Unit1_LeeK

toc = = = =

= Lesson 1: =

__ Summary- Introduction to the Language of Kinematics: __ Mechanics is the study of the motion of objects such as projectiles. It is very diverse, leading to many subtopics. One such subtopic is kinematics which is the study of describing the motion of objects using words, diagrams, numbers, graphs, and equations. Thus, kinematics is more of a "description" of mechanics.
 * Classwork:**

= Lesson 2: =

**Classwork (Kinematics):** __Important Terms:__
 * = **Terms** ||= **Definition** ||= **Units** ||= **Symbol** ||= **Type of Quantity** ||
 * = //distance// ||= how far an object travels in total ||= m ||= d ||= scalar (quantitative, no direction) ||
 * = //displacement// ||= net/change in position relative to an origin ||= m ||= d ||= vector (quantitative, direction) ||
 * = //speed// ||= rate of change of position; how fast an object is moving ||= m/s ||= v ||= scalar (quantitative, no direction) ||
 * = //velocity// ||= speed that includes direction; based on displacement ||= m/s ||= v ||= vector (quantitative, direction) ||
 * = //acceleration// ||= how fast the speed is changing ||= m/s^2 ||= a ||= vector (quantitative, direction) ||

__Units:__
 * = **Prefix** || **Symbol** ||= **Factor** ||
 * tera || T || 1,000,000,000,000 ||
 * giga || G || 1,000,000,000 ||
 * mega || M || 1,000,000 ||
 * kilo || k || 1,000 ||
 * hecto || h || 100 ||
 * deca || da || 10 ||
 * — || — || 1 ||
 * deci || d || 0.1 ||
 * centi || c || 0.01 ||
 * milli || m || 0.001 ||
 * micro || μ || 0.000,001 ||
 * nano || n || 0.000,000,001 ||
 * pico || p || 0.000,000,000,001 ||
 * femto || f || 0.000,000,000,000,001 ||

__Four Types of Motion:__
 * 1) //at rest-// no motion, not moving <--average speed = d / t (constant)
 * 2) //constant velocity/speed-// no change in speed, covering same distance in same time <--average speed = d / t (constant); Vav = delta d/delta t
 * 3) //increasing velocity/speed-// starting slower and steadily getting faster
 * 4) //decreasing velocity/speed-// starting fast and steadily getting slower

__Types of Speed/Velocity:__
 * 1) //instantaneous-// speed/velocity at a very specific time or instant
 * 2) //average-// the mean or the overall speed/velocity

__Lab- Position of A Tennis Ball Traveling At A Constant Speed:__
 * //Objective:// What does the position graph of a tennis ball traveling at a constant speed look like?
 * //Hypothesis:// The graph will be linear.
 * //Rationale://A constant speed insures a constant gain in distance within equal units of time within anytime the ball is traveling. Thus, the slope (speed) of the graph is the same throughout, creating a linear relationship. //Data and Observations goes here... and you need to insert a pic of your data table and your graph. You also need your spreadsheet, but I will not generally open it unless you ask me for help, or if I'm checking your formulas.//
 * //Conclusion:// The position graph of an object traveling at a constant speed is linear. Analyzing the slope, the average speed within a specified interval, is crucial to this experiment. Since the speed of the tennis ball is constant throughout the experiment, the slopes of the graph should be consistent throughout the entire graph. In other words the instantaneous speed within any position equals the average speed. Thus, the average speed and the instantaneous speed can be denoted with the equation s = d/t. The speed/slope of the shallow incline was .1154 m/s and the speed/slope of the steep incline was .1435 m/s. Secondly, analyzing the variance (R^2), a value between -1 and 1, is also important. A negative variance describes a negative association and a positive variance describes a positive association. The closer the variance is to -1 or 1, the more linear-like the association among the scatter plots are. A negative variance denotes a negative velocity and a positive variance denotes a positive velocity. In the lab, the variance for the shallow incline was .97901 and the variance for the steep incline was .97791. Because the two variances are very close to 1, the association of the scatterplots can be described as a linear relationship. The acceleration of an object traveling at a constant speed is also constant because acceleration is the rate at which speed/velocity changes. Because the speed/velocity is constant, the acceleration is also constant (a = 0).
 * Data/Observations: [[file:Physics_Lab1.xlsx]]

__Summary- Scalars and Vectors:__ The words "going fast", "stopped", "moving slowly", etc describes motion. Yet, we can also describe motion through mathematical quantities divided into two subtopics: vectors and scalars. Scalars are quantities described alone by a numerical value (magnitude) while vectors are described both by a numerical value (magnitude) and by a direction.
 * Describing Motions With Words:**

__Summary- Distance and Displacement:__ Distance and displacement are two mathematical quantities that describe motion. Distance is scalar that describes the total distance an object has traveled. Displacement is a vector that describes the net distance an object has traveled. It is important to note that distance is only 0 if the object is at rest while displacement can be 0 even if the object moves. For example, if a car drives north for 5 miles and drives back south for 5 miles, it traveled a net distance of 0.

__Summary- Speed and Velocity:__ Speed and velocity are two mathematical quantities that describe motion. Speed is a scalar quantity that describes how fast an object is moving; the rate at which an object covers distance. On the other hand, velocity is a vector that refers to the rate at which an object changes its position. When referring to velocity, a direction always needs to be given. Average speed is given in the equation form distance traveled/time = d/t. Average velocity is given in the form change in distance/time = delta d/t. Furthermore, it is important to distinguish from two different types of speeds: instantaneous and average. instantaneous speed is the speed at any given instant in time and average speed is the average of all instantaneous speeds (distance/time ratio).



__Summary- Accelerations:__ Another important mathematical quantity that describes motion is acceleration, a vector quantity that describes the rate which an object changes its velocity. If an object is changing its velocity, it is considered to be accelerating. A constant acceleration is when the velocity changes at a constant rate. A constant acceleration should not be confused with a constant velocity! The average velocity is given in equation form a = delta v/t or (vt - vi) / t. Because acceleration is a vector quantity, it has a direction which depends on whether the object is speeding or slowing down and whether it is moving towards or away from an origin. If an object is speeding up, the acceleration is positive. If an object is slowing down, then its acceleration is in the opposite direction of its motion [GENERAL RULE OF THUMB].



__Summary- The Meaning of Shape for a p-t Graph:__ Here are the position graphs for different accelerations:
 * Describing Motions With p-t Graphs:**

Positive Velocity ||~ Positive Velocity Changing Velocity (acceleration) || a = constant; a = positive
 * ~ Constant Velocity
 * = [[image:http://www.physicsclassroom.com/Class/1DKin/U1L3a5.gif width="170" height="152"]] ||= [[image:http://www.physicsclassroom.com/Class/1DKin/U1L3a10.gif width="176" height="154"]] ||

Constant Velocity ||~ Fast, Rightward(+) Constant Velocity || a = constant; a = constant
 * ~ Slow, Rightward(+)
 * = [[image:http://www.physicsclassroom.com/Class/1DKin/U1L3a7.gif width="191" height="137"]] ||= [[image:http://www.physicsclassroom.com/Class/1DKin/U1L3a11.gif width="184" height="136"]] ||

Constant Velocity ||~ Fast, Leftward(-) Constant Velocity || a = constant; a = constant
 * ~ Slow, Leftward(-)
 * = [[image:http://www.physicsclassroom.com/Class/1DKin/U1L3a8.gif width="184" height="138"]] ||= [[image:http://www.physicsclassroom.com/Class/1DKin/U1L3a14.gif width="186" height="137"]] ||

Slow to Fast ||~ Leftward (-) Velocity Fast to Slow || a = negative (becoming more negative); a = positive (becoming more positive/resisting negativity)
 * ~ Negative (-) Velocity
 * = [[image:http://www.physicsclassroom.com/Class/1DKin/U1L3a16.gif width="177" height="150"]] ||= [[image:http://www.physicsclassroom.com/Class/1DKin/U1L3a17.GIF width="183" height="150"]] ||

__Summary- The Meaning of Slope for a p-t Graph:__ The slope of a p-t graph gives us a lot information about an object's velocity. For example, a small slope means a small velocity; a large slope means a large velocity; constant slope means a constant velocity a changing slope means a changing velocity; a negative slope means a negative velocity; a positive slope means a positive velocity. Overall, the slope of the graph = the velocity.

__Summary- Determining the Slope of a p-t Graph:__ To find the slope/velocity when given a position graph of an object, follow the following instructions below:
 * 1) Pick two points on the line and determine their coordinates.
 * 2) Determine the difference in y-coordinates of these two points (//rise//).
 * 3) Determine the difference in x-coordinates for these two points (//run//).
 * 4) Divide the difference in y-coordinates by the difference in x-coordinates (rise/run or slope).

//So much of this summary is NOT in your own words... parts are lifted almost verbatim from the PC. Because of this, it is very repetitive and there is information that you don't really need (you don't know how to find slope at your level?) The entire exercise of summarizing is to force you to decide what is important! If you don't consciously filter the material, you are not accomplishing the task I have set before you.//

[|Position/Velocity/Acceleration] //This is a good site and I have used it often. Did you play with it at all?//
 * Applet:**



= **Lesson 3:** =

__Motion Diagrams:__
 * Classwork:**
 * qualitative representations
 * represents relative size and directions of velocity and accelerations
 * can change size of velocity vector and direction of vectors
 * examples of vectors: --->; ->; <--; <-

__Four Types of Motion:__ **IMPORTANT: DECREASE IN SPEED AND INCREASE IN SPEED ARE BOTH TYPES OF ACCELERATIONS!**
 * 1) //Rest-// v = 0; a = 0
 * 2) //Constant:// a = 0; --> --> --> -->
 * 3) //Increasing:// -> --> ---> > ->; a -->; example: v = -2 to v = -8
 * 4) Decreasing: > ---> --> ->; <-- a; velocity and acceleration point in opposite directions

__MD Practice:__ -> > ---> --> ->; <-- a --> --> --> --> -->; a --> --> ---> > > ->; a --> <- <-- <--- < <-; a -->
 * //A ball thrown up into the air until highest point.//
 * //Car driving down a hill at constant speed.//
 * //Skier going downhill race.//
 * //A person landing on a trampoline, sinking until stopping.//

__Summary- Introduction to Diagrams:__ To understand 1-D kinetics, we need to use means of representation. Two means of representation of 1-D kinetics are ticker tape diagrams and vector diagrams.
 * Describing Motion With Diagrams:**

__Summary- Ticker Tape Diagrams:__ By using ticker tape diagrams, we can observe many common motions in physics. A long tape is attached to a moving object and threaded through a device that places a tick upon the tape at regular intervals of time As the object moves, it drags the tape through the "ticker," thus leaving a trail of dots. The trail of dots provides a history of the object's motion and therefore a represents the object's motion. The distance between the dots represent the change in object's position during the time interval. The more spread apart the dots, the faster the object was moving. The closer the dots, the slower the object was moving. A change in the distance between dots illustrates an acceleration as velocity is changing. If the dots are spread apart evenly, the velocity is constant. If the dots start out clustered then gradually spreads out, the velocity is increasing and vice versa. It is important to note that ticker tape diagrams are scalar representations.



__Summary- Vector Diagrams:__ Vector diagrams depict the magnitude and the direction of vectors. In a vector diagram, the magnitude of the vector quantity is represented by the length of the arrows. If the length of the arrows are the same, the speed of the object is constant. If the length of the vector increases in length, the object is accelerating. To depict direction, the arrows point towards a direction.



__Summary- The Meaning of Shape for a v-t Graph:__ v-t has two types of motion- constant velocity and changing velocity (acceleration). The two types of v-t graphs can be shown below: Zero Acceleration ||~ Positive Velocity Positive Acceleration || constant velocity has an a = 0; changing velocity is does not have a = 0
 * Describing Motion With v-t Graphs:**
 * ~ Positive Velocity
 * = [[image:http://www.physicsclassroom.com/Class/1DKin/U1L4a4.gif width="215" height="165"]] ||= [[image:http://www.physicsclassroom.com/Class/1DKin/U1L4a5.gif width="215" height="165"]] ||

__Summary- The Meaning of Slope for a v-t Graph:__

The slope the line of a v-t graph is the acceleration. For example, if the acceleration is zero, then the velocity-time graph is a horizontal line (the slope is zero). If the acceleration is positive, then the line is an upward sloping line (the slope is positive). If the acceleration is negative, then the velocity-time graph is a downward sloping line (the slope is negative). If the acceleration is great, then the line slopes up steeply (the slope is great).

__Summary- Relating the Shape to the Motion:__ The velocity and time graphs gives us a lot of information about motion. Here are some examples.



__Summary- Determining the Slope on a v-t Graph:__ The slope of the line on a velocity graph is equal to the acceleration. To find the slope of a v-t graph, follow the directions below:
 * 1) Pick two points on the line and determine their coordinates.
 * 2) Determine the difference in y-coordinates for these two points (//rise//).
 * 3) Determine the difference in x-coordinates for these two points (//run//).
 * 4) Divide the difference in y-coordinates by the difference in x-coordinates (rise/run or slope).

__Summary- Determining the Are on a v-t Graph:__ The method used to find the area under a line on a velocity-time graph depends upon whether the section bound by the line and the axes is a rectangle, a triangle or a trapezoid. Area formulas for each shape are given below:
 * //Area of a Rectangl//e = b x h
 * //Area of a Trapezoid// = [b x (h1 + h2)] / 2
 * //Area of a Triangle// = (b x h) / 2

= Lesson 4: =

__Important Equations:__ use subscripts and superscripts to make the equations easier to read... go to 4th button from left (in Edit Bar) and select "Vertical Spacing" to select one of these options. Also Option-j is a delta symbol on a Mac.
 * Classwork:**
 * vav= ∆ d/∆t
 * aav= ∆ v/ ∆ t
 * vf2= vo2 + 2a∆d
 * ∆d = ½ (vf + vo)t
 * ∆d = vot + ½ at2
 * vf= vo + at

__Lab- Free-fall Lab:__
 * //Objective:// What is acceleration due to ?
 * //Hypothesis:// The acceleration of gravity is 9.81 m/s^2.
 * //Procedure//: A spark-timer was hung on the ceiling and a spark-tape was cut to length equal to the height of the room. A 200g cylinder was then attached to one end of the spark-tape. After, the tape ran through a spark-timer as the 200g cylinder was dropped to a free-fall (air resistance negligible) from the ceiling, resulting in dotted marks on the spark-tape, which represents the cylinders position at 1/60 of a second. At the end, the distances between the first dot and all the other were measured and recorded.
 * Data/Observations: I suppose this is misplaced? Where is your graph? If you upload a picture but it has the same name as a previous file, then it will replace the earlier one as well. Be sure to name each screenshot an original name.


 * //Analysis:// The association of the plots results in a polynomial trend-line. The equation of this trend-line is y = 4.5842x^2 + .495x. This equation takes the form of the equation d = .5at^2 + v0t. Thus, a = 4.5842 x 2 = 9.1684 m/s^2 and the initial velocity (v0) is .495 m/s. Given an R^2 vale of .99993, the trend-line accurately depicts the association of the data plots. When comparing the acceleration 9.1684 m/s^2 to the class average, the % difference of 0.9736%. Therefore, the data is accurate within the limits of our experimentation. However, the actual acceleration of gravity is 9.81 m/s^2. When comparing the acceleration 9.1684 m/s^2 to 9.81 m/s^2, the % error is 6.5403%. What accounts for such a high difference between the theoretical acceleration and the experimental acceleration? The answer is friction. Although there are air resistance and some human error involved, these are negligible. Friction is created when the spark-tape runs through the spark-timer, slowing down the free-fall of the 200g cylinder. Thus, the experimental acceleration is much lower than the theoretical acceleration. **(//Calculations://** ** P ercent Error: |4.905 - 4.5842)| / 4.905 = .0654026504 = 6.5403%;** **Percent Difference: |4.5842 - 4.54| / 4.54 =0.009735682819 = 0.9736%)**

__Summary- Introduction of Free Fall:__ Any object acted upon only by the force of gravity is said to be in free fall. Free fall objects to not encounter air resistance and accelerate towards gravity at a rate of 9.81m/s^2, called the acceleration of gravity.
 * Free Fall and Acceleration of Gravity:**

__Summary- The Acceleration of Gravity:__ The acceleration of gravity is important in physics. It is so important that it is denoted with the symbol "g". Although the acceleration of gravity is given as 9.81 m/s^2, it varies slightly depending on altitude.

__Summary- Representing Free Fall By Graphs:__ The line of p-t graph for a free-falling object is curved. It starts off with a low velocity (shallow slope) then gradually becomes steeper as velocity increases (steeper slope). Furthermore, the slopes of the graph are negative as the the line indicates a negative velocity (object falling). On the other hand, the line of a v-t graph for an object in free fall is a straight, diagonal line. Because a free-falling object is undergoing an acceleration downward at a rate of -9.81 m/s^2, the line of the acceleration graph is expected to have a negative slope since acceleration is equal to the slope of the velocity graph line. Looking at the graph, the object starts with a initial velocity of 0 and finishes with a large negative velocity.



__Summary- How Fast? and How Far?:__ To calculate the velocity of an object in free fall after t seconds, use the following formula below: **vf = g * t** To calculate the distance of an object in freefall after t seconds, use the following formula below: **d = 0.5 * g * t^2**

__Summary- The Big Misconception:__ Regardless of mass, all objects accelerate in free-fall at the same rate.

__Summary- The Kinematics Equations:__ Kinematic equations are equations that describe the motion of objects. Kinematic equations are a set of four equations that can be utilized to predict unknown information about an object's motion if other information is known. It is important to note that all four kinetic equations have an initial velocity. (d = displacement, vi = initial velocity, vf = final velocity, a = acceleration, t = time)
 * Describing Motion With Equations:**



__Summary- Kinematic Equations and Problem-Solving:__ To solve kinematic problems, use the following approach:
 * 1) Construct an informative diagram of the physical situation.
 * 2) Identify and list the given information in variable form.
 * 3) Identify and list the unknown information in variable form.
 * 4) Identify and list the equation that will be used to determine unknown information from known information.
 * 5) Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown information.
 * 6) Check your answer to insure that it is reasonable and mathematically correct.

Example: Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s2, then determine the displacement of the car during the skidding process. (Note that the direction of the velocity and the acceleration vectors are denoted by a + and a - sign.) The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identificat ion and listing of known information in variable form. Note that the vf value can be inferred to be 0 m/s since Ima's car comes to a stop. The initial velocity ( vi ) of the car is +30.0 m/s since this is the velocity at the beginning of the motion (the skidding motion). And the acceleration ( a ) of the car is given as - 8.00 m/s2. (Always pay careful attention to the + and - signs for the given quantities.) The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the un known quantity. The results of the first three steps are shown in the table below. vf = 0 m/s
 * ~ Diagram: ||~ Given: ||~ Find: ||
 * = [[image:http://www.physicsclassroom.com/Class/1DKin/U1L6b1.gif width="137" height="71"]] ||= vi = +30.0 m/s

a = - 8.00 m/s2 ||= d = ?? || The next ste p of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vf, vi , a , and d. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top right contains all four variables. Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below.

(0 m/s)2 = (30.0 m/s)2 + 2*(-8.00 m/s2)*d 0 m2/s2 = 900 m2/s2 + (-16.0 m/s2)*d (16.0 m/s2)*d = 900 m2/s2 - 0 m2/s2 (16.0 m/s2)*d = 900 m2/s2 d = (900 m2/s2)/ (16.0 m/s2) d = (900 m2/s2)/ (16.0 m/s2) d = 56.3 m

__Summary- Kinematic Equations and Free Fall:__ For free fall kinetics problems, we use the four kinetics equations mentioned before. However, there are some important characteristics of free fall motion when solving free problems. These characteristics are listed below.
 * 1) An object in free fall experiences an acceleration of -9.8m/s^2.
 * 2) If an object is merely dropped from an elevated height, then the initial velocity of the object is 0 m/s.
 * 3) If an object is projected upwards in a perfectly vertical direction, then it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s and its acceleration is still -9.81 m/s^2.
 * 4) If an object is projected upwards in a perfectly vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity that it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have a downward velocity of -30 m/s when it returns to the same height.

__Summary- Kinematic Equations and Graphs:__ If kinematic graphs have lines with more than two distinct shapes, isolate the two shapes and evaluate them separately. After evaluating the different intervals separately, piece the data together.

**Example:** Consider an object that moves with a constant velocity of +5 m/s for a time period of 5 seconds and then accelerates to a final velocity of +15 m/s over the next 5 seconds. Such a verbal description of motion can be represented by a velocity-time graph. The graph is shown below.

The horizontal section of the graph depicts a constant velocity motion, consistent with the verbal description. The positively sloped (i.e., upward sloped) section of the graph depicts a positive acceleration, consistent with the verbal description of an object moving in the positive direction and speeding up from 5 m/s to 15 m/s. The slope of the line can be computed using the rise over run ratio. Between 5 and 10 seconds, the line rises from 5 m/s to 15 m/s and runs from 5 s to 10 s. This is a total rise of +10 m/s and a total run of 5 s. Thus, the slope (rise/run ratio) is (10 m/s)/(5 s) = 2 m/s2. Using the velocity-time graph, the acceleration of the object is determined to be 2 m/s2 during the last five seconds of the object's motion. The displacement of the object can also be determined using the velocity-time graph. The area between the line on the graph and the time-axis is representative of the displacement; this area assumes the shape of a trapezoid. As discussed in Lesson 4, the area of a trapezoid can be equated to the area of a triangle lying on top of the area of a rectangle. This is illustrated in the diagram below.

The total area is then the area of the rectangle plus the area of the triangle. The calculation of these areas is shown below. Area = (10 s) * (5 m/s)
 * ~ Rectangle ||~ Triangle ||
 * = Area = base * height

Area = 50 m ||= Area = 0.5 * base * height Area = 0.5 * (5 s) * (10 m/s)

Area = 25 m || The total area (rectangle plus triangle) is equal to 75 m. Thus the displacement of the object is 75 meters during the 10 seconds of motion.

The above discussion illustrates how a graphical representation of an object's motion can be used to extract numerical information about the object's acceleration and displacement. Once constructed, the velocity-time graph can be used to determine the velocity of the object at any given instant during the 10 seconds of motion. For example, the velocity of the object at 7 seconds can be determined by reading the y-coordinate value at the x-coordinate of 7 s. Thus, velocity-time graphs can be used to reveal (or determine) numerical values and relationships between the quantities displacement (d), velocity (v), acceleration (a) and time (t) for any given motion.

Now let's consider the same verbal description and the corresponding analysis using kinematic equations. The verbal description of the motion was: > An object that moves with a constant velocity of +5 m/s for a time period of 5 seconds and then accelerates to a final velocity of +15 m/s over the next 5 seconds

Kinematic equations can be applied to any motion for which the acceleration is constant. Since this motion has two separate acceleration stages, any kinematic analysis requires that the motion parameters for the first 5 seconds not be mixed with the motion parameters for the last 5 seconds. The table below lists the given motion parameters. vf = 5 m/s
 * ~ t = 0 s - 5 s ||~ t = 5 s - 10 s ||
 * = vi = 5 m/s

t = 5 s

a = 0 m/s2 ||= vi = 5 m/s vf = 15 m/s

t = 5 s ||

Note that the acceleration during the first 5 seconds is listed as 0 m/s2 despite the fact that it is not explicitly stated. The phrase //constant velocity// indicates a motion with a 0 acceleration. The acceleration of the object during the last 5 seconds can be calculated using the following kinematic equation. vf = vi + a*t The substitution and algebra are shown here. 15 m/s = 5 m/s + a*(5 s) 15 m/s - 5 m/s = a*(5 s)  10 m/s = a*(5 s)  (10 m/s)/(5 s) = a  a = 2 m/s2 This value for the acceleration of the object during the time from 5 s to 10 s is consistent with the value determined from the slope of the line on the velocity-time graph.  The displacement of the object during the entire 10 seconds can also be calculated using kinematic equations. Since these 10 seconds include two distinctly different acceleration intervals, the calculations for each interval must be done separately. This is shown below. d = (5 m/s)*(5 s) +0.5*(0 m/s2)*(5 s)2 d = 25 m + 0 m d = 25 m ||= d = ((vi + vf)/2)*t d = ((5 m/s + 15 m/s)/2)*(5 s) d = (10 m/s)*(5 s) d = 50 m ||
 * ~ t = 0 s - 5 s ||~ t = 5 s - 10 s ||
 * = d = vi*t + 0.5*a*t2

The total displacement during the first 10 seconds of motion is 75 meters, consistent with the value determined from the area under the line on the velocity-time graph. Way too much detail! I feel like I pretty much just read the PC sections in entirety!

cool applet, thanks!
 * Applet:**
 * [|Free-Fall] **

= = = **Lesson 5:** =

__Newton's 1st Law of Motion: Inertia:__
 * Classwork:**
 * An object at rest will stay at rest unless an outside force intervenes.
 * An object in motion will stay in motion at constant speed in a straight line unless an outside force intervenes.
 * At rest => static equilibrium; v = 0; a = 0; forces balanced.
 * Galileo's Incline Experiment- A ball dropped at any incline will seek to reach the same height it started from.
 * Dynamic Equilibrium: v =/= 0, a = 0

__Mechanical Forces:__
 * A "push" or a "pull".
 * Mechanical forces are external to the system (object).
 * There must be contact (1 exception).
 * Mechanical forces cannot be transferred or carried.
 * There are four types of mechanical forces (vectors).
 * Units: Newtons (N)


 * = **Force** ||= **Definition** ||= **Direction** ||= **Symbol** ||= **Equation** ||
 * = //weight// ||= pull of earth on mass ||= points straight down ||= w, Fg ||= w = m * g ||
 * = //friction// ||= force exhibited when 2 surfaces rub together ||= parallel surfaces; opposite direction of motion ||= f, Ff ||= f = u * N ||
 * = //normal// ||= "support" force whenever 2 surfaces touch ||= points perpendicular to the surface and through the system ||= N, Fn ||=  ||
 * = //tension// ||= rope or chain (pull) ||= runs along rope/chain away from object ||= T, Ft ||=  ||

__Summary- Newton's First Law:__ Newton's first law is also referred to as "the law of inertia". It states that an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force. When an object is at rest, a = 0 m/s^2 and v = 0 m/s unless disturbed by some external force. When an object is in motion, a = 0 m/s^2 and the velocity is constant as the object travels in a straight direction unless disturbed by some external force. If an object is moving, a force is not needed to keep the object from moving. Also, it is another force that stops an object from moving.
 * Newton's First Law of Motion:**

__Summary- Inertia and Mass:__ Inertia is the tendency to resist change when an object is in motion or the resistance an object has to change in its state of motion. An object's mass is solely dependent on the inertia of the object. Galileo experimented with inertia, believing that objects eventually stopped from motion due to friction and other forces.



__Summary- State of Motion:__ Within motion, inertia can be defined as the tendency of an object to resist motion or changes in velocity and acceleration.

__Summary- Balanced and Unbalanced Forces:__ According to Newton's 1st Law, an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an UNBALANCED force. If two forces are in opposite directions and have the same magnitude, they balance each other and are said to be in equilibrium. Balanced forces don't cause accelerations. Unbalanced forces cause accelerations and is the result of not having two forces in opposite directions with the same magnitude. Below is an example of an unbalanced force.



In this example, there is no force to balance friction. Thus, the book accelerates to the left and slowly decreases its velocity and eventually stops. This exhibits Newton's First Law as it takes an outside force (friction) to stop an object in motion.

= Lesson 6: =

__Newton's 2nd Law of Motion:__
 * Classwork:**
 * Acceleration is directly proportional to force.
 * Mass is indirectly proportional to acceleration.
 * Acceleration = force / mass

__Free Body Diagrams:__
 * Representation of all forces acting on a system on the system.
 * All forces are shown with straight arrows and are labeled with symbols.

__Lab- Newton's Second Law:__ watch sig figs in calculations... remind me to show you how to adjust this in Excel.
 * //Objective:// What is the relationship between the net force and the acceleration of an object? What is the relationship between the mass of an object and its acceleration?
 * //Hypothesis:// Net force and acceleration are directly proportional as force induces motion to an object in rest. Mass and acceleration are indirectly proportional as objects with more mass require more force to induce movement.
 * //Data://


 * //Analysis://Newton's Second Law, which states that acceleration = force / mass is illustrated by this lab. As seen in the equation, force and acceleration are directly proportional, producing a linear graph. The trend-line from the force vs. acceleration graph had an equation of y = .5459x +.0313. This equation is in the form force = mass * acceleration, a variation of Newton's second law. Thus, the equation y = .5459x + .0313 can be translated as F = .5459a + .0313. The coefficient of "a" is the net mass. The y intercept, .0313, is error mostly due to friction from the movement of the cart and the pulley as an object with no force should have an acceleration of 0. With a .5459 kg experimental net mass and a theoretical net mass of .53 kg, the % error is 3%. Furthermore, the R^2 value (how well the data fit the trend line) is .99617, supporting the fact that a linear relationship between force and acceleration is indeed valid. On the other hand, the trend-line from the mass vs. acceleration graph depict an inversely proportional relationship as the trend line is a negative power function. The equation of the trend-line is y = 0.2285x ^ -1.277, which can be translated into a = force / mass. In this case, .2285 N is the force. With a theoretical force of (9.81 * .03) = .2943 N the % error is 22.36%. In addition, the variance which has a value of .97159 supports the fit of the trend-line. The high % error results mostly from friction due to the movement of the cart and the pulley as great masses exponentially increase the effects of friction. ** (//Calculations:// ** **Percent Error of Mass: |.53 - .5459)| / .53 = .03 = 3%;** **Percent Error of Force: |.2943 - .2285| / .2943 = 0.2235813795 = 22.36%)**

__Summary- The Meaning of Force:__ A force is a push or pull upon an object resulting from the object's interaction with another object. A force is present upon two objects if there is an interaction between them. Forces only result from interactions. Forces can be divided into two groups: contact forces and forces resulting from action-at-a-distance. Contact forces are forces created when two objects physically contact each other (friction, air resistance, etc). Action-at-a-distance forces result even when two interacting objects are not in physical contact with each other (gravity, electrical forces, etc). To measure force, we use Newtons (N). One Newton is the amount of force required to give a 1 kg mass an acceleration of 1 m/s^2. Although forces can be measured in Newtons, they are a vector, requiring a direction.
 * Force and Its Representation:**

__Summary- Types of Forces:__
 * < **Contact Forces** ||< **Action-at-a-Distance Forces** ||
 * < Frictional Force ||< Gravitational Force ||
 * < Tension Force ||< Electrical Force ||
 * < Normal Force ||< Magnetic Force ||
 * < Air Resistance Force ||<  ||
 * < Applied Force ||<  ||
 * < Spring Force ||


 * = **Force** ||= **Definition** ||= **Direction** ||= **Symbol** ||= **Equation** ||
 * = //weight// ||= pull of earth on mass ||= points straight down ||= w, Fg ||= w = m * g ||
 * = //friction// ||= force exhibited when 2 surfaces rub together ||= parallel surfaces; opposite direction of motion ||= f, Ff ||= f = u * N ||
 * = //normal// ||= "support" force whenever 2 surfaces touch ||= points perpendicular to the surface and through the system ||= N, Fn ||=  ||
 * = //tension// ||= rope or chain (pull) ||= runs along rope/chain away from object ||= T, Ft ||=  ||


 * Mass is the measurement of how much matter there is and weight is the measurement of the force of gravity acted upon a mass.
 * Sliding friction results when an object slides across a surface.
 * Static friction results when the surfaces of two objects are at rest relative to one another and a force exists on one of the objects to set it into motion relative to the other object.
 * Ffrict-static ≤ μfrict-static• Fnorm
 * The symbol μfrict-static represents the coefficient of static friction between the two surfaces. Like the coefficient of sliding friction, this coefficient is dependent upon the types of surfaces that are attempting to move across each other. In general, values of static friction coefficients are greater than the values of sliding friction coefficients for the same two surfaces. Thus, it typically takes more force to //budge// an object into motion than it does to maintain the motion once it has been started.

__Summary- Drawing Free-Body Diagrams:__ Free-body diagrams display the relative magnitude and direction of all forces acting upon an object. A free-body diagram is special example of a vector diagram. The arrows in a free-body diagram represent the forces that are acting upon an object. They point in the direction of the force and their lengths represent the magnitude of the force. It is important to label these arrows with the type of force.



__Summary- Force and Its Representation: __ It is important to figure out the net force of free-body diagrams (sigma F = net force).



very cute! :)
 * Cartoon:**
 * [|Newton's Second Law]**



= Lesson 7: =

**Classwork:** __Newton's Second Law:__
 * force = mass * acceleration
 * force = net force (what's causing the mass to accelerate)
 * mass = system mass (what is being accelerated)
 * All drops have accelerations between 0 m/s^2 to 9.8 m/s^2.
 * Drop and pull situation where mass 1 is on the table and mass 2 is hanging: a = (m2 * g) / (m1 + m2)

__Friction:__
 * Coefficient of Friction- Number that represents how two surfaces interact when they try to slide across each other.
 * A ratio; no units (almost always between 0 and 1)
 * symbol = µ
 * µ = force of friction / normal of surface
 * friction force and normal force increase proportionally

__Summary- Newton's Second Law:__ According to Newton's Second Law, the acceleration of an object depends on the net force and the mass of the object. Acceleration and the net force are directly proportional while the acceleration and the mass is inversely proportional.
 * Newton's Second Law of Motion:**
 * **a = Fnet / m**
 * **Fnet = m * a**

__Summary- The Big Misconception:__ Sustaining motion does not require a continuous force.

__Summary- Finding Acceleration:__ To find the acceleration of an object, the mass and the net force needs to be known. If the mass and the net force are known, then the acceleration is determined by the use of the equation below.



**Example:** **Fnorm = 80 N; m = 8.16 kg; Fnet = 40 N, right; a = 4.9 m/s/s, right** ( Fnorm = 80 N; m = 8 kg; Fnet = 40 N, right; a = 5 m/s/s, right )

Since there is no vertical acceleration, normal force = gravity force. The mass can be found using the equation Fgrav = m • g. The Fnet is the vector sum of all the forces: 80 N, up plus 80 N, down equals 0 N. And 50 N, right plus 10 N, left = 40 N, right. Finally, a = Fnet / m = (40 N) / (8.16 kg) = 4.9 m/s/s.

__Summary- Finding Individual Forces:__ By using the equations for friction (Ffrict = μ•Fnorm), net force (Fnet = ma), and gravity (Fgrav = m•g), the values of all individual forces can be determined.

**Example:**



**Fnet = 0 N; Fgrav = 58.8 N; Fnorm = 58.8 N; Fapp = 15 N** When the velocity is constant, a = 0 m/s/s and Fnet = 0 N Since the mass is known, Fgrav can be found: Fgrav = m • g = 6 kg • 9.8 m/s/s = 58.8 N Since there is no vertical acceleration, the normal force equals the gravity force. Since there is no horizontal acceleration, Ffrict = Fapp = 15 N

[|Newton's Second Law]
 * Cartoon:**

= **Lesson 8:** =

__Lab- The Coefficient of Friction__ //Data:// //Analysis:// The coefficient of friction is the ratio between the friction force and the normal force. The coefficient of friction differs depending on the surface used. The surface that induces the least friction is the table top with a µ of .205 and the surface that induces the greatest friction is the concrete with a µ of .487. Even with different masses, when the same surface is used, the µ is still the same. This is because normal force = weight (proportional to mass) and as weight increases, friction increases proportionally as well.
 * Classwork:**
 * = Surface ||= Trial 1: µ ||= Trial 2: µ ||= Trial 3: µ ||= Average µ ||
 * = Carpet ||= 0.463 ||= 0.417 ||= 0.39 ||= 0.423333333 ||
 * = Gym Floor ||= 0.28 ||= 0.279 ||= 0.3 ||= 0.286333333 ||
 * = Concrete ||= 0.49 ||= 0.49 ||= 0.48 ||= 0.486666667 ||
 * = Floortile ||= 0.283 ||= 0.275 ||= 0.275 ||= 0.277666667 ||
 * = Table Top ||= 0.199 ||= 0.213 ||= 0.202 ||= 0.204666667 ||
 * = Paving Stone ||= 0.36 ||= 0.36 ||= 0.36 ||= 0.36 ||

**//Calculations://** **Net Force (y) = ma(y) = normal - weight** **a = 0 (not moving vertically)** **0 = normal - weight** **normal force = weight = 9.81 * (mass of bars + bucket)**

**Net Force (x) = ma(x) = tension - friction** **a = 0 (constant speed)** **0 = tension - friction** **friction = tension** good!


 * Applet:**

[|µ Applet]



=** Lesson 9/10: **=

__Air Resistence:__ Friction caused by air particles If enough air particles build up on a surface, the object reaches terminal velocity/dynamic equilibrium in which velocity is constant Terminal Velocity/Dynamic Equilibrium: Fa.r = W
 * Classwork:**

__Newton's Third Law:__ Every action has an equal opposite reaction Analogy: One cannot touch something without it touch one back All forces come in pairs; equal in size; pointing in opposite directions; acting on 2 separate systems

__Question:__ What weight is necessary to drag a 500 g physics cart 0.8 meters in 2.1 seconds?
 * Lab- Drop and Pull Challenge Problem:**

__Calculations:__

D = 0.8 m Vo = 0 m/s t = 2.1 s m2 (cart) = 0.5 kg m1 (weight needed) = ?

D = VoF +0.5At^2 0.8 = 0(F) + (0.5)A(2.1^2) 0.8 = 0 + 2.205A A = .3628 m/s^2

A = F / Total Mass F = m1 * g Total Mass = m1 + m2 A = (m1 * g) / (m1 + m2) .3628 = (m1 * 9.81) / (m1 + 0.5) .3628 * (m1 + 0.5) = m1 * 9.81 .3628m1 + .1814 = 9.81m1 m1 = .0192 kg = 19.2 g

__Results:__

Analysis: your results % error List cause of error

2.12 seconds 2.16 seconds 2.10 seconds


 * Air Resistance:**
 * When an object falls, it experiences air resistance (friction due to air particles)
 * Air resistance is determined by the speed and the surface area of the object
 * If enough air particles build up, object will no longer increase its velocity, reaching terminal velocity/dynamic equilibrium
 * In terminal velocity/equilibrium, force of air resistance equals the weight of the object


 * Newton's Third Law:**
 * For each individual action, there is an equal and opposite reaction
 * All forces come in pairs
 * Each of the forces in a pair act in different systems


 * Vectors, Fundamentals, and Operations:**
 * Forces are vectors and therefore need directional values
 * ** Directions are represented by angle measurements (out of 360 degrees) in free body diagrams **
 * When adding the net magnitude of vectors in 1D, you simply add the magnitudes because the vectors are pointing in the same direction or opposite directions
 * To find the net magnitude of vectors that don't have opposite or similar direction, triangles need to be made
 * The pythagorean theorem and trig. rules (SOH-CAH-TOA) are applied to find the magnitude
 * Resultant is the vector sum
 * In 2-D vector diagrams, each vector is called the vector component
 * 2 vector components make a triangle in which the hypotenuse is the resultant
 * Viy = sum of Vi sin theta
 * Vix = sum of vi cose theta



Projectiles: x and y axis are indep. ax = 0 net force = 0 77.8

41.5 41.5 41.4 41.4 40.8 42.3 42.2 42.8

avg = 41.7375

.0135 .0135 .0145 .013

1.6cm / t

data calc cup 10cm photogte analysis: comparing photogate theoretical independence and x and y --- hypoth what is velo? Answer obj iwth evidence experimental error and ways to eliminate if redo lab

gsintheta = a theoretical