Ali and Caroline 9/8 Objective:What is the speed of a Constant Motion Vehicle (CMV)? Hypotheses: A CMV moves 20 ft/15 secs --> 609.6 cm/15 secs --> 40.64 cm/sec. Distances can be measured to the nearest millimeter, but then, an educated guess must be used for the next decimal place. A time position graph tells the time relative to the distance away from zero (which is also known as the position). This, hopefully, should be a linear function.
Available Materials: Constant Motion Vehicle, Tape measure and/or metersticks, spark timer and spark tape Data:
Analysis Essentially, the graph of the CMV shows that it is moving at a constant speed, because of the slope of the line (positive direction). At each interval, it is moving approximately the same distance away from the point before it. Discussion questions
Why is the slope of the position-time graph equivalent to average velocity?
Slope is the measure of Δy/ Δx
in this case, the Δy=Δd (distance) and the Δx= Δt (time)
therefore; Δy/ Δx = Δd/ Δt = V
Why is it average velocity and not instantaneous velocity? What assumptions are we making?
Instantaneous velocity is at one point, so it only accounts for one point. Average velocity takes all of the points into account; therefore, it is more accurate. We are assuming that all of the values will be relatively close, and not many outliers will exist.
Why was it okay to set the y-intercept equal to zero?
It is okay to set the y-intercept equal to zero because since the y-axis is the position, if it’s at zero, it will not be moving. This was the starting point from which or average velocity was graphed
What is the meaning of the R2 value?
R2 is the percentage of points described by the trend line. It is a tool to help us know how “fit” our data is compared to the trend line. If the value is around .99… the data is accurate.
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?
If it were to be slower, the line would lie below our trend line. It is still traveling in the same amount of time, but the distance traveled won’t be nearly as much.
Conclusion My results showed that the average speed of a CMV was 37.962 cm/s. It seems that as the time sped up, the distance the CMV traveled lessened. My hypothesis was that the CMV would travel 40.64 cm/s. Although it did not travel as far as I had originally thought, the CMV still traveled in a 5 cm range to my guess. Therefore, I consider it fairly accurate. However, whenever performing an experiment, there is always room for error. This error could be found in the instruments that we used for measuring the length between the dots on the spark tape. The meter stick may have slid while we were recording down the measurements; the shape of the meter stick also was thick, so it made it difficult to make out the exact numbers. Anotheraspect of error was that the measurements weren’t as precise as they could have been. If we had more effective, digital tools, we could have gotten the exact measurement. In order to minimize error, next time, we could use an instrument that does not slide and is flatter, therefore making the numbers easier to read. Something like measurement tape could be more effective. Rather, a digital tool could be used for better accuracy.
CMV Lab Part 2
Ali, Caroline, Dani, Julia 9/21
Objectives: Both algebraically and graphically, solve the following 2 problems. Then set up each situation and run trials to confirm your calculations.
Find another group with a different CMV speed. Find the position where both CMV’s will meet if they start at least 600 cm apart, move towards each other, and start simultaneously.
According to our calculations, the blue car should travel 372.321 cm and then crash into the yellow car (which would have moved 229.214 cm). Below is a video of what ACTUALLY happened. The data table below it shows each trial and the distance each CMV actually moved.
Here, the two cars crash at a certain point. We then proceeded to mark the place where they met, proving our calculations to be approximately correct.
Find the position where the faster CMV will catch up with the slower CMV if they start at least 1 m apart, move in the same direction, and start simultaneously.
According to our calculations, the blue car should travel 260.156 cm and then become parallel with the yellow car. Below is a video of what ACTUALLY happened. The data table below shows each trial and the distance it took for the blue to catch up.
The faster of the two cars (blue) is catching up to the yellow car at a certain point. We then proceeded to mark the place where they met, proving our calculations to be approximately correct.
Analysis Part A: crashing
Percent error- based on theoretical values-average experimental values
percent difference
Based on this percent error of .57%, I can conclude that our results were very accurate for the case of the blue car striking the yellow car. The smaller percent that we have, the better our results are in actuality. We chose to take an average of or experimental values because all of our experimental values were very close to each other. For percent difference of the crash, we took the average experimental value and subtracted the individual value. Then, we divided it by the average value and multiplied by 100 to get a percentage. Our percentages were very small, thus demonstrating precision. Overall, between these two methods, we can conclude that our results were very accurate and precise.
Part B: catching up
percent error
percent difference
Once again, we used percent error again to find out how accurate the results for Part B were. As shown above, our results turned out to be extremely accurate, within .03% of the theoretical results. For percent difference of the catching up, we took the average experimental value and subtracted the individual value. Then, we divided it by the average value and multiplied by 100 to get a percentage. Our percentages were very small, thus demonstrating precision. There was one outlier where the blue car caught up at 230.96 cm, resulting in a percentage of 11.25%. This outlier may have been due to slow reaction time or improper readings of the meter stick. However,it can still be concluded that our results were very accurate and precise. Discussion questions
Where would the cars meet if their speeds were exactly equal?
The cars should realistically meet at 300 meters given that their speeds were exactly equal; this is the half way point.
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.
crashing
catching up
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?
this is when the cars crash
this is when the blue car surpasses the yellow car
Conclusion In conclusion, our results turned out to be accurate. Each time that we performed a trial, the results ended up to be in some range of what we had originally predicted.For part A, our calculations showed that the two cars would meet and crash once the blue traveled 372. 321 cm and the yellow traveled 229.214 cm.For part B, our calculations showed that the faster, blue car would catch up with the slower, yellow car at the 260.156 cm mark. Since we used ramps, the cars were able to move in a straight line, thus giving us more accurate results. One of our sources of error found in our calculations was that for part A, instead of the two distances adding up to 600, it added up to 601. This was due to the fact that we used significant figures, thus truncating any "useless" decimals. Error could have also been found in the timing and reaction time. If we used another type of method to measure, like a USB to DataStudio, it could have been more accurate. Our percent error for part A was 0.57%. The percent difference ranged from -0.2% to 0.3% for the blue car and for the yellow car percents ranged from -0.55% to 0.55%. These values as well were very small. For Part B, percent error was 0.03 % and percent difference ranged from -0.2% to 1.95%, with an outlier at 11.35%. However, I believe the outlier was a human error. Overall, these percentages are very small, but could be improved. There are other evident ways to improve our data, but yet again our data is very precise. I believe our calculations couldn't have been that much closer to the actual values. If we were to improve the lab, I would say that it were required for each group to use the lab and a USB to get the most accurate results. Also, the lab could note that each car should be using a fresh pack of batteries. Therefore, everyone would be on the same page.
Accelerating/Decelerating Car Lab
Ali and Caroline 9/14 Objectives:
What does a position-time graph for increasing speeds look like?
What information can be found from the graph?
Hypotheses: A position-time (x-t) graph for increasing speed looks like a curve upwards, starting less steep and becoming steeper as time goes on. This is because it is covering a greater distance in less time. From a position-time graph, we can tell the velocity, because we can figure out the slope of the line.
Available Materials Spark tape, spark timer, track, dynamics cart, ruler/meterstick/measuring tape Procedure Acceleration We placed the car on the ramp, which was equipt with an inclination (due to one text book, shown). Therefore, this would make the car accelerated down the ramp. We attached a piece of spark tape to the car and put it through the spark timer. We then released the car until it was at the end of the ramp. Then, we collected the data by measuring the distance between the initial point to the final point, for each dot. The result is below. Deceleration We placed the car on the bottom of the the ramp, which was equipt with an inclination (due to one text book, shown). Therefore, this would make the car decelerate up the ramp (values would decrease because it went uphill). We attached a piece of spark tape to the car and put it through the spark timer. We then pushed the car up the ramp until it stopped. Then, we collected the data by measuring the distance between the initial point to the final point, for each dot. The result is below. Data and Graph
Accelerating Data:
a=time(s) b=distance (cm)
Decelerating Data:
c= time (s) d=distance (cm)
Analysis Interpret the equation of the line (slope, y intercept) and the R2
y= Ax2+Bx
y= distance& x= time
change in distance= 1/2at2 + vit
accelerating
Δd= 15.541t2 + 11.795t
R2= 0.99996
decelerating
Δd= -25.32t2 + 90.698t
R2= 0.9999
Both of our R values are VERY close to 100%. But, keep in mind, we used a polynomial trendline, rather than a linear trendline. Originally, when we used the linear trendline, our results were for increasing and decreasing were .92 and .88, respectively. Therefore, the polynomial trendline was a better choice because it was a curve.
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:
halfway= 6 cm/s
end= 5cm/s
Decreasing:
halfway= 5.49 cm/s
end= 1.5 cm/s
Find the average speed for the entire trip
Accelerating
Δd/Δt= 38.84 cm/s
Decelerating
Δd/Δt= 8.9375 cm/s
Discussion Questions
What would your graph look like if the incline had been steeper?
It would be a steeper curve, because the points would travel a greater distance in a shorter amount of time.
What would your graph look like if the cart had been decreasing up the incline?
The cart would start out going fast, with a steeper slope. It would then curve the opposite way, and gradually have a less steep slope.
Compare the instantaneous speed at the halfway point with the average speed of the entire trip.
Accelerating instantaneous speed at the halfway point is 60 cm/s. For the average accelerating speed we had 38.84 cm/s. I believe that there is such a difference because the line that we used curved. There most likely was an error in calculation, as our values should have been close to that of the instantaneous speed.
Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense?
This makes sense because instantaneous speed is the speed that the object is moving during the specified time. Here, the tangent line intersects with one point out of the eighteen on the graph. Thus, by taking this point, we are figuring out velocity, which should be consistent throughout.
In class today we learned that the formula for some polynomial is y=Ax2+ Bx. We learned that the A variable is 1/2 the acceleration (or in our case, 15.421, accelerating), while the B variable is initial velocity (again, in our instance, 11.795. )If you replace the variables with those from the acceleration equation, they turn out to be the same. The x axis can be exchanged with Δt, or the time value. The y variable can be replaced with Δd because it is position.
Draw a v-t graph of the motion of the cart. Be as quantitative as possible.
Conclusion The results of this experiment proved our hypothesis to be successful. We hypothesized that the result of data would be plotted as a curve, and so it was. It seemed to be that as the car traveled along the ramp, speed was gained, and thus traveling greater distances in a smaller amount of time. Likewise, decelerating started off fast, but gradually slowed down, covering a less distance in a longer time. In this lab, there were a few sources of error. To start off, me and my partner did not start measuring at the first dot made by the spark timer. Thus, we had a false "initial velocity". This definitiley skewed our data, without us even thinking it done harm. It seems as if the car was already in motion, therefore accounting for this high initial velocity. Instantaneous velocity is supposed to be similar to the average speed, but as stated above, this was not accomplished by our data. Another source of error was most likely in our mathematical calculations; we may have calculated wrong or messed up numbers. Lastly, although our data was accurate, it was not as precise as it could have been. If we had been using something to hook up to the computer, our measurements would have been far more precise. However, this was not the case. Overall, from this lab I learned that in a position-time graph lab, a curve will result from acceleration. The faster the object moves, the more distance covered in shorter periods of time, and the "steeper" the curve will become.
Lab Acceleration of a Falling Body
Ali, Caroline 10/5 Objective:What is the acceleration of a falling body?Hypothesis:The acceleration of a falling body should be 9.8 m/s2 or 981 cm/s2. Hypothetically, the v-t graph should look like the following . Hopefully, the g from this graph will be 981 cm/s2 (referring back to the acceleration of a falling body). It should appear in our position time graph as the A value in Ax2+bx. Materials: Ticker Tape Timer, Timer tape, Masking tape, Mass, clamp, meterstick. Procedure For our lab, we took a spark timer, spark tape, masking tape, a weight, and dropped it off from a height. Dots were plotted for each interval of a second (.1,.2,...1). We then laid out the spark tape on the floor and measured the distance between each dot made by the timer by using a measuring tape. A picture of how we measured is shown below.
Data and Graphs
position-time graph data
position time graph
Above is the data for the position time graph, which compares the amount of time passed as the weight fell to the ground (measured in cm).
velocity-time graph data
velocity-midtime graph
The two images above show the data for the instantaneous velocity v. the mid time. Below is an example of how we found the instantaneous velocity. cm/s/s. We found the mid-time by taking the final and initial value and dividing it by 2. For example, MT= (0 s +0.1 s)/2 = 0.5 s.
Sample Calculations Below is the percent difference of -5.98%. This is comparing the class average of 839.417 cm to our average of 889.65 cm. This percentage is very small and is within the "20 percent range" concept we created for accuracy.
Below is the percent error of 9.31. It is comparing the theoretical value of 981 cm/s/s to our experimental value of 889.65 cm/s/s. The percent is very small and is within the "20 percent range" concept we created for accuracy. Below is an example of how we found the velocity for a given point.
93.2 cm/s/s
Analysis
To begin, we found that the polynomial equation y=Ax2+Bx happens to come from the equation d=1/2at2+vit; therefore, A=1/2acceleration and B=initial velocity. Additionally, y=mx+b can be derived from the equation Vf= at +Vi. These factors will later be discussed throughout the analysis. For the position-time graph, the trend is a polynomial one, indicated by the equation of the line (y=445.35x2+50.138x). In this case, the acceleration is 445.35 and the initial velocity was 50.138. Our R2 value was extremely accurate at 1, showing that we had great results. The velocity time graph, unlike the p-t graph, was a linear function (y=889.65x+50.91). In this case, slope=889.65 and B=50.91. Our R2 value was extremely accurate with a value of 0.99994. For both graphs, we got an initial velocity that was not zero as expected because we may have had a delayed reaction when dropping the weight, after the ticker tape was already turned on. According to the original equation, 1/2A equals 445.35. Technically, this should mean that the value 445.35 cm from the p-t graph, doubled, should equal our slope of the velocity time graph. Although not an exact fit, the value almost equals 889.65 cm/s/s, as it equals 890.7 cm/s/s. Thus, we can conclude it is accurate. Discussion Questions
Does the shape of your v-t graph agree with the expected graph? Why or why not?
No, the expected graph was a constantly sloped line in the negative direction moving away from the origin. This is because we omitted the negatives when counting our results.
Does the shape of your x-t graph agree with the expected graph? Why or why not?
Yes, we expected the graph to be a polynomial--gradually increasing as time passed. It completely matches up with our expected graph, a J curve.
How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.)
Our results confirm that our data is very accurate. Again, our data was in the "20 percent range". The class average was 839.417 cm/s/s, which clearly is a lower value than ours. As you can see in the data set, most values resided in the seven and eight hundreds. However, there were a few outliers at 1225.4 cm/s/s and 659.39 cm/s/s, which skewed our average.
Did the object accelerate uniformly? How do you know?
Yes, because in the V-T graph there is a constant slope. As shown in the graphs in the data section above, acceleration is almost exactly uniform as there is almost a constant slope shown in the linear trendline. We know this because the slope of the velocity time graph infact equals acceleration.
What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?
In our lab, it was hard to make gravity the only factor. Being that we are in a high school, equipment such as a vacuum is not easily accessible. In this experiments, factors that could have been a cause to increase acceleration to be higher than it should be were the density of the weight and the factor of air pressure. Contrary, the effect of friction could have caused acceleration to be lower than it should be.
Conclusion Our results turned out to be very accurate. Our hypothesis was partially correct. We believed that the value of our accelerating object should be about 981 cm/s/s. As we found from our percent error, our calculation was relatively accurate, with a percent of 9.31%. However, we originally thought that the v-t graph would ride in the fourth quadrant and be negatively sloped away from the origin. However, since we truncated the negative values, the v-t graph was in a positive direction. We were correct in believing that the accelerating "g" value would come from the A value in the position-time graph, which is shown in the data section above. Our lab, although very accurate, contained some sources of error. Again, or value of percent difference from the class average was -5.98%, while the percent error from the g value of 981 cm/s/s turned out to be 9.31%. This error could have been found in rounding and determining measurements. Since we used a tape measure "taped" to the ground (as was the ticker tape), it may have shifted slightly as we were recording. Another source of error was the fact that the spark tape was rubbing through the spark timer, causing friction. This may have delayed the amount of time between each dot and skewed our data. A way to change the lab in favor of all of the lab groups would be to alter the tactic of tracking the time between each dot. If there were a more accurate way to record the position of a weight over time, such as an electronic device that rids free of friction, it should be used. In conclusion, our experiment was successful!
Activities
9/12
Constant Motion (slow and Fast) and at Rest Graphs
Graph 1 X-T: where are you located? slope=speed Graph 2 V-T: how fast are you going? negative= going towards/backwards; slope=acceleration area=displacement Graph 3 A-T: what is the change in velocity?= area under the graph at rest: standing still. constant speed: covers a distance at a rate. fast constant speed: covers large distances. slow constant speed: covers small distances. accelerating graph: distance increases between the points. decelerating graph: distance decreases between the points.
How can you tell that there is no motion on a…
position vs. time graph
it is at the starting point
velocity vs. time graph
horizontal line at zero
acceleration vs. time graph
horizontal line at zero
How can you tell that your motion is steady on a…
position vs. time graph
it will be a constant slope
velocity vs. time graph
it will be a horizontal line
acceleration vs. time graph
it will be a horizontal line
How can you tell that your motion is fast vs. slow on a…
position vs. time graph
steeper slope indicates a faster moving object
velocity vs. time graph
y value of the line would be greater for the faster moving object than the slower one
acceleration vs. time graph
the speed stays the same so there is no acceleration --> keeps it at zero
How can you tell that you changed direction on a…
position vs. time graph
the slope will become negative once you turn around, but positive if it were previously negative
velocity vs. time graph
reaches zero and decreases under x-axis
acceleration vs. time graph
only shows the change in speed; does not show direction
What are the advantages of representing motion using a…
position vs. time graph
it shows direction, gives intervals of time, and has a slope
velocity vs. time graph
slope shows the increase/decrease in velocity
acceleration vs. time graph
given that there is a change in speed, slope shows this change
What are the disadvantages of representing motion using a…
position vs. time graph
acceleration is not shown
velocity vs. time graph
this takes into account displacement, not total distance
acceleration vs. time graph
only shows changes within the speed but doesn't show velocity
Define the following:
No motion: when the object is at rest; velocity, position, and speed are at zero
Constant speed:when the object is moving at a constant rate; a constant distance is covered in the same amount of time, no matter the interval.
Notes
Kinematics: study of motion
Constant Speed
Types of Speed average speed= total distance/ total time==> can also be found if you know two different speeds.. you avereage them together! constant speed--> not changing, the instantaneous speed is always the same for the time interval (time is really small) Types of Motion
at rest
constant speed
increasing speed
decreasing speed
acceleration: changing speed; involves increasing and decreasing speeds
Motion and Ticker Tape Diagrams
show direction of velocity
the direction of the arrows change depending on the direction (right left up down)
when at rest (v=0; a=0) when at constant speed
---> ---> ---> ---> (a=0)
when increasing speed
--->------>----------->
---> (a)
when decreasing speed
-------------->------->--->
<--------(a) (negative value)
Ticker Tape {.} at rest
X= constant Y= increasing Z=decreasing
*draw back to this*: ticker tapes dont show directions; but can be easily used when you need to take measuresments Signs are Arbitrary
depends on the situation, depends which way it points
Graphs at Rest and at Constant Speed Examples
Acceleration
(a) m/ s (squared)
V= Δd/ Δt (only for average or constant speeds)
V= (Vi +Vf)/2 (only for average speed)
FIVE KINEMATICS EQUATIONS
a= Vf- Vi/ Δt Vf= Vi +aΔt (no displacement Δd= 1/2( Vi + Vf)t Δd= Vit + 1/2at(squared) Vf(squared)=Vi(squared)+2aΔd
Increasing and Decreasing Speed Graphs
increasing speed position time graph
increasing speed velocity time graph
increasing speed acceleration time graph
decreasing speed position time graph
decreasing speed velocity time graph
decreasing speed acceleration time graph
Interpreting Position Time Graphs (E,F,G)
E. F. G.
Free Fall
only gravity is acting on object and no other forces
ignoring air resistance
ie: dropping something, the instant it is dropped to the time it hits the ground is free fall
EXAMPLE
example of freefall
Homework
9/8 (Lesson 1: 1D Kinematics)
What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
Displacement was one of the topics that I understood well from our class discussion. It simply is the combination of distance traveled from the starting point, plus direction—aka the overall change of position
Speed v. Velocity. This concept I understood because to me, velocity : displacement :: speed : distance. Speed is the rate at which something moves, to matter the distance. Velocity, however, takes into account displacement. If an object moves ten feet forward and ten feet back in 20 seconds, the velocity will be zero; however, the speed will be 1 ft/sec.
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.
I was not clear about scalars when I first read it. I didn’t even know what the term was, so it struck me. But now, I can tell that it is just the magnitude alone (which I know because of Precalc last year). It is simply one quantitative value.
What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
I do understand everything that I read.
What (specifically) did you read that was not gone over during class today?
I did not know of scalars. I only knew of vectors from last year in Precalc.
9/9 (Lesson 1: 1D Kinematics)
What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
Ticker tape. I understood the concept, that the faster an object is moving, the more space there is in between each dot. Vice Versa.
Motion diagram. I understood this concept in class. It is simply a visual aid to depict how fast/slow an object is moving, as well as what direction it is moving in.
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.
I was unclear about how to show an accelerating object. But after the reading, I understood that accelaration could be clearly shown by a cluster of dots that spreads out, or vice versa. Below is a picture that demonstrates this concept.
What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
I do understand everything that I read.
What (specifically) did you read that was not gone over during class today?
I knew of everything, as we went over it in class.
9/13 (Lesson 2 1D Kinematics)
What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
Acceleration is a vector quantity that is defined as the rate at which an object changes velocity. If it's velocity changes, then it is accelerating.
Constant Acceleration is when it changes its amount by the same amount, each second.An object that is constantly changing its velocity is an accelerating object--but not at constant acceleration.
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.
I was not clear about calculating the Average Acceleration. But, the equation below from the physics classroom helped me understand more easily.
What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
I do understand everything that I read.
What (specifically) did you read that was not gone over during class today?
I did not know of Average Acceleration.
9/15 (Lesson 3&4: 1D Kinematics)
Lesson 3
What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
Ticker Tape Diagrams I already understood well from all of the labs that we have completed recently. As the object is accelerating, the spaces increase between the dots. Likewise, the spaces decrease between the dots when the object in motion are decelerating.
Vector Diagrams include an aspect that ticker tape diagrams cannot. They show direction of the object AND direction of acceleration. Vector diagrams are used more with velocity, whereas ticker tape diagrams are used with speed.
What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.
To be honest, there was nothing in this section that I did not understand. The concept is fairly simple, especially since we have gone over it multiple times.
What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
I do understand everything that I read.
What (specifically) did you read that was not gone over during class today?
:) Nada!
Lesson 4
What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
I already understood the meaning of the shape of a position-time class. The activity that we did in class involving the motion sensor helped me understand that when an object is at a constant, positive velocity, the p-t graph will look like a straight line. However, if the velocity changes, it will become a curved line.
Essentially, slope= velocity, in ALL circumstances when looking at a p-t graph. Therefore, displacement is definitely accounted for. As the slope goes, so goes velocity!
What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.
There really wasn't anything in the reading that wasn't gone over thoroughly in class. I feel confident in this topic.
What () did you read that you still don’t understand? Please word these in the form of a question.
I do understand everything that I read.
What (specifically) did you read that was not gone over during class today?
Everything has been covered in the past few days or so.
10/3 Lesson 5 Method 1
Free fall is an object that falls, with the only force acting upon it being gravity. All objects, despite the mass, will fall at the same rate given that it is in a vacuum. Introduction to Free Fall A free falling object is an object that is falling under the sole influence of gravity. There are 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 (10 m/s/s)
The dot diagram at the right depicts the acceleration of a free-falling object. If distance is increasing it is a sign that the ball is speeding up as it falls downward. The Acceleration of Gravity 9.8 m/s/s is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. The symbol g denotes it. Representing Free Fall by Graphs A position versus time graph for a free-falling object is shown below. A velocity versus time graph for a free-falling object is shown below.
A diagonal line on a velocity versus time graph signifies an accelerated motion.
*The velocity of a free-falling object that has been dropped from a position of rest is dependent upon the time that it has fallen*
The formula for determining the velocity of a falling object after a time of t seconds is vf = g * t Example Calculations: At t = 6 s vf (9.8 m/s2) * (6 s) 58.8 m/s
At t = 8 s vf (9.8 m/s2) * (8 s) 78.4 m/s The distance fallen after a time of t seconds is given by the formula= d = 0.5 * g * t2 Example Calculations: At t = 1 s d = (0.5) * (9.8 m/s2) * (1 s)2 = 4.9 m At t = 2 s d = (0.5) * (9.8 m/s2) * (2 s)2 = 19.6 m At t = 5 s d = (0.5) * (9.8 m/s2) * (5 s)2 = 123 m (rounded from 122.5 m) The Big Misconception
Free-fall is the motion of objects that move under the sole influence of gravity; free-falling objects do not encounter air resistance. More massive objects will only fall faster if there is an appreciable amount of air resistance present. All objects accelerate at the same rate involves the concepts of force and mass. 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.
Egg Drop Project
our project
"cone shaped capsule"
Description Our final project had a parachute attached by a string to a cone, made of newspaper. Inside, there were strips of crumpled paper. Along the sides to provide cushioning were straws. Also, at the tip of the cone, we stuffed newspaper and two paperclips. This, hopefully, would provide some weight and some distance between the egg and the crash area. Results Our egg cracked a tiny bit from the crash. It was mostly protected from the force (stopping) because of the cushioning at the bottom of the cone. Analysis Originally, our project had been a square shape crate that held the egg in the center. Since the egg jumped out of this contraption, we decided to start over. We then moved onto the cone model, and added straws to provide cushioning. On our final prototype, we added two more straws for cushioning. Finally, when we dropped our project, we had hoped that the parachute would provide some air resistance so that it would be an easy fall. As it fell, the parachute gave way. But, in result, our egg was mostly safe, aside from a tiny crack. The cushioning at the bottom of the cone most likely created a better crash zone, preventing the egg from smashing. Our result was that our cone moved at a velocity of 5.55 m/s. It's acceleration was 7.26 m/s2. Below is our calculation of how we got acceleration. What would we do to fix it? To fix our project, we most likely would have made a larger surface area on the parachute in ration to the size of the cone. We could have also taken out some of the paper clips to reduce the weight. In my opinion, was a smart move for us to go from a box to a cone.
Table of Contents
Kinematics
Constant Speed
CMV Lab
Ali and Caroline 9/8Objective:What is the speed of a Constant Motion Vehicle (CMV)?
Hypotheses: A CMV moves 20 ft/15 secs --> 609.6 cm/15 secs --> 40.64 cm/sec. Distances can be measured to the nearest millimeter, but then, an educated guess must be used for the next decimal place. A time position graph tells the time relative to the distance away from zero (which is also known as the position). This, hopefully, should be a linear function.
Available Materials: Constant Motion Vehicle, Tape measure and/or metersticks, spark timer and spark tape
Data:
Analysis
Essentially, the graph of the CMV shows that it is moving at a constant speed, because of the slope of the line (positive direction). At each interval, it is moving approximately the same distance away from the point before it.
Discussion questions
- Why is the slope of the position-time graph equivalent to average velocity?
- Slope is the measure of Δy/ Δx
- in this case, the Δy=Δd (distance) and the Δx= Δt (time)
- therefore; Δy/ Δx = Δd/ Δt = V
- Why is it average velocity and not instantaneous velocity? What assumptions are we making?
- Instantaneous velocity is at one point, so it only accounts for one point. Average velocity takes all of the points into account; therefore, it is more accurate. We are assuming that all of the values will be relatively close, and not many outliers will exist.
- Why was it okay to set the y-intercept equal to zero?
- It is okay to set the y-intercept equal to zero because since the y-axis is the position, if it’s at zero, it will not be moving. This was the starting point from which or average velocity was graphed
- What is the meaning of the R2 value?
- R2 is the percentage of points described by the trend line. It is a tool to help us know how “fit” our data is compared to the trend line. If the value is around .99… the data is accurate.
- 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?
- If it were to be slower, the line would lie below our trend line. It is still traveling in the same amount of time, but the distance traveled won’t be nearly as much.
ConclusionMy results showed that the average speed of a CMV was 37.962 cm/s. It seems that as the time sped up, the distance the CMV traveled lessened. My hypothesis was that the CMV would travel 40.64 cm/s. Although it did not travel as far as I had originally thought, the CMV still traveled in a 5 cm range to my guess. Therefore, I consider it fairly accurate.
However, whenever performing an experiment, there is always room for error. This error could be found in the instruments that we used for measuring the length between the dots on the spark tape. The meter stick may have slid while we were recording down the measurements; the shape of the meter stick also was thick, so it made it difficult to make out the exact numbers. Anotheraspect of error was that the measurements weren’t as precise as they could have been. If we had more effective, digital tools, we could have gotten the exact measurement.
In order to minimize error, next time, we could use an instrument that does not slide and is flatter, therefore making the numbers easier to read. Something like measurement tape could be more effective. Rather, a digital tool could be used for better accuracy.
CMV Lab Part 2
Ali, Caroline, Dani, Julia 9/21Objectives: Both algebraically and graphically, solve the following 2 problems. Then set up each situation and run trials to confirm your calculations.
Available Materials: Constant Motion Vehicle, Tape measure and/or metersticks, Masking tape (about 30 cm/group), Stop watch, spark timer and spark tape
According to our calculations, the blue car should travel 372.321 cm and then crash into the yellow car (which would have moved 229.214 cm). Below is a video of what ACTUALLY happened. The data table below it shows each trial and the distance each CMV actually moved.
Here, the two cars crash at a certain point. We then proceeded to mark the place where they met, proving our calculations to be approximately correct.
According to our calculations, the blue car should travel 260.156 cm and then become parallel with the yellow car. Below is a video of what ACTUALLY happened. The data table below shows each trial and the distance it took for the blue to catch up.
The faster of the two cars (blue) is catching up to the yellow car at a certain point. We then proceeded to mark the place where they met, proving our calculations to be approximately correct.
Analysis
Part A: crashing
Based on this percent error of .57%, I can conclude that our results were very accurate for the case of the blue car striking the yellow car. The smaller percent that we have, the better our results are in actuality. We chose to take an average of or experimental values because all of our experimental values were very close to each other. For percent difference of the crash, we took the average experimental value and subtracted the individual value. Then, we divided it by the average value and multiplied by 100 to get a percentage. Our percentages were very small, thus demonstrating precision. Overall, between these two methods, we can conclude that our results were very accurate and precise.
Part B: catching up
Once again, we used percent error again to find out how accurate the results for Part B were. As shown above, our results turned out to be extremely accurate, within .03% of the theoretical results. For percent difference of the catching up, we took the average experimental value and subtracted the individual value. Then, we divided it by the average value and multiplied by 100 to get a percentage. Our percentages were very small, thus demonstrating precision. There was one outlier where the blue car caught up at 230.96 cm, resulting in a percentage of 11.25%. This outlier may have been due to slow reaction time or improper readings of the meter stick. However,it can still be concluded that our results were very accurate and precise.
Discussion questions
- Where would the cars meet if their speeds were exactly equal?
- The cars should realistically meet at 300 meters given that their speeds were exactly equal; this is the half way point.
- 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.
crashing
catching up
- 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?
this is when the cars crash
this is when the blue car surpasses the yellow car
ConclusionIn conclusion, our results turned out to be accurate. Each time that we performed a trial, the results ended up to be in some range of what we had originally predicted.For part A, our calculations showed that the two cars would meet and crash once the blue traveled 372. 321 cm and the yellow traveled 229.214 cm.For part B, our calculations showed that the faster, blue car would catch up with the slower, yellow car at the 260.156 cm mark. Since we used ramps, the cars were able to move in a straight line, thus giving us more accurate results. One of our sources of error found in our calculations was that for part A, instead of the two distances adding up to 600, it added up to 601. This was due to the fact that we used significant figures, thus truncating any "useless" decimals. Error could have also been found in the timing and reaction time. If we used another type of method to measure, like a USB to DataStudio, it could have been more accurate. Our percent error for part A was 0.57%. The percent difference ranged from -0.2% to 0.3% for the blue car and for the yellow car percents ranged from -0.55% to 0.55%. These values as well were very small. For Part B, percent error was 0.03 % and percent difference ranged from -0.2% to 1.95%, with an outlier at 11.35%. However, I believe the outlier was a human error.
Overall, these percentages are very small, but could be improved. There are other evident ways to improve our data, but yet again our data is very precise. I believe our calculations couldn't have been that much closer to the actual values. If we were to improve the lab, I would say that it were required for each group to use the lab and a USB to get the most accurate results. Also, the lab could note that each car should be using a fresh pack of batteries. Therefore, everyone would be on the same page.
Accelerating/Decelerating Car Lab
Ali and Caroline 9/14Objectives:
- What does a position-time graph for increasing speeds look like?
- What information can be found from the graph?
Hypotheses: A position-time (x-t) graph for increasing speed looks like a curve upwards, starting less steep and becoming steeper as time goes on. This is because it is covering a greater distance in less time. From a position-time graph, we can tell the velocity, because we can figure out the slope of the line.Available Materials
Spark tape, spark timer, track, dynamics cart, ruler/meterstick/measuring tape
Procedure
Acceleration
We placed the car on the ramp, which was equipt with an inclination (due to one text book, shown). Therefore, this would make the car accelerated down the ramp. We attached a piece of spark tape to the car and put it through the spark timer. We then released the car until it was at the end of the ramp. Then, we collected the data by measuring the distance between the initial point to the final point, for each dot. The result is below.
Deceleration
We placed the car on the bottom of the the ramp, which was equipt with an inclination (due to one text book, shown). Therefore, this would make the car decelerate up the ramp (values would decrease because it went uphill). We attached a piece of spark tape to the car and put it through the spark timer. We then pushed the car up the ramp until it stopped. Then, we collected the data by measuring the distance between the initial point to the final point, for each dot. The result is below.
Data and Graph
Accelerating Data:
Analysis
Interpret the equation of the line (slope, y intercept) and the R2
- y= Ax2+Bx
- y= distance& x= time
- change in distance= 1/2at2 + vit
- accelerating
- Δd= 15.541t2 + 11.795t
- R2= 0.99996
- decelerating
- Δd= -25.32t2 + 90.698t
- R2= 0.9999
Both of our R values are VERY close to 100%. But, keep in mind, we used a polynomial trendline, rather than a linear trendline. Originally, when we used the linear trendline, our results were for increasing and decreasing were .92 and .88, respectively. Therefore, the polynomial trendline was a better choice because it was a curve.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:
- halfway= 6 cm/s
- end= 5cm/s
- Decreasing:
- halfway= 5.49 cm/s
- end= 1.5 cm/s
Find the average speed for the entire trip- Accelerating
- Δd/Δt= 38.84 cm/s
- Decelerating
- Δd/Δt= 8.9375 cm/s
Discussion Questions- What would your graph look like if the incline had been steeper?
- It would be a steeper curve, because the points would travel a greater distance in a shorter amount of time.
- What would your graph look like if the cart had been decreasing up the incline?
- The cart would start out going fast, with a steeper slope. It would then curve the opposite way, and gradually have a less steep slope.
- Compare the instantaneous speed at the halfway point with the average speed of the entire trip.
- Accelerating instantaneous speed at the halfway point is 60 cm/s. For the average accelerating speed we had 38.84 cm/s. I believe that there is such a difference because the line that we used curved. There most likely was an error in calculation, as our values should have been close to that of the instantaneous speed.
- Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense?
- This makes sense because instantaneous speed is the speed that the object is moving during the specified time. Here, the tangent line intersects with one point out of the eighteen on the graph. Thus, by taking this point, we are figuring out velocity, which should be consistent throughout.
- In class today we learned that the formula for some polynomial is y=Ax2+ Bx. We learned that the A variable is 1/2 the acceleration (or in our case, 15.421, accelerating), while the B variable is initial velocity (again, in our instance, 11.795. )If you replace the variables with those from the acceleration equation, they turn out to be the same. The x axis can be exchanged with Δt, or the time value. The y variable can be replaced with Δd because it is position.
- Draw a v-t graph of the motion of the cart. Be as quantitative as possible.
ConclusionThe results of this experiment proved our hypothesis to be successful. We hypothesized that the result of data would be plotted as a curve, and so it was. It seemed to be that as the car traveled along the ramp, speed was gained, and thus traveling greater distances in a smaller amount of time. Likewise, decelerating started off fast, but gradually slowed down, covering a less distance in a longer time. In this lab, there were a few sources of error. To start off, me and my partner did not start measuring at the first dot made by the spark timer. Thus, we had a false "initial velocity". This definitiley skewed our data, without us even thinking it done harm. It seems as if the car was already in motion, therefore accounting for this high initial velocity. Instantaneous velocity is supposed to be similar to the average speed, but as stated above, this was not accomplished by our data. Another source of error was most likely in our mathematical calculations; we may have calculated wrong or messed up numbers. Lastly, although our data was accurate, it was not as precise as it could have been. If we had been using something to hook up to the computer, our measurements would have been far more precise. However, this was not the case.
Overall, from this lab I learned that in a position-time graph lab, a curve will result from acceleration. The faster the object moves, the more distance covered in shorter periods of time, and the "steeper" the curve will become.
Lab Acceleration of a Falling Body
Ali, Caroline 10/5Objective:What is the acceleration of a falling body?Hypothesis:The acceleration of a falling body should be 9.8 m/s2 or 981 cm/s2. Hypothetically, the v-t graph should look like the following
Materials:
Ticker Tape Timer, Timer tape, Masking tape, Mass, clamp, meterstick.
Procedure
For our lab, we took a spark timer, spark tape, masking tape, a weight, and dropped it off from a height. Dots were plotted for each interval of a second (.1,.2,...1). We then laid out the spark tape on the floor and measured the distance between each dot made by the timer by using a measuring tape. A picture of how we measured is shown below.
Data and Graphs
Above is the data for the position time graph, which compares the amount of time passed as the weight fell to the ground (measured in cm).
The two images above show the data for the instantaneous velocity v. the mid time. Below is an example of how we found the instantaneous velocity.
We found the mid-time by taking the final and initial value and dividing it by 2. For example, MT= (0 s +0.1 s)/2 = 0.5 s.
Sample Calculations
Below is the percent difference of -5.98%. This is comparing the class average of 839.417 cm to our average of 889.65 cm. This percentage is very small and is within the "20 percent range" concept we created for accuracy.
Below is the percent error of 9.31. It is comparing the theoretical value of 981 cm/s/s to our experimental value of 889.65 cm/s/s. The percent is very small and is within the "20 percent range" concept we created for accuracy.
Below is an example of how we found the velocity for a given point.
Analysis
To begin, we found that the polynomial equation y=Ax2+Bx happens to come from the equation d=1/2at2+vit; therefore, A=1/2acceleration and B=initial velocity. Additionally, y=mx+b can be derived from the equation Vf= at +Vi. These factors will later be discussed throughout the analysis.
For the position-time graph, the trend is a polynomial one, indicated by the equation of the line (y=445.35x2+50.138x). In this case, the acceleration is 445.35 and the initial velocity was 50.138. Our R2 value was extremely accurate at 1, showing that we had great results.
The velocity time graph, unlike the p-t graph, was a linear function (y=889.65x+50.91). In this case, slope=889.65 and B=50.91. Our R2 value was extremely accurate with a value of 0.99994.
For both graphs, we got an initial velocity that was not zero as expected because we may have had a delayed reaction when dropping the weight, after the ticker tape was already turned on. According to the original equation, 1/2A equals 445.35. Technically, this should mean that the value 445.35 cm from the p-t graph, doubled, should equal our slope of the velocity time graph. Although not an exact fit, the value almost equals 889.65 cm/s/s, as it equals 890.7 cm/s/s. Thus, we can conclude it is accurate.
Discussion Questions
- Does the shape of your v-t graph agree with the expected graph? Why or why not?
- No, the expected graph was a constantly sloped line in the negative direction moving away from the origin. This is because we omitted the negatives when counting our results.
- Does the shape of your x-t graph agree with the expected graph? Why or why not?
- Yes, we expected the graph to be a polynomial--gradually increasing as time passed. It completely matches up with our expected graph, a J curve.
- How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.)
-
- Our results confirm that our data is very accurate. Again, our data was in the "20 percent range". The class average was 839.417 cm/s/s, which clearly is a lower value than ours. As you can see in the data set, most values resided in the seven and eight hundreds. However, there were a few outliers at 1225.4 cm/s/s and 659.39 cm/s/s, which skewed our average.
- Did the object accelerate uniformly? How do you know?
- Yes, because in the V-T graph there is a constant slope. As shown in the graphs in the data section above, acceleration is almost exactly uniform as there is almost a constant slope shown in the linear trendline. We know this because the slope of the velocity time graph infact equals acceleration.
- What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?
- In our lab, it was hard to make gravity the only factor. Being that we are in a high school, equipment such as a vacuum is not easily accessible. In this experiments, factors that could have been a cause to increase acceleration to be higher than it should be were the density of the weight and the factor of air pressure. Contrary, the effect of friction could have caused acceleration to be lower than it should be.
ConclusionOur results turned out to be very accurate. Our hypothesis was partially correct. We believed that the value of our accelerating object should be about 981 cm/s/s. As we found from our percent error, our calculation was relatively accurate, with a percent of 9.31%. However, we originally thought that the v-t graph would ride in the fourth quadrant and be negatively sloped away from the origin. However, since we truncated the negative values, the v-t graph was in a positive direction. We were correct in believing that the accelerating "g" value would come from the A value in the position-time graph, which is shown in the data section above.
Our lab, although very accurate, contained some sources of error. Again, or value of percent difference from the class average was -5.98%, while the percent error from the g value of 981 cm/s/s turned out to be 9.31%. This error could have been found in rounding and determining measurements. Since we used a tape measure "taped" to the ground (as was the ticker tape), it may have shifted slightly as we were recording. Another source of error was the fact that the spark tape was rubbing through the spark timer, causing friction. This may have delayed the amount of time between each dot and skewed our data.
A way to change the lab in favor of all of the lab groups would be to alter the tactic of tracking the time between each dot. If there were a more accurate way to record the position of a weight over time, such as an electronic device that rids free of friction, it should be used.
In conclusion, our experiment was successful!
Activities
9/12
Constant Motion (slow and Fast) and at Rest Graphs
Graph 1 X-T: where are you located? slope=speedGraph 2 V-T: how fast are you going? negative= going towards/backwards; slope=acceleration area=displacement
Graph 3 A-T: what is the change in velocity?= area under the graph
at rest: standing still.
constant speed: covers a distance at a rate.
fast constant speed: covers large distances.
slow constant speed: covers small distances.
accelerating graph: distance increases between the points.
decelerating graph: distance decreases between the points.
How can you tell that there is no motion on a…
- position vs. time graph
- it is at the starting point
- velocity vs. time graph
- horizontal line at zero
- acceleration vs. time graph
- horizontal line at zero
How can you tell that your motion is steady on a…- position vs. time graph
- it will be a constant slope
- velocity vs. time graph
- it will be a horizontal line
- acceleration vs. time graph
- it will be a horizontal line
How can you tell that your motion is fast vs. slow on a…- position vs. time graph
- steeper slope indicates a faster moving object
- velocity vs. time graph
- y value of the line would be greater for the faster moving object than the slower one
- acceleration vs. time graph
- the speed stays the same so there is no acceleration --> keeps it at zero
How can you tell that you changed direction on a…- position vs. time graph
- the slope will become negative once you turn around, but positive if it were previously negative
- velocity vs. time graph
- reaches zero and decreases under x-axis
- acceleration vs. time graph
- only shows the change in speed; does not show direction
What are the advantages of representing motion using a…- position vs. time graph
- it shows direction, gives intervals of time, and has a slope
- velocity vs. time graph
- slope shows the increase/decrease in velocity
- acceleration vs. time graph
- given that there is a change in speed, slope shows this change
What are the disadvantages of representing motion using a…- position vs. time graph
- acceleration is not shown
- velocity vs. time graph
- this takes into account displacement, not total distance
- acceleration vs. time graph
- only shows changes within the speed but doesn't show velocity
Define the following:Notes
Kinematics: study of motionConstant Speed
Types of Speedaverage speed= total distance/ total time==> can also be found if you know two different speeds.. you avereage them together!
constant speed--> not changing, the instantaneous speed is always the same for the time interval (time is really small)
Types of Motion
- at rest
- constant speed
- increasing speed
- decreasing speed
acceleration: changing speed; involves increasing and decreasing speedsMotion and Ticker Tape Diagrams
- show direction of velocity
- the direction of the arrows change depending on the direction (right left up down)
when at rest (v=0; a=0)when at constant speed
- ---> ---> ---> ---> (a=0)
when increasing speed- --->------>----------->
- ---> (a)
when decreasing speed- -------------->------->--->
- <--------(a) (negative value)
Ticker Tape{.} at rest
*draw back to this*: ticker tapes dont show directions; but can be easily used when you need to take measuresments
Signs are Arbitrary
Graphs at Rest and at Constant Speed Examples
Acceleration
FIVE KINEMATICS EQUATIONS
a= Vf- Vi/ ΔtVf= Vi +aΔt (no displacement
Δd= 1/2( Vi + Vf)t
Δd= Vit + 1/2at(squared)
Vf(squared)=Vi(squared)+2aΔd
Increasing and Decreasing Speed Graphs
Interpreting Position Time Graphs (E,F,G)
E.F.
G.
Free Fall
Homework
9/8 (Lesson 1: 1D Kinematics)
9/9 (Lesson 1: 1D Kinematics)
9/13 (Lesson 2 1D Kinematics)
9/15 (Lesson 3&4: 1D Kinematics)
Lesson 3- What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
- Ticker Tape Diagrams I already understood well from all of the labs that we have completed recently. As the object is accelerating, the spaces increase between the dots. Likewise, the spaces decrease between the dots when the object in motion are decelerating.
- Vector Diagrams include an aspect that ticker tape diagrams cannot. They show direction of the object AND direction of acceleration. Vector diagrams are used more with velocity, whereas ticker tape diagrams are used with speed.
- What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.
- To be honest, there was nothing in this section that I did not understand. The concept is fairly simple, especially since we have gone over it multiple times.
- What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
- I do understand everything that I read.
- What (specifically) did you read that was not gone over during class today?
- :) Nada!
Lesson 410/3 Lesson 5 Method 1
Free fall is an object that falls, with the only force acting upon it being gravity. All objects, despite the mass, will fall at the same rate given that it is in a vacuum.Introduction to Free Fall
A free falling object is an object that is falling under the sole influence of gravity. There are two important motion characteristics that are true of free-falling objects:
The dot diagram at the right depicts the acceleration of a free-falling object. If distance is increasing it is a sign that the ball is speeding up as it falls downward.
The Acceleration of Gravity
9.8 m/s/s is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. The symbol g denotes it.
Representing Free Fall by Graphs
A position versus time graph for a free-falling object is shown below.
A velocity versus time graph for a free-falling object is shown below.
*The velocity of a free-falling object that has been dropped from a position of rest is dependent upon the time that it has fallen*
The formula for determining the velocity of a falling object after a time of t seconds is vf = g * t
Example Calculations:
At t = 6 s vf
(9.8 m/s2) * (6 s)
58.8 m/s
At t = 8 s vf
(9.8 m/s2) * (8 s)
78.4 m/s
The distance fallen after a time of t seconds is given by the formula= d = 0.5 * g * t2
Example Calculations:
At t = 1 s d = (0.5) * (9.8 m/s2) * (1 s)2 = 4.9 m
At t = 2 s d = (0.5) * (9.8 m/s2) * (2 s)2 = 19.6 m
At t = 5 s d = (0.5) * (9.8 m/s2) * (5 s)2 = 123 m (rounded from 122.5 m)
The Big Misconception
Free-fall is the motion of objects that move under the sole influence of gravity; free-falling objects do not encounter air resistance. More massive objects will only fall faster if there is an appreciable amount of air resistance present.
All objects accelerate at the same rate involves the concepts of force and mass. 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.
Egg Drop Project
Description
Our final project had a parachute attached by a string to a cone, made of newspaper. Inside, there were strips of crumpled paper. Along the sides to provide cushioning were straws. Also, at the tip of the cone, we stuffed newspaper and two paperclips. This, hopefully, would provide some weight and some distance between the egg and the crash area.
Results
Our egg cracked a tiny bit from the crash. It was mostly protected from the force (stopping) because of the cushioning at the bottom of the cone.
Analysis
Originally, our project had been a square shape crate that held the egg in the center. Since the egg jumped out of this contraption, we decided to start over. We then moved onto the cone model, and added straws to provide cushioning. On our final prototype, we added two more straws for cushioning. Finally, when we dropped our project, we had hoped that the parachute would provide some air resistance so that it would be an easy fall. As it fell, the parachute gave way. But, in result, our egg was mostly safe, aside from a tiny crack. The cushioning at the bottom of the cone most likely created a better crash zone, preventing the egg from smashing.
Our result was that our cone moved at a velocity of 5.55 m/s. It's acceleration was 7.26 m/s2. Below is our calculation of how we got acceleration.
What would we do to fix it?
To fix our project, we most likely would have made a larger surface area on the parachute in ration to the size of the cone. We could have also taken out some of the paper clips to reduce the weight. In my opinion, was a smart move for us to go from a box to a cone.