Ch2_HallR

=Lesson 1: Describing Motion with Words=

9/7/11 - Constant Speed
toc

Lab: A Crash Course in Velocity (Part 1), 9/7/11, Ryan Hall and Sarah Malley
__ Objective: __ What is the speed of a Constant Motion Vehicle (CMV)? __Hypotheses:__ 1. The CMV will go between 2-3 mph 2. A position is able to be measured to the thousandth of a centimeter 3. A position-time graph shows the change in an object's position over a period of time. __Data:__ Time vs. Position of a CMV Table and Graph __Analysis:__ In class, we discussed that the slope of the line on a Position vs. Time Graph is equal to the object's average velocity. This is because the formula for the line's slope is the change of y divided by the change in x. Since the y value represents distance and the x value represents time, this formula is equivalent to the change in distance divided by the change in time. This is the same formula as the one for average velocity, meaning that the slope of the line and the average velocity are equal. The line is linear because the CMV is moving at constant motion, hence it's name. Therefore, it is moving a certain distance away from the origin during a certain time period, neither of which change. __ 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 formula for the slope is the change in x over the change in y, which in this case is the change in distance over the change in time. This is also the formula for an object’s velocity. Since the slope is of the average of the change of distance over the change in time, it is equal to the average velocity.


 * 1) Why is it average velocity and not instantaneous velocity? What assumptions are we making?
 * 2) It is average velocity because it is measuring the total change in distance over the total change in time, not at a single moment in time. We are assuming that the CMV is moving at a constant velocity, as its name implies.


 * 1) Why was it okay to set the y-intercept equal to zero?
 * 2) It was okay to set the y-intercept equal to zero, because at zero seconds, the car had not yet begun moving, and thus it had not yet moved position, which is the y-axis.


 * 1) What is the meaning of the R2 value?
 * 2) The meaning of the R2 value is the percentage of how accurate the trendline is compared to the points. The closer the R2 value is to one, the more accurate it is.


 * 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?
 * 2) If the graph of a slower CMV was plotted on the same graph, it would have the same y-intercept (zero), since it too would not have started moving at zero seconds. However, since it is slower, it’s slope would be lower, and thus the points on the graph would be much lower on the y-axis because it would take longer to change position.

__ Conclusion: __ In the end, we discovered that the speed of a CMV is 46.703 centimeters per second, or 1.045 miles per hour. My original hypothesis was that it would move between two to three miles per hour, which is a little faster that it did in our experiment. My hypothesis that positions can be measured to the thousandths of a centimeter was also off as well, while in reality, we were only able to measure to the hundredth of a centimeter, with the last digit being an educated guess. Of course, it would vary depending on the instrument used, with digital ones being the most accurate. However, I was correct in my hypothesis that a position-time graph tells the change of position over a period of time, which is equivalent to an object’s velocity. In our experimentation, there could have been many possible instances of human error, such as the shifting of the meter stick while measuring the spark tape, not starting off the spark time at a point of constant speed, and incorrect estimation of hundredths of centimeters. Also, equipment limitations of using a meter stick to measure could also be a cause of a mistake. In order to minimize these issues if I had to redo this lab, I would make sure that the CMV would have fresh batteries to ensure that it travels at a constant speed, and I would use a longer strip of spark tape to ensure that we would be able to measure ten dots on the tape during a point of constant speed.

9/8/11 - Summary of Lesson 1: Describing Motion with Words
__Notes:__ a.) Introduction to the Language of Kinematics b.) Scalars and Vectors There are two categories in which quantities used to describe the motion of objects can be divided into c.) Distance and Displacement d.) Speed and Velocity __Summary:__
 * Mechanics:** the study of the motion of objects
 * Kinematics:** the science of describing the motion of objects using words, diagrams, numbers, graphs, and equations; a branch of mechanics
 * Scalar:** a quantity that is completely described by a magnitude, or number value, alone
 * Vector:** a quantity that is completely described by both a magnitude and a description.
 * Distance:** a scalar quantity which represents how far an object has moved while in motion
 * Displacement:** a vector quantity which shows how far out of place an object is (an object's total change in position)
 * Speed:** a scalar quantity that shows how fast an object is moving; the rate an object covers a distance
 * Velocity:** a vector quantity that shows that rate an object changes position
 * Instantaneous Speed:** an object's speed at any given instant in time
 * Average Speed:** the average of all of an object's instantaneous speeds (distance/time ratio)
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) Speed: Speed is the measurement of how fast an object is moving It is a scalar quantity, so it does not matter which direction the object is going in. Speed is found by the equation of the amount of distance traveled divided by the time it took to travel that distance. In other words, it is the rate in which an object covers a certain distance.
 * 3) Distance: Distance is a measurement of how far an object has moved while in motion. Since it is not a vector quantity, the direction that the object travels in does not affect the amount of distance that it covers.
 * 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) By reading the text, I was able to clarify the meaning of velocity. During class, I thought that speed and velocity had the same meaning. However, after reading, I discovered that velocity was a vector quantity, and depended on which direction the object moved in. For example, if an object moved to the right for 3 meters and then to the left 5 meters, taking a total of 10 seconds, the speed would be .4 meters per second, while the velocity would be .1 meters per second, to the left.
 * 6) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 7) What would the velocity/displacement be of an object moving upwards or downwards vertically?
 * 8) What (specifically) did you read that was not gone over during class today?
 * 9) Scalars: a category of quantities in which quantities are able to be completely described by a magnitude, or number value. Ex. Speed & Distance
 * 10) Vectors: the other category of quantities in which quantities are completely described by both a magnitude and a direction. Ex. Displacement & Velocity.

Class Notes:
__Constant Speed:__ Speed is never changing, always the same pace. Therefore, the instantaneous speed is always the same as well. Distance/Time works for average and instantaneous speed Types of Motion: __Acceleration:__ Changing speed; can be both increasing or decreasing speed. Motion Diagrams are used for direction, Ticker Tape diagrams are used for measurments Signs are arbitrary. __Motion Diagrams__ At rest: v=0, a=0 Constant speed: Increasing speed: Decreasing speed __Ticker Tape Diagrams__ At rest: Constant speed: Increasing speed: Decreasing speed: =Lesson 2: Describing Motion with Diagrams= __Animation:__ []
 * At rest (not moving)
 * Moving at constant speed
 * Increasing speed
 * Decreasing speed
 * __Summary:__**
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) Ticker-tape graphs: Also called oil drop graphs, they measure the speed of an object. They are put through a ticker tape machine, and attached to a moving object. At even time intervals, the machine leaves a mark on the paper. By measuring the distance of the dots and knowing the time interval in which they were left, one can find out how quickly an object moved over the distance between the dots. For example, if there is a single large dot, that means that the object is not moving. If the dots are evenly spaced apart, that indicated that the object is moving at a constant speed. Also, If the dots are close together but get further and further apart, the object is speeding up, and vice versa.
 * 3) Position vs. Time graphs: Also called p-t graphs, they plot an object's position during a certain time as a point on a graph. The time is on the x-axis, while the position is on the y-axis. By plotting more than one point, the slope of the line can be found, which is the average speed for the object. If the line is pointing upwards to the right, the object is either accelerating or moving at a constant pace. If the line is pointing down to the left, the object either has a leftward velocity or is decreasing in speed. If the object is speeding up or slowing down, the line will be curved, while if it was moving at a constant speed, it would be straight.
 * 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) In class, I did not realize that a p-t graph could show that an object had a leftward velocity. Now I know that this can be shown by a line with a negative slope. If the line is straight, it has a constant velocity, while if it curved upwards, it is moving speeding up, and vice versa.
 * 6) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 7) How do you plot a p-t graph with an upward or downward velocity?
 * 8) What (specifically) did you read that was not gone over during class today?
 * 9) I read that p-t graphs can show that an object has a leftward velocity; I originally assumed that all p-t graphs went to the right.

Activity: Graphical Representations of Motion
__**Data:**__ __No movement:__ __Constant Fast:__ __Constant Slow:__ __Change in direction, constant speed:__
 * Objectives:**
 * What is the difference between static and dynamic equilibrium?
 * How is “at rest” represented on a position vs. time graph? On a velocity vs. time graph? On an acceleration vs. time graph?
 * How is constant speed represented on a position vs. time graph? On a velocity vs. time graph? On an acceleration vs. time graph?
 * How are changes in direction represented on a position vs. time graph? On a velocity vs. time graph? On an acceleration vs. time graph?
 * __Discussion Questions:__**
 * 1) How can you tell that there is no motion on a…
 * 2) position vs. time graph
 * 3) There is no change in position at all.
 * 4) velocity vs. time graph
 * 5) There is no change in velocity at all.
 * 6) acceleration vs. time graph
 * 7) There is no change in acceleration at all.


 * 1) How can you tell that your motion is steady on a…
 * 2) position vs. time graph
 * 3) There is a constant change in position over time, with the line sloping upwards.
 * 4) velocity vs. time graph
 * 5) The velocity is following a straight line just above the x-axis, meaning that it is positive, since the object (the person walking) is moving away.
 * 6) acceleration vs. time graph
 * 7) The acceleration stays on the x-axis since the velocity is not changing over time.


 * 1) How can you tell that your motion is fast vs. slow on a…
 * 2) position vs. time graph
 * 3) You can tell that your motion is fast vs. slow on a position vs. time graph based on the slope of the line formed. If the slope is steeper, that means that an object is moving faster since it is covering more distance in less time. Conversely, if the slope is flatter, the object is moving slower since it takes more time to cover as much distance.
 * 4) velocity vs. time graph
 * 5) You can tell that your motion is fast vs. slow on a velocity vs. time graph based on how far off the x-axis the velocity is. If it is further away, it means that the object is traveling faster since it is being displaced in a quicker time (assuming it is not turning around). If the velocity is closer to the x-axis, that means that the object is taking a longer time to be displaced as much (again assuming that the object is not turning around).
 * 6) acceleration vs. time graph
 * 7) The acceleration stays on the x-axis for both, since because the object is traveling at a constant speed, the velocity is not changing at all over time.


 * 1) How can you tell that you changed direction on a…
 * 2) position vs. time graph
 * 3) You can tell that an object has changed direction on a position vs. time graph because the line suddenly stops its positive upwards slop and stays constant, before beginning a negative slope downwards. This shows that the object stops moving away, turns around, and comes back to the motion detector. Since the line ends up back on the same y-value that it started on, it means that the object's displacement is zero, since the object has returned back to it's starting spot as well.
 * 4) velocity vs. time graph
 * 5) You can tell that an object has changed direction on a velocity vs. time graph because the line stops following its constant horizontal slope above the x-axis, and crosses the x-axis to begin a now-negative horizontal slope. This is because the object is no longer moving away, and instead is returning back to its starting spot. When it returns back, it will not have been displaced, so the line ends up back on the y-value that it started on, which is zero.
 * 6) acceleration vs. time graph
 * 7) The acceleration stay on the x-axis since the velocity is not changing until the object turns around, which is when the velocity line begins its negative slope. Due to this change in velocity, the acceleration line has a spike off the x-axis, and then returns.


 * 1) What are the advantages of representing motion using a…
 * 2) position vs. time graph
 * 3) It is able to show how far a distance an object moves in a certain period of time. Also, the slope of the line is the object's average speed.
 * 4) velocity vs. time graph
 * 5) It is able to show how much an object is displaced in a certain period of time. Also, the slope of the line is the object's average velocity.
 * 6) acceleration vs. time graph
 * 7) It is able to show how much an object's velocity changes during a period of time. This would only be important for object's that are changing velocity, and not moving at a constant rate.

>>>
 * 1) What are the disadvantages of representing motion using a…
 * 2) position vs. time graph
 * 3) If an object moves a large distance in a period of time, the graph would be very large, or large intervals would have to be used on the axises. Also, if an object is moving back and forth from the motion detector, the position would keep going from having a positive slope to a negative one and vice versa, which might make it hard to comprehend.
 * 1) velocity vs. time graph
 * 2) If an object moves a large distance in a period, the graph would be very large, or large intervals would have to be used on the axises. Also, if an object is moving back and forth from the motion detector, the velocity would keep on going from positive to negative and vice versa, which might make it hard to comprehend.
 * 3) acceleration vs. time graph
 * 4) Acceleration only changes when there is a change in velocity, so it is not useful for objects that are only moving at a constant motion.


 * 1) Define the following
 * 2) No motion: An object is not covering any distance in any period of time.
 * 3) Constant speed: An object's speed and velocity are not changing, meaning that the object's acceleration is zero.

Class Notes




Summary of Lesson 1e, Acceleration
__**Acceleration:**__ a vector quantity that is the rate of the change in velocity. (If an object changes velocity, it is accelerating) __**Constant Acceleration:**__ when velocity changed by a constant amount every second; different than constant velocity. __**Non-Constant Acceleration:**__ when velocity changes by different amounts every second When objects have constant acceleration, the distance of its travel is directly proportional to the square of the time traveled. Units = velocity per time; such as: m/s/s, mi/hr/s, km/hr/s, and m/s2 Since acceleration is a vector quantity, it involves direction, which depends on two things. 1. if the object is increasing or decreasing speed. 2. if the object is moving in a positive or negative direction. If an object is slowing down, then its association is in the opposite direction of its motion. __**Positive Acceleration:**__ When either 1. object is speeding up and its acceleration is in the same direction as its velocity, or when an object is slowing down in a negative direction and its acceleration is in the opposite direction of the velocity. __**Negative Acceleration:**__ When either 1. object is slowing down and its acceleration is in the same direction as its velocity, or when an object is speeding up in a negative direction and its acceleration is in the same direction of the velocity. =9/14/11=
 * __Notes:__**
 * __Questions:__**
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) Acceleration: The rate of an object's change in velocity over a certain period of time. It does not matter what the object's speed is, only its velocity. Also, since it is a vector quantity, it has a direction associated with it as well. The direction depends on if the object is speeding up or slowing down and if the object is moving in a positive or negative direction.
 * 3) Constant Acceleration: When an object's velocity is changing by a constant amount each second. This is different than constant velocity. In this case, the object's velocity changes constantly, while in constant velocity, the velocity is staying the same.
 * 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) During class, I did not think that acceleration was a vector quantity, since it seemed to just change when velocity changed. Now I understand that acceleration also depends on the direction the object is moving as well as if it is speeding up or slowing down. For example, when an object is speeding up in a negative direction and its acceleration is in the same direction of the velocity, its acceleration will be negative.
 * 6) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 7) How would acceleration be affected if an object was moving in either upwards or downwards, instead of in a compass direction?
 * 8) What (specifically) did you read that was not gone over during class today?
 * 9) In the reading, we read about free-falling objects, which experience constant acceleration, and continuously gain a constant amount to their velocity.

Lab: Acceleration Graphs, 9/14/11, Ryan Hall and Sarah Malley
__**Objectives:**__ 2. From the graph you can fine the change in velocity, and by using that over the change in distance, the acceleration can also be found. __**Procedure:**__ __**Data:**__ __**Analysis:**__ a) Interpret the equation of the line (slope, y-intercept) and the R2 value. i. Equation: 12.288x2 + 1.8806x i. Equation: y = -28.682x2 + 114.9x
 * What does a position-time graph for increasing speeds look like?
 * What information can be found from the graph?
 * __Hypotheses:__**
 * 1) The graph for increasing speeds on a position-time graph will be a curve going left-upward, since the object is not traveling at a constant speed, and is traveling more distance in less time.
 * 1) Using educated guesses, write hypotheses to the objectives
 * 2) Obtain all needed materials, including: spark tape, a spark timer, a track, a dynamics cart, and a ruler.
 * 3) Place the book on a flat surface, and lean the track on the top of it, so that it’s slanting down, with the other end on the table.
 * 4) Place the spark timer on top of the book, next to the edge of the ramp. Thread the spark tape through the timer, with the ‘shiny’ end facing up. Tape the spark tape to the dynamics cart, and make sure that the spark timer is set to 10 Hz.
 * 5) Turn on the spark timer, and release the cart. The cart will roll down the track; stop it at the bottom, and turn off the spark timer.
 * 6) Untape the spark tape from the cart, and lay the tape down on a flat surface so you can measure it.
 * 7) Using the metric side of a ruler or tape measure, measure the dots on the tape, and input them with the corresponding time into an Excel spreadsheet.
 * 8) Using the points of position vs. time in the spreadsheet, form a graph of the cart’s acceleration.
 * 1) Increasing Speed Line:
 * 1) The A coefficient is equal to the half of the acceleration of the dynamics cart. By doubling this coefficient, the average acceleration of the cart can be found, which would be 12.288 times 2, or 24.576. The B coefficient is equal to the object’s initial velocity, which means how fast the cart was going when it first started, which would be 1.8806 meters per second.
 * 2) Decreasing Speed Line:
 * 1) The A coefficient is equal to the half of the acceleration of the dynamics cart. By doubling this coefficient, the average acceleration of the cart can be found, which would be -28.682 times 2, or -57.364. The acceleration is negative in this line because the cart is slowing down. The B coefficient is equal to the object’s initial velocity, which means how fast the cart was going when it first started, which would be 114.9 meters per second.

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.)



Increasing speed: Halfway point: .444 cm/s2 End point: .645 cm/s2

Decreasing speed: Halfway point: 4 cm/s2 End point: 2.5 cm/2

c) Find the average speed for the entire trip. __**Discussion Questions:**__ 1. What would your graph look like if the incline had been steeper? a. If the incline had been steeper, the cart would have accelerated at a faster rate. Therefore, it would travel a distance in a faster time, so the points would be at higher up on the y-axis at and further to the left on the x-axis, making the curve sharper, and the slope greater.

2.What would your graph look like if the cart had been decreasing up the incline? a. If the cart had been decreasing up the incline, the graph would be red line on the graph. It starts off with a steep slope since when first pushed it travels a distance in a short period of time, but as gravity slows it down, the line curves, and the slope gets shallower. The points stop being so far away from each other in regards to the y-axis, since the cart travels less distance during a period of time.

3. Compare the instantaneous speed at the halfway point with the average speed of the entire trip.
 * 1) Increasing speed: As the time goes on, the cart is accelerating, and covers more distance in a period of time. Therefore, the instantaneous speed is faster at the end point than at the halfway point, since the object has sped up.
 * 2) Decreasing speed: As time goes on, the cart is slowing down, so it takes longer to cover a certain distance. Due to this, the instantaneous speed at the end point is less than that of the halfway point, since the object has slowed down.

4. Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense? a. Since the tangent line is a linear line, its slope is equal to the velocity of an object, which is moving at a constant speed. Since this line passes through the one point it is tangent to, its slope, or the speed, is also the instantaneous speed at that point.

5. Draw a v-t graph of the motion of the cart. Be as quantitative as possible. __**Conclusion:**__ After our tests, we plotted our results on a graph. The line for increasing speed turned out to be curved, curving upwards and to the right. This is because as the cart increases in speed, it takes less time to cover distance. This was the opposite of what I guessed in my hypothesis, which was that “it will be a curve going left-upward.” Also, on the graph, we could find the change in velocity to graph a v-t graph, as well as the object’s acceleration, which is just as I guessed in my second hypothesis. For example, since the cart had traveled 11.7 cm in .9 seconds, the cart’s velocity can be found, which is equal to 13 cm/s. However, there are many possible sources of error that could have created inaccuracies in our data, such as the cart being accidently pushed down the ramp instead of rolling down by itself, the spark timer not starting until after the cart has already started moving, and the ramp being accidently moved while leaning on the textbook. In order to minimize these issues if we redid the lab, we could attach a ramp to a ring stand, so that everyone’s tests would have the same amount of incline. Also, one should make sure that the spark timer has started before the cart is moving, and just let go of the cart instead of pushing it at all. =Lesson 3: Describing Motion with Position vs. Time Graphs=

9/15/11 - Summary
__The Meaning of Shape for a p-t Graph__ If the velocity is constant, then the slope is constant; if it is changing, then the slope is changing A curved line of changing slope is a sign of accelerated motion __The Meaning of Slope for a p-t Graph__ The shape of the line on the graph (straight, curving, steeply sloped, mildly sloped, etc.) is descriptive of the object's motion. The slope of the line on a position-time graph is equal to the velocity of the object. __Determining the Slope of a p-t Graph__ =Lesson 4: Describing Motion with Velocity vs. Time Graphs=
 * __Summary:__**
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) Slope of a p-t graph: Slope is found by the change in the y-coordinates divided by the change in x-coordinates. Since the y-coordinates is position and the x's are time, this formula would be the change in position over the time, which is also the equation for velocity. Therefore, the slope of the line equals the object's velocty.
 * 3) Meaning of the Shape of a p-t Graph: If the line is straight, then the velocity is constant since the slope is not changing. Also, if the line is pointing upwards, then the object has a positive velocity, and vice versa. Furthermore, if the line is curved, an object's velocity is changing, and it takes a different amount of time for an object to travel a certain distance. If an object has a curve to the right before going up, it is gaining speed, while if the line curves up and then moves right, it is losing speed.
 * 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) None
 * 6) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 7) How would a p-t graph look like if an object moved upwards or downwards vertically?
 * 8) What (specifically) did you read that was not gone over during class today?
 * 9) None

9/15/11 - Summary
__The Meaning of Shape for a V-T Graph__ The slope of the line on a velocity-time graph relates to the object's acceleration If the slope is zero, positive, or negative, than the acceleration is zero, positive, or negative, respectfully The acceleration can be negative or positive depending on the direction of the velocity. __The Meaning of the Slope for a V-T Graph__ The slope of the line on a velocity-time graph is equal to the acceleration of the object __Relating the Shape to the Motion__

__Determining the Slope on a V-T Graph__ __Determining the Area on a V-T Graph__ A velocity vs time graph can also be used to determine the displacement of an object, which is the area bound by the line and the axes.
 * __Summary:__**
 * 1) What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 2) Plotting points on a v-t graph: Since the y-axis is velocity and the x-axis is time, a point is plotted at the object's velocity at that time period. If the object's velocity is, the plots will be on a line with a positive slope, and vice versa. Also, if an object stops moving, the velocity remains constant at whichever y-value (velocity) it was last at.
 * 3) Increasing and Decreasing Velocity on V-T Graphs: If the line is moving away from the origin left to right, than the object is speeding up since is moving further away from where it started. Contrastingly, if the line is moving towards the origin left to right, it is slowing down, because it is returning to the spot where it started moving.
 * 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) During class, I was confused on how the acceleration was found using the line on the v-t graph. However, after reading, I know see that since the line's slope is the change in velocity of change in time, it has the same formula that is used for acceleration.
 * 6) What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * 7) How would you find the amount of displacement on a v-t graph is the line was curved?
 * 8) What (specifically) did you read that was not gone over during class today?
 * 9) Using Velocity-Time Graphs to find the amount of displacement. This is done by finding the area of a shape expanding from the line of velocity to the x-axis.

Lab: A Crash Course in Velocity (Part II) - Ryan Hall, Sarah Malley, Jenna Malley, Nicole Kloorfain
__Objective:__ Both algebraically and graphically, solve the following 2 problems. Then set up each situation and rune trials to confirm your calculations. To find it graphically, we used the velocities of both CMVs (found in the previous lab), to find the distance that they would travel after 1, 2, 3, etc. seconds. By plugging the desired time into the formula V=d/t as, and with the corresponding velocity found last week used for V, we easily found how far each CMV would travel after a certain number of seconds. However, one CMV was starting at 0 cm, and the other was going in the opposite direction towards it starting from 600 cm. Therefore, we had to take the distances found for the CMV starting from 600 cm, and subtract them from 600. This was because the CMV is moving away from 600 cm, and going towards 0 cm. Therefore, while the slower CMV has a positive slope and velocity, the faster CMV, going towards 0 from 600, has a negative slope and velocity. After plotting these points, we found that the two CMV’s would crash at about 164.21 cm away from 0, or 440 cm away from 600 cm. It was closer to 0 cm than 600 cm where they would meet since the faster CMV would be placed at 600cm, and the slower one at 0cm.
 * __A).__**

**__B).__**

Similarly to in part A, we found how far each CMV would travel after 1,2,3, etc. seconds by using the velocities found for them in the previous lab, and then by plugging them as well as the desired time into the formula V=d/t. However, in this case, the slower CMV is placed 100cm ahead of the faster CMV, which starts at 0cm. Therefore, we added 100cm to each of the slower CMV’s distances, because it was already positioned 100cm further than the other CMV. Since each CMV is moving away from 0cm, they each have a positive slope and velocity. After plotting the points, we found that the faster CMV would overtake the slower CMV after around 160cm.

__Interpreting Your Results:__ **Crashing CMV's:** media type="file" key="Movie on 2011-09-21 at 08.57.mov" width="300" height="300"

As shown in the data and video, the faster CMV would crash into the slower one, if it were going in a straight path, at around 160 cm away from zero (the starting point for the slower CMV), or 440 com away from 600 (where the faster CMV started at)**.** This matches our graphically and algebraical answers from the beginning of the lab.

media type="file" key="Movie on 2011-09-21 at 09.06.mov" width="300" height="300" As shown in the data and video, the faster CMV would pass the slower one at around 161 cm away from zero**.** This is 161 cm away from where the faster CMV started, but only 61 cm away from the starting point of the slower CMV, since the slower CMV started 100 cm further away than the faster one did. This matches our graphically and algebraical answers from the beginning of the lab.
 * Overtaking CMV:**

__Discussion Questions:__ 1. Where would the cars meet if their speeds were exactly equal? In the crashing situation, the cars would meet at halfway, or 300 cm, since they both would travel the equal distance during equal time. In the catching up situation, they would never meet, since the car starting 100 cm behind the first one 2. Sketch position-time graphs to represent the catching up and crashing situations. Show the point where they are at the same place at the same time.

3. Sketch velocity-time graphs to represent the catching up situation. Is there any way to find the points when they are at the same place at the same time? No, it is not possible to find the points where the lines intersect using a velocity time-graph. This is because the CMV's are continuously moving at a constant velocity, so the lines are horizontal throughout the graph. Therefore, the lines do not intersect. It is due to the fact that one CMV has a faster velocity than the other that the two lines intersect in a position-time graphs, not because of any change in velocity.

__Percent Error:__ __Percent Difference:__ __Conclusion:__ In the end, our experimental answers were very close to our theoretical answers. For the crashing experiment, we thought that the two CMV's would crash at 164.21 cms away from zero, while in reality, they did so at an average of 160.37 cms. Also, we estimated that in the other experiment, the faster CMV would overtake the slower one at about 160.6 cms, while it was about 161.24 in the actual experiment. These results are extremely close to our answers that we found graphically and algebraically, meaning that we performed the correct equations while trying to accomplish the lab's objective. For the crashing part, we had 2.34% error, and for the catching up, we had .134% error. In this lab, there could have been a large amount of sources of errors, such as the two CMV's not starting moving at the same time, the tape measure not being laid out straight, and the CMV's changing velocity due to a drain in batteries. In order to minimize these issues if this lab was to be redone, I would suggest using a machine to release the CMV's to eliminate human reaction time, taping down the measuring tape in more than two places to ensure that it is straight, and to replace the batteries of the CMV's right before they are used to ensure that they go at a constant velocity.
 * Crashing CMV's: **
 * Catching Up: **

=Egg Drop Project= __Pictures:__ __Results:__ In both prototype tests and in the final drop, the egg remained intact in my device. My final contraption weighed just shy of 150 grams with the the egg inside, but did not change much since the original try. Since my first prototype had worked, the only modifications that I made to it were to add hot glue to the 'basket' to strengthen it and prevent it from breaking. Although I probably could have taken some excess weight off of the device, I chose not to since I wanted to make sure that it landed straight up, and was not blown by wind during the drop. Also, for the second test and the final drop, I wrapped the egg in aluminum foil, so that if the egg did break, the yolk would not get all over the contraption. __Acceleration:__ __Analysis:__ My egg drop device worked due to a variety of reasons. First, the rectangular shape of the contraption's bottom gave it much air resistance which helped slow its fall. Also, the top of the device, which utilized rubber bands around straws to hold the egg in place, allowed the egg slightly shift, but made sure that it did not fall out nor hit anything. However, the 'secret' of the contraption was the 'basket' full of small pieces of cut up straws, which were stuffed beneath the egg. These absorbed most of the force of the impact, and were soft so they did not break the egg. If I were to redo the project, the only thing that I would do different would be to remove a few of the top straws on the 'platform' of my contraption, to remove unneeded weight. This would be all I would do because I fear that any more changes in weight would compromise its tendency to land straight up, and other modifications might also allow it to be blown in the wind while falling. =Lesson 5: Free Fall and Acceleration of Gravity=
 * Without Egg:**
 * With Egg:**

10/3/11: Summary
=10/5/11=

Lab: Falling Object, Ryan Hall and Sarah Malley
__Purpose:__ We are attempting to verify that a free-falling object will accelerate due to the rate of gravity. __Hypotheses:__ 1. What results do you think your experiment will produce? I think that the experiment will produce a velocity-time graph of a line with a constant slope upwards. I believe this since a constant upwards slope would signify constant acceleration, which would make sense because the weight would keep on increasing speed due to gravity. Also, I think that it will form a position-time graph with a negative slope increasing away downwards, since the weight would constantly getting get further and further away at a faster rate, but will fall going down to form a negative line. Finally, I think that the weight will accelerate at a rate of -980.00 cm/s/s. 2. What do you think the v-t graph will look like? I think that it will be a constant slope upwards, since the weight was constantly increasing speed while it was falling due to acceleration by gravity. 3. How will you find “g” from this graph? "G," signifying gravity, is equal to the negative of the amount of acceleration due to gravity. This will be found by finding the slope of the velocity-time graph, which will be the object's acceleration, and then multiplying it by negative one.

__Data__ To find instantaneous speed, the equation is v = the change in distance divided by the change in time, or v = (final position-initial position) divided by (final time-initial time). To find mid-time, the equation is mid-time = (final time-initial time)/2, which is the average of the two times.
 * Group Data:**

__Analysis (including Simple Calculations and Graphs)__ Velocity-Time Graph: In this graph, the trendline shown is equal to the weight's acceleration as it was falling, since the acceleration is equal to change in velocity over change in time, and the trendline's slope is the change in y over the change in x, which in this case is also the change in velocity over the change in time. It is a positive constantly increase line, because the weight was constantly gaining velocity due to the rate of acceleration by gravity, which is 9.8 meters per second, or 980.00 centimeters per second. Therefore, since the rate of acceleration is equal to the slope of the trendline (which represents acceleration), in a perfect world, the m value (the one before the x, as seen in the equation y=mx+b) would be equal to 980.00. However, due to many possible errors in our experiment, our m value is greater (1225.40), which is technically impossible normally, unless other factors caused it to fall at a quicker rate. In addition, the b value in the trendline's formula of y=mx+b represents initial velocity of the weight from where we started measuring. This makes sense because if the object was starting from at rest, at zero seconds, the object's velocity would be zero, and the line would begin from the origin. However, since 31.455 (the b value) is added, the trendline starts at the y-intercept of (0, 31.455), which means that at zero seconds, the weight had the initial velocity of 31.455 meters per second. The R squared value is the percent of how close the trendline is to exactly hitting all of the points. To find the percent, the R squared value must be multiplied by 100. Therefore, in this case, the percentage would be 99.765%, meaning that it is very very close to being exact.
 * Class Data:**
 * Velocity Equation**
 * Percent Error and Difference:**
 * Graphs:**

Position Time Graph: In this graph, the trendline shown is equal to the weight's velocity as it was falling, since the velocity is equal to change in position over change in time, and the trendline's slope is the change in y over the change in x, which in this case is also the change in position over the change in time. It is a positive line curving upwards, because the weight is constantly moving further away in less time, since its velocity is increasing because of the rate of acceleration due to gravity. Since the the trendline is symbolizing change in distance over time, it's equation of y=Ax squared + Bx + C is equal to the equation Change of Distance = 1/2at squared + vit. Therefore, 1/2at squared + vit = Ax squared +Bx (The C in the trendline's equation represents the weights initial position at 0 seconds, which should be 0 as well. However, it is a little off due to an error in where we started measuring the ticker-tape). Since those two equations are equal, the A in the trendline's equation is equal to 1/2a, or one half of the weight's acceleration. Furthermore, B from the trendline's equation is equal to vi, or initial velocity. The R squared value is the percent of how close the trendline is to exactly hitting all of the points. To find the percent, the R squared value must be multiplied by 100. Therefore, in this case, the percentage would be 99.997%, meaning that it is extremely close to being exact.

__Discussion Questions__ 1. Does the shape of your v-t graph agree with the expected graph? Why or why not?

It agrees in the fact that it is a constant line, because the weight's velocity is constantly increasing as it falls due to the rate of acceleration by gravity. However, I initially thought that the line would be negative since the weight was falling downwards. However, because we entered in positive position amounts instead of negative ones, the line is positive and not negative. Besides for that, the graph is as expected.

2. Does the shape of your x-t graph agree with the expected graph? Why or why not?

Yes, it agrees with the expected graph. This is because, as expected, the weight was moving further away from the origin (the spark timer on top of the railing) in less and less time, because the acceleration by gravity was constantly increasing its velocity. Therefore, as time went on, the weight covered more distance in less time, as show on the graph.

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

Our results were much different than that of the class. Compared to the class's average rate of acceleration of 839.47 cm/s/s, we found an amount of 1225.40 cm/s/s. This is almost 1 and a half times the class's average, and 45.98% different. Technically, our result is almost impossible, since it is higher than the rate of acceleration by gravity, which is 980.00 cm/s/s. However, many errors could have been present that may have resulted in our weight accelerating faster than being acted upon solely by gravity.

4. Did the object accelerate uniformly? How do you know?

No, our weight did not accelerate uniformly. We know that because our rate of acceleration, 1225.40 cm/s/s, is greater than the rate of acceleration due to gravity, 980.00 cm/s/s. Therefore, it accelerated faster than only gravity would have allowed it to. This could have been caused by many errors that may have forced our weight to accelerate faster than being pulled down only due to gravity.

5. What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?

Acceleration due to gravity could have been higher than it should have been due to it being thrown down the railing rather than dropped, or someone accidentally pulling down the weight faster while it was falling. Also, acceleration due to gravity could've been lower than it should have been due to the ticker tape experiencing friction while going through the spark time, someone stepping on or holding onto the ticker tape while it was going through, and the weight hitting the side of the railing/ledge on the way down or something on the ground.

__Conclusion:__ In my hypotheses, I was correct on what the position-time graph would look like, but was incorrect on what the velocity-time graph would look like, and how to find g. Since I thought we would put in negative position values because the weight was dropping downwards, I thought that the acceleration would also be negative, meaning that the line in the velocity-time graph would be negative. However, we instead but in the position values as positive, so the acceleration and line in the velocity-time graph were positive. Since the acceleration was positive, there was no need to multiply it by negative one to find g. Instead, the rate of acceleration due to gravity was direclty equal to g. Also, I was wrong on what the accelerate rate of the weight should have been. I had said -980.00 cm/s/s, since I again thought the positive position points we put in would be negative since the weight was falling downwards. Even though the weight SHOULD have accelerated at a rate of 980.00 cm/s/s, errors caused ours to fall faster than the rate of acceleration due to gravity; instead at 1225.40 cm/s/s. While our class had a percent error of 14.35%, our group had a percent error of 25.04%. This error cannot be definitely located, but the most likely source is the weight not being able to fall freely from the railing, such as someone was standing on the ticker tape that was going through the spark timer, or the weight was thrown and not dropped. Other possible causes of error include the weight hitting into the side of the railing/ledge while falling or hitting something on the floor, the ticker tape experiencing friction while going through the spark timer, and incorrect measurement of the dots on the ticker tape. These would have all either changed the weight's acceleration as it was falling, or accidentally modify the distance between two or more dots on the ticker tape dramatically. If this I were to redo this lab, I would attach a string to the weight to hold before dropping it, so the weight would not be accidentally thrown down, and instead start from rest when the string is released. Also, I would do the experiment in a less crowded area, so that nobody could accidentally slow or prevent the ticker tape from being pulled threw the spark timer. Lastly, I would have held the weight further away from the edge of the railing, so that it could not hit into it and slow its descent.