, the slopes should have been 981, the acceleration due to gravity in cm/s 2. We had a percent error of 12.97, which isn't bad for our circumstances. Sources of error that led to this are discussed in the conclusion.



__Discussion__:


 * 1) Does the shape of your v-t graph agree with the expected graph? Why or why not?
 * 2) Yes. I expected the v-t graph to be a straight line with a constant slope because the acceleration due to gravity is always the same, resulting in the velocity changing at the same rate each second.
 * 3) Does the shape of your x-t graph agree with the expected graph? Why or why not?
 * 4) Yes. I expected a curve in the x-t graph because the change in position of a freely falling increases with each second due to a constant acceleration g (constant change in velocity).
 * 5) How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.)
 * 6) Our results were very close to the average, being 19.69 cm/s 2 away. Our percent difference was 2.36. Our results were a bit more accurate because our acceleration was closer to the theoretical value of 981 cm/s 2
 * 7) Did the object accelerate uniformly? How do you know?
 * 8) Yes. The acceleration is the change in velocity, and the change in velocity was constant. We know this because the v-t graph is linear and has a constant slope, so the mass's velocity was increasing at the same rate each second.
 * 9) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;">What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?
 * 10) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;">If the acceleration was higher than g, it would mean that another force was aiding in the fall of the mass pushing or pulling it downwards. If it was lower, it would mean that another force was acting against gravity, preventing it from pulling the mass downwards as smoothly as it normally would.

__Conclusion:__

Although the acceleration due to gravity is theoretically 981 cm/s 2 (9.8 m/s 2 ), our results show that the acceleration is 853.72 cm/s 2. Although the v-t graph is theoretically supposed to start at (0,0) because the mass's velocity is 0 at t=0, we did not set the y-intercept to 0 because the first spark on the spark paper was most likely not made at exactly t=0. We probably dropped the mass a tiny bit after that spark was made, so setting the intercept to 0 would throw off the trendline. We were correct in predicting the shape of the v-t graph because we knew that a constant acceleration due to gravity meant a constant slope of the v-t graph. We were also correct in saying that acceleration could be found by calculating the slope of the v-t graph. The graph shows velocity compared to time. The change in velocity over the change in time is equal to acceleration. One main source of error came from the friction between the spark tape and timer. This unwanted force slowed down the speed of the mass, and since an extra force was introduced into the situation, the mass was no longer freely falling. This is why every group had an acceleration that was below the theoretical value of 981 cm/s 2. This source of error could be minimized by holding the spark tape and timer in a way to reduce the friction, such as holding both vertically, allowing the tape to run through the timer smoothly. Another source of error could be the way we arrived at the data for position at each 0.1s interval. Although we tried to measure as accurately as possible, it may very well be that we misread the measuring tape. The only way to avoid this with the materials that we have is to make sure that we are reading the tape correctly, double-check our readings, and confirm our partner's readings.

Our percent of error was 12.97, and our percent difference was 2.36. While our acceleration turned out to be 853.72 cm/s 2, the class's accelerations ranged from 708.97 to 887.79 cm/s 2 , the average being 834.03. We used a 100g mass while other groups may have used different masses, but the mass of an object does not affect its acceleration, so this should not have affected the results.