Ch2_LefflerM

Class Notes 9/7/11
Position- the place in which you stand, defined by coordinates. Distance - ( Scaler) the total amount traveled. Displacement - (Vector) the change in your position, using direction and numerical values. Distance is a scaler property. Displacement is a vector; which is anything that can be represented with an arrow.

ALL UNITS ARE METERS! Speed (Scaler) - how fast Velocity (Vector) - rate of change of position



= Lab: A Crash Course Into Velocity (Part 1) = September 8th, 2011 Lindsay Marella and Magna Leffler

**Objectives:** 1. What is the speed of a Constant Motion Vehicle (CMV)? 2. What does a position-time graph tell us? 3. How precisely are you required to measure?

**Hypothesis:** 1. (Q) What is the speed of a CMV? (A) About 2mph 2. (Q) What does a position time graph tell us? (A) A position time graph tell us at what time the CMV is at point A, and when it travels Y amount of meters away from the origin that it took X amount of time. 3. (Q) How precisely are you required to measure? (A) You should measure to the extent the tool will tell you plus 1 decimal place. Length (cm): 24.5__X__ <--- estimated value, you always have a "guess" value. Precision tell us how often we hit the same spot, while accuracy is how close we get to a target.

**Data Points:** **Graph:**

**Discussion questions** 1. Why is the slope of the position-time graph equivalent to average velocity? 2. Why is it average velocity and not instantaneous velocity? What assumptions are we making? 3. Why was it okay to set the y-intercept equal to zero? 4. What is the meaning of the R2 value? 5. If you were to add the graph of another CMV that moved more slowly on the same axes as your current graph, how would you expect it to lie relative to yours?
 * The slope is equal to the rise over run, rise is the change in y, and run is the change in x. Y is position in this graph and x is time, so change in position divided by the change in time is equal to average velocity.
 * It takes time to build up to speed, we are making the assumption that once it builds up to it’s peak, its speed will stay constant.
 * It was okay to set the y-intercept to zero because, that was the point at which the line started, zero seconds and zero centimeters.
 * How close the line is to a perfect fit.
 * I would expect the line of the slower moving CMV to lie below the line of the faster moving CMV because it will cover a smaller amount of space for the same amount of time and the points will be closer together.

**Conclusion** The ending result was that our blue constant motion vehicle’s speed was 40.196 centimeters/second, as opposed to our hypothesis, 2mph, or 89.408 centimeters/second. The speed of the constant motion vehicle was actually closer to 1mph. There are countless sources or error in this particular lab and some ways to minimize these issues would be to put brand-new batteries in each one, find a flat floor area and use it for every experiment, make sure that all the CMVs go straight; no turning, and use a flatter meter-stick, or a tape measure for easier, more precise measuring.

**Home Work: Notes 9/8/11 and 9/14/11**
**One Dimentional Kinematics - Lesson 1** **Summarize the reading using technique 2a.**
 * I really had a good grasp on Distance vs. Displacement, distance is the total amount that you traveled; while displacement is how far you are from the origin. Displacement is a Vector and Distance is Scaler. I also took from todays lesson how to calculate both.
 * Vector vs. Scalar measurement I was a little shaky on, but after reading and having multiple examples of the two types I feel better about it. The example with the football coach was extremely helpful, because I can really visualize it. Vectors are measured with directions and numbers while Scalar is merely number based.
 * When finding acceleration do you need to find displacement and velocity? Does every changing motion have an acceleration?
 * We didn't go over acceleration, which is probably why I am having a little trouble understanding what exactly defines the change in acceleration. ( I later found out in class that this wasn't assigned... Oops sorry Ms. Burns...)

**Class Notes 9/9/11**
1. **Definitions**
 * Average Speed: average of all your speeds
 * Constant speed: always has the same instantaneous speed
 * Instantaneous Speed: the speed at the very moment
 * velocity = change in distance/ change in time miles/second

2. **Types of Motion**
 * at rest: a=0, v=0
 * constant speed: ---v> ---v> ---v> ---v> {a=o} there is not acceleration.
 * increasing speed -v--> --v---> ---v>{ ---a>}velocity and acceleration point in the same direction.
 * decreasing speed ---v> --v---> -v--> {<a---} acceleration points in the opposite direction from velocity.

**Signs are arbitrary always label!!**

**IF** the signs are the **SAME speed is increasing**

**IF** the signs are **DIFFERENT speed is decreasing**

**Home Work 9/10/11**
**One Dimensional Kinematics - Lesson 2:**
 * I didn't really have a good grasp on anything from class, I was very distracted and had a hard time copying things down. After the reading I have a better understanding of things.
 * Both types of diagrams (ticker tape and vector)also how to read them, I was shaky on the arrow directons and the inverse relation of deceleration and velocity.
 * When the vector diagram is pointing downwards what would that look like in real life? A person parachuting off a building?
 * I think that we covered everything in class. I was trying to get adjusted to my wiki.

= **Lab 9/12/11: Motion Sensors** = **Objectives:** Static equilibirum is unchanging, while dynamic changes. Position- Time Graph: a strait line that starts at the y-axis(postition) Velocity-Time Graph: a strait line at the origin Acceleration -Time Graph: a strait line at the origin > **Graphs:** Run # 1 shows person at rest about a meter away from the motion sensor. Run # 1 Shows a person walking slowly towards the motion sensor (origin) Run # 2 Shows a person walking slowly away from the origin.
 * 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?

Run # 3 represents a person walking quickly towards the origin. Run # 4 represents a person walking quickly away from the origin.

**Discussion Questions:** 1. How can you tell that there is no motion on a… A. position vs. time graph B. velocity vs. time graph C. acceleration vs. time graph
 * The line will be strait and non-changing.
 * The line will be strait and non-changing.
 * The line will be strait and non-changing.

2. How can you tell that your motion is steady on a… A. position vs. time graph B. velocity vs. time graph C. acceleration vs. time graph
 * The position will increase and be at a constant increase (slope).
 * The velocity should be a strait-ish non-changing line.
 * The acceleration should be a constant un-changing line.

3. How can you tell that your motion is fast vs. slow on a… A. position vs. time graph B. velocity vs. time graph C. acceleration vs. time graph
 * The line should go from high to low in a short distance, while if it were slow it would take a larger area to decrease.
 * The line is strait but shorter on a fast moving motion graph.
 * The line should be strait as well because there is no acceleration because it is constant motion, but the line is shorter.

4. How can you tell that you changed direction on a… A. position vs. time graph B. velocity vs. time graph C. acceleration vs. time graph
 * The line would have the negative slope of the first line.
 * You wouldn’t be able to because it is not changing.
 * The acceleration would be opposite of what it was on the run going forward.

5. What are the advantages of representing motion using a… A. position vs. time graph B. velocity vs. time graph C. acceleration vs. time graph
 * you can see if motion is constant.
 * you can see if the motion has an unchanging velocity or if it is changing every step
 * you can tell which direction based on negative or positive acceleration.

6. What are the disadvantages of representing motion using a… A. position vs. time graph B. velocity vs. time graph C. acceleration vs. time graph
 * It doesn’t take into consideration the shift of weight from one leg to another or a hesitation.
 * If you were to mis-step or wobble just a little, it would throw off your results and make the velocity line choppy
 * If you were to speed up just a little bit or if you slow down, the sensor isn’t pointing at your midsection or you sneeze while running a test you could mess up the entire run.

7. Define the following: A. __No motion__: B. __Constant speed__:
 * stationary with a set position, no velocity or acceleration.
 * always at the same velocity and acceleration, increases by the same number each second.

= Cart On An Incline Lab = 9/13/11 Magna Leffler and Lindsey Marella **Objectives:**

1. What does a position-time graph for increasing speeds look like? 2. What information can be found from the graph?
 * A upward, changing, non-strait line.
 * You can find the place where the cart is as well as what the amount of time it took to get there.

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

**Analysis:** a) Interpret the equation of the line (slope, y-intercept) and the R2 value. Y=12.319x^2+1.3126x, R^2 = 0.99964  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.) c) Find the average speed for the entire trip. add equations  13.47 cm/s

**Discussion Questions:**

1. What would your graph look like if the incline had been steeper? 2.What would your graph look like if the cart had been decreasing up the incline? 3.Compare the instantaneous speed at the halfway point with the average speed of the entire trip. 4.Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense? 5. Draw a v-t graph of the motion of the cart. Be as quantitative as possible. Conclusion?
 * The line would have an increased slope, and the curve would be steeper.
 * It would be in a downward curve.
 * Average speed = 13.47 cm/s
 * Instantaneous speed = 12.5 cm/s
 * The average speed was greater because you reached a much higher speed at some points towards the end. Also the instantaneous speed was only halfway through the data and not at a particularly fast point.
 * This makes sense because the slope of the line equals velocity, and if the line is strait the velocity is the same through the entire line.

Class Work 9/15/11
Interpreting Graphs: A. D. E. F. G.

Home Work: Notes 9/15/11
Kinematics Lesson 3: H.W. Summarize 2a

What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. 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. What (specifically) did you read that you still don’t understand? Please word these in the form of a question. What (specifically) did you read that was not gone over during class today?
 * That the slope of a position time graph is the velocity. How the graphs looked based on where the object was coming from, and speed.
 * I didn't really not understand anything that we did in class. The reading for lesson 3 helped me fortify my understanding
 * Will we be using the slope formula later in class to help find the slope and then later the velocity of a moving object?
 * We covered a lot today in class and I think that everything in lesson 3 was gone over in class.

Kinematics Lesson 4: H.W. Summarize 2a (same night)

Finding slope of a line, and the meaning of the slope of the line in a v-t graph. I was good with what we learned in class, the reading for lesson 4 confused a little bit. How does finding the area help you in any way? When the line is under the x-axis is it always negative even if it breaches the x-axis, or are they treated differently? The area of the v-t graph also the relating the graph to the motion(maybe).
 * What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
 * 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.
 * What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
 * What (specifically) did you read that was not gone over during class today?

= Lab CMV CRASH! = Partners: Lindsey marrela, Rachel Knaple, Molly Lambert **Objectives:** Both algebraically and graphically, solve the following 2 problems. Then set up the situation and run trials to confirm your calculations. Hypothesis: 1. It won't take long for the Faster CMV to catch up to the slower one, I think it will only be one and a half meters before the fast one overpasses the slow one. 2. The CMV's will crash halfway through the 600m.

Calculations for part 1, overtaking.



The calculations pictured above show that the blue CMV should move about 1.4 m, or 135.416 cm until it overtakes the yellow CMV. In the table are the actual meeting points from our trials. Our experimental values for the first part we not that close to what our theoretical point is. Here is a Movie of our blue CMV over taking our yellow CMV. The blue car starts out 1m behind the yellow and then we start them at the same time. When the Blue overcomes the yellow we marked the spot. media type="file" key="Movie on 2011-09-22 at 11.14.mov" width="330" height="330"

Calculations part 2, CRASH The calculations pictured above show where the cars should meet, the blue CMV should travel 475.75 cm while the yellow CMV should travel 124.626 cm. In the cart below are trials that we did, and the results we got for each trial.

CMV's crashing media type="file" key="Movie on 2011-09-22 at 11.21.mov" width="330" height="330"



Discusion Questions: If their speeds were exactly equal, then they would meet exactly halfway, in this case they would meet at 3 meters. The yellow car would travel -3m and the blue car would travel +3m (each from their starting spot).
 * 1. Where would the cars meet if their speeds were exactly equal?**
 * 2. Sketch position-time graphs to represent the catching up and crashing situations. Show the points 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?**

//Conclusion:// //Our results were not what we were expecting. Our Blue CMV was supposed to overtake the yellow at 135.461 cm. In reality, on average it took 256 cm for the blue to overtake the yellow. For the crash part of the lab the blue car was supposed to crash into the yellow car after it had traveled 475.75 cm (yellow traveling -124.626 cm). In our experiments the blue traveled much less than what we had anticipated, normally by more than 100 cm. The reason why our experimental data was so different that what we had anticipated was caused by a few things, the main problem with our experiment was the lack off control that we exerted over the CMV's. Once we let them go there was no stopping which direction (left, right, or forward) they were going. That was what I think was our main error. Other than that it could have been that one person had released the CMV to early. One thing that did worry me, was the fact that the yellow CMVs batteries were constantly falling out, which may have had a slight effect on the lab. If I could do the lab over, I would make a track for the CMV's to travel on while we run the experiment. We did use the long metal "tracks" that are seen in the video, but they didn't offer enough structure. Over all our percent error was really high, meaning that the experiment wasn't very accurate. The percent difference on the other hand was very low which means that our experiment was very precise.//

=Egg Drop Lab= Dropped on 9/28/11 Lab partner Katie Dooman Pictured above is the drawing of the final prototype. Our design started with full-length straws held together with tiny rubber bands, and we made the straws into a cylinder. Then, inside that, we placed two layers of shorter straws, also held together with tiny rubber bands. Inside the two layers of rubber bands, we placed the egg. The egg was wrapped in newspaper, and we placed straws inside the newspaper as we wrapped the egg, in an attempt to create a barrier. Then we used larger rubber bands to secure the egg (which was now inside the newspaper and two smaller cylinders of straws) at the top of the large cylinder. This held it suspended, which when tested, prevented the egg from taking the shock of the impact. Into the bottom of the large cylinder, we threaded additional small straws in order to increase the surface area of the bottom, hoping that it would make for a more stable landing, also absorbing more of the shock, instead of having the shock go to the egg. Our acceleration was 12 m/s^2 which is much to high to be possible for the highest possible acceleration is 9.8 m/s^2. The reason why our acceleration is so off is because the timing of the drop and impact of the egg wasn't properly done. If we were to do it again we would count down from three to zero and then let go. We would also have more than two people time the experiment giving us a higher range to average from. If we had more runs it would have been possible to get an even more accurate average time, which would result in a more correct acceleration. Next time because our design didn't work I would try and make a cone shaped one that all the impact was taken by the tip of the cone. The egg would sit about five inches from the tip so there would be no impact on the egg. Also possibly packing the egg with paper shreds or string

Home Work: 10/3/11
Lesson 5 a,b,c,d,e summarized using summery time 1a Any object that is being acted upon only by the force of gravity is said to be in a state of ** free fall **. There are two important motion characteristics that are true of free-falling objects:
 * Introduction to Free Fall **
 * 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 (often approximated as 10 m/s/s for //back-of-the-envelope// calculations)



Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s, a [|ticker tape trace] or dot diagram of its motion would depict an acceleration. The dot diagram at the right depicts the acceleration of a free-falling object. The position of the object at regular time intervals - say, every 0.1 second - is shown. The fact that the distance that the object travels every interval of time is increasing is a sure sign that the ball is speeding up as it falls downward. If an object travels downward and speeds up, then its acceleration is downward. ** g = 9.8 m/s/s, downward ( ~ 10 m/s/s, downward) **
 * The Acceleration of Gravity **
 * acceleration of gravity ** - the acceleration for any object moving under the sole influence of gravity shown as the symbol ** g ** (9.8 m/s/s)There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude

It is the ratio of velocity change to time between any two points in an object's path. To accelerate at 9.8 m/s/s means to change the velocity by 9.8 m/s each second. If the velocity and time for a free-falling object being dropped from a position of rest were tabulated, then one would note the following pattern. Another way to represent this acceleration of 9.8 m/s/s is to add numbers to our dot diagram that we saw [|earlier in this lesson]. The velocity of the ball is seen to increase as depicted in the diagram at the right A position versus time graph for a free-falling object is shown below.
 * Representing Free Fall by Graphs **

A curved line on a position versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9.8 m/s/s), it would be expected that its position-time graph would be curved. A further look at the position-time graph reveals that the object starts with a small velocity (slow) and finishes with a large velocity (fast). Since the slope=velocity of the object, small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity. Negative slope of the line indicates a negative (i.e., downward) velocity

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

The line on the graph is a straight, diagonal line = accelerated motion (g = 9,8 m/s/s, downward). The object starts with a zero velocity (as read from the graph) and finishes with a large, negative velocity; that is, the object is moving in the negative direction and speeding up. An object that is moving in the negative direction and speeding up is said to have a negative acceleration. This analysis of the slope on the graph is consistent with the motion of a free-falling object - an object moving with a constant acceleration of 9.8 m/s/s in the downward direction.


 * How Fast? and How Far? **



Free-falling objects are in a state of [|acceleration]. The velocity of a free-falling object that has been dropped from a position of rest is dependent upon the time that it has fallen. The formula for determining the velocity of a falling object after a time of ** t ** seconds is  ** v **** f **** = g * t **

is also dependent upon the time of fall. This distance can be computed by use of a formula; the distance fallen after a time of ** t ** seconds is given by the formula. ** d = 0.5 * g * t **** 2 **

All objects free fall at the same rate of acceleration, regardless of their mass.
 * The Big Misconception **

=**Free Fall Lab**= 4 October 2011 Magna Leffler & Lindsey Marella **Objective:**  What is the acceleration of a falling body? What is acceleration due to gravity? 9.8 m/s 2 What would a v/t graph look like? A diagonal line going away from the origin. How would you get acceleration due to gravity from making a v/t? Acceleration due to gravity is represented on a v/t graph by the slope. (slope = acceleration due to gravity)
 * Hypothesis: **

100 g weight, ticker tape, spark timer, masking tape, measuring tape
 * Materials: **

**Procedure:** We took a 100g weight, attached ticker tape to it with a piece of masking tape, and then dropped it off of the balcony to the lobby to find the acceleration of a free falling object. After we collected our data we made graphs using excel. Below are our P/T graphs and our V/T graphs which we used to find the acceleration. DATA: When we made the line of best fit for the P/T graph we used a quadratic line. The equation for our line is y=407.06x 2 - 45.123x, its formula is y=Ax^2 + Bx. We went on to explain that A is really one half of the acceleration and that B is the initial velocity. So the equation of the line is d=v i t+1/2at 2. Our acceleration (2A/100) is 8.1412m/s and our Initial velocity is (B/100) 0.44123 m/s. When we made the line of best fit for the V/T graph we used a linear line. The equation for our line is y=804.55x - 41.792, its formula is y=Mx + B. We went on to explain that M is the slope of our line which is the acceleration and that B is the initial velocity. So the equation of the line is v f =v i t+at. Our acceleration (M) is 8.0455 m/s/s and our Initial velocity is (B/100) 0.41792 m/s.

Discussion Questions: Does the shape of your v-t graph agree with the expected graph? Why or why not? Does the shape of your x-t graph agree with the expected graph? Why or why not?
 * It doesn’t agree with what I thought it would be. I thought that the v-t graph would be a diagonal line going away from the origin with a negative slope. What we got though was a diagonal line with a positive slope going away from the origin. This is because we left the negatives out of our data the graph was flip-flopped.
 * I had thought that the x-t graph would be curved. I thought this because our experiment was accelerating downward which would produce a graph similar to the one was made.

How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.
 * Our results were spot on with the average of the class. Our percent difference was only 0.17%, which I think is a miniscule difference.

Did the object accelerate uniformly? How do you know?
 * The object accelerated fairly uniformly. I know this because the R 2 value is very close to one it was 0.996.

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 be higher due to the person releasing the object putting force on it. The reason why it would be lower than 9.8 m/s/s is because of friction, either from the tape going through the ticker or it tape having to run through someone’s hands so that it goes through strait.

Conclusion: We had to answer three questions for our hypothesis. Two of them I answered them correctly, the last one I answered most of it correctly, the second part, which I left out was the direction of the line. What I would have said is that it would be negative. The way we did the calculations the line was positive because we didn’t switched the positions of the start and finish. Instead of the drop being the final position it was the origin. We had a fairly high percent error (17.99%) it was most likely caused by the friction of the tape going through the ticker box. Also the dots which we measured were light on the tape and hard to see, we could have miss measured the dots. Our percent difference was very low (.17%), which showed that we were very close to the rest of the class. When redoing the lab I would use a broad-spectrum motion detector so there would be no friction. Also I would do multiple tests to make sure that we didn't get any outliers.