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9/3 Do now • Give at least three words to describe motion. • All do now count as extra credit for the next quiz. • If you are absent or late for the do now, you will not get the credit. • You have 5 min to finish the Do now Homework – Due Tue. 9/7 • Posted on school website – go to school website, click on teacher, click on LaBarbera, click on regents physics homework. One Dimensional Kinematics - Chapter Outline 1. Describing Motion with Words 2. Describing Motion with Diagrams 3. Describing Motion with Position vs. Time Graphs 4. Describing Motion with Velocity vs. Time Graphs 5. Free Fall and the Acceleration of Gravity 6. Describing Motion with Equations Lesson 1 : Describing Motion with Words objectives 1. Introduction to the Language of Kinematics 2. Differentiate Scalars and Vectors 3. Describe motion in terms of • frame of reference, • Displacement and distance, • speed and velocity • Acceleration Introduction to the Language of Kinematics • A typical physics course concerns itself with a variety of broad topics. One such topic is mechanics - the study of the motion of objects. • Kinematics is the science of describing the motion of objects using words, diagrams, numbers, graphs, and equations. Kinematics is a branch of mechanics. Motion takes place over time and depends upon the frame of reference • http://www.teachersdomain.org/resource/ls ps07.sci.phys.fund.frameref/ • http://www.phy.ntnu.edu.tw/ntnujava/index .php?topic=140 • motion is relative • A frame of reference is a background against which you can measure changes in position. • Unless stated otherwise, we choose the Earth as our frame of reference. Scalars and Vectors • The motion of objects can be described by words. Words and phrases such as going fast, stopped, slowing down, speeding up, and turning provide a sufficient vocabulary for describing the motion of objects. In physics, we use these words and many more. We will be expanding upon this vocabulary list with words such as distance, displacement, speed, velocity, and acceleration. • Scalars are quantities which are fully described by a magnitude (or numerical value) alone. • Vectors are quantities which are fully described by both a magnitude and a direction. example Quantity Category a. 5 m Scalar b. 30 m/sec, East vector c. 5 mi., North vector d. 20 degrees Celsius Scalar e. 256 bytes Scalar f. 4000 Calories Scalar Distance and Displacement • Distance and displacement are two quantities which may seem to mean the same thing yet have distinctly different definitions and meanings. • Distance is a scalar quantity which refers to "how much ground an object has covered" during its motion. • Displacement is a vector quantity which refers to "how far out of place an object is"; it is the object's overall change in position. Distance and displacement A B example • A car is driven from Buffalo to Albany and on to New York City, as shown in the diagram. • Compared to the magnitude of the car's total displacement, the distance driven is 1. shorter 2. longer 3. the same example • A physics teacher walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North. Distance: 12 m Displacement: 0 Example • A student walk east 3.0 meters, then walk west 4.0 meters, 1. What is the student’s displacement? N 2. What is the student’s distance traveled? 1. Use scales to draw the diagram: 1 cm = 1 m 4mW 3mE resultant 1. The displacement is 1 meter west - vector 2. Distance traveled is 7.0 meters - scalar example • The diagram below shows the position of a cross-country skier at various times. At each of the indicated times, the skier turns around and reverses the direction of travel. In other words, the skier moves from A to B to C to D. • Use the diagram to determine the resulting displacement and the distance traveled by the skier during these three minutes. The skier covers a distance of (180 m+140 m+100 m) = 420 m and has a displacement of 140 m, rightward. example • A football coach paces back and forth along the sidelines. The diagram below shows several of coach's positions at various times. At each marked position, the coach makes a "U-turn" and moves in the opposite direction. In other words, the coach moves from position A to B to C to D. • What is the coach's resulting displacement and distance of travel? Click the button to see the answer. The coach covers a distance of (35 yds+20 yds+40 yds) = 95 yds and has a displacement of 55 yards, left. Check Your Understanding 1. What is the displacement of the cross-country team if they begin at the school, run 10 miles and finish back at the school? The displacement of the runners is 0 miles. 2. What is the distance and the displacement of the race car drivers in the Indy 500? The displacement of the cars is 0 miles. The successful cars have covered a distance of 500 miles 1st period • For your safety, please take off all the chairs from all the desks. Thank you! Do now • What is a scalar? • What is a vector? • You may refer your notes for answers. You have 5 min. objectives • Differentiate between speed and velocity • Calculate speed and velocity Speed and Velocity • Speed is a scalar quantity which refers to "how fast an object is moving." An object with no movement at all has a zero speed. • Velocity is a vector quantity which refers to "the rate at which an object changes its position.“ • Velocity is a vector quantity. • For example: It would not be enough to say that an object has a velocity of 55 mi/hr. One must include direction information in order to fully describe the velocity of the object. You must describe an object's velocity as being 55 mi/hr, east. • This is one of the essential differences between speed and velocity. Speed is a scalar quantity and does not keep track of direction; velocity is a vector quantity and is direction aware. Direction of the velocity • The direction of the velocity vector is simply the same as the direction which an object is moving. It would not matter whether the object is speeding up or slowing down. Calculating Average Speed and Average Velocity example • While on vacation, Lisa Carr traveled a total distance of 440 miles. Her trip took 8 hours. What was her average speed? example • During a race on level ground, André runs with an average velocity of 6.02 m/s to the east. What distance does André cover in 137 s? v=d/t 6.02 m/s = d / 137 s d = 825 m Note: the calculator answer is 824.74 m, but both the values for velocity and time have three significant figures, so the displacement must be reported as 825 m Average Speed versus Instantaneous Speed • Instantaneous Speed - the speed at any given instant in time. Average Speed - the average of all instantaneous speeds; found simply by a distance/time ratio. Constant speed vs. changing speed example • Now let's consider the motion of that physics teacher again. The physics teacher walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North. The entire motion lasted for 24 seconds. Determine the average speed and the average velocity. her average speed is 0.50 m/s. However, her average velocity is 0 m/s. example • The diagram below shows the position of a cross-country skier at various times. At each of the indicated times, the skier turns around and reverses the direction of travel. In other words, the skier moves from A to B to C to D. Determine the average speed and the average velocity of the skier during these three minutes. The skier’s average speed is (420 m) / (3 min) = 140 m/min The skier’s average velocity is (140 m, right) / (3 min) = 46.7 m/min, right example • A football coach paces back and forth along the sidelines. The diagram below shows several of coach's positions at various times. At each marked position, the coach makes a "U-turn" and moves in the opposite direction. In other words, the coach moves from position A to B to C to D. What is the coach's average speed and average velocity? The coach’s average speed is (95 yd) / (10 min) = 9.5 yd/min The coach’s average velocity is (55 yd, left) / (10 min) = 5.5 yd/min, left In conclusion • speed and velocity are kinematics quantities which have distinctly different definitions. Speed, being a scalar quantity, is the rate at which an object covers distance. The average speed is the distance (a scalar quantity) per time ratio. Speed is ignorant of direction. On the other hand, velocity is a vector quantity; it is direction- aware. Velocity is the rate at which the position changes. The average velocity is the displacement or position change (a vector quantity) per time ratio. homework • Online School website Practice •Text book page 44 – practice 2A #4-6 4. Simpson drives his car with an average velocity of 48.0 km/h to the east. How long will it take him to drive 144 km on a straight highway? Given: v = 48.0 km/h east d = 144 km east unknown: t = ? v=d/t 48.0 km/h = 144 km / t t = 144 km/ (48.0 km/h) = 3 hr = 3.00 hr Note: since the significant figures of given values is 3, so the displacement must be reported as 3 significant figures which is 3.00 hr 9/8 do now • Text book – page 45 – conceptual challenge • Homework is due You may come post session to finish the homework. • Lab safety letter is due 1. Displacement = 0; average velocity = 0; average speed = d / t = (2 x 1.75 m + 2 x 2.25 m) / 23 s = 0.35 m/s 2. No, the velocities are not the same because they are traveling in different directions. Objective - Section review • Text book page 47 – 1, 2, 4, 5, 6 • castle learning • Finish the lab • Homework is due 6. A bus travels 280 km south along a straight path with an average velocity of 88 km/h to the south. The bus stops for 24 min, then it travels 210 km south with an average velocity of 75 km/h to the south. a. How long does the total trip last? b. what is the average velocity for the total trip? Given: v1 = 88 km/h; d1 = 280 km south unknown: ttotal = ? t2 = 24 min = (24/60) hr = 0.4 hr v3 = 75 km/h; d3 = 210 km south v=d/t 88 km/h = 280 km / t1 t1 = 280 km/ (88 km/h) = 3.18 hr t3 = 210 km/ (75 km/h) = 2.80 hr ttotal = t1 + t2 + t3 = 3.18 hr + 0.4 hr + 2.80 hr = 6.38 hr = 6.4 hr Note: since the significant figures of given values is 2, so the displacement must be reported as 2 significant figures which is 6.4 hr 6. A bus travels 280 km south along a straight path with an average velocity of 88 km/h to the south. The bus stops for 24 min, then it travels 210 km south with an average velocity of 75 km/h to the south. a. How long does the total trip last? b. what is the average velocity for the total trip? Given: v1 = 88 km/h; d1 = 280 km south unknown: vavg = ? t2 = 24 min = (24/60) hr = 0.4 hr v3 = 75 km/h; d3 = 210 km south v = d / t (total displacement / total time) v = (280 km S. + 210 km S.) / 6.4 hr = 76.56 km/hr south V = 77 km/h south Note: since the significant figures of given values is 2, so the displacement must be reported as 2 significant figures which is 77 km/h south Lab 1: intro to physics objective: to acquire training in scientific methods of observation and hypothesis • Make your own group with 3 or less people in a group. • Go to each of 7 stations – Follow directions and describe your observations – Try to give a reason for your observation 9/10 do now • A bus travels 300 km south along a straight path with an average velocity of 100 km/h to the south. The bus stops for 30 min, then it travels 150 km south with an average velocity of 75 km/h to the south. [you need to convert the min. to hr.] • How long does the total trip last? • [write equation, substitute number and units, answer with unit] objective • Finish the lab • Homework correction • Define and calculate acceleration Do now Class work • Practice packet: page 1-6 1. Describing Motion Verbally with Distance and Displacement 2. Describing Motion Verbally with Speed and Velocity 9/13 do now • Write in your own words the definition of acceleration. objective • Define acceleration • Calculate acceleration Acceleration • Acceleration is a vector quantity which is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity. • Acceleration has to do with changing how fast an object is moving. If an object is not changing its velocity, then the object is not accelerating. The velocity is changing by a constant amount - 10 m/s - in each second of time. Anytime an object's velocity is changing, the object is said to be accelerating; it has an acceleration. The Meaning of Constant Acceleration • Sometimes an accelerating object will change its velocity by the same amount each second. This is referred to as a constant acceleration since the velocity is changing by a constant amount each second. The data tables below depict motions of objects with a constant acceleration and a changing acceleration. Note that each object has a changing velocity. Calculating the Average Acceleration • Acceleration values are expressed in units of velocity/time. The units on acceleration are velocity units divided by time units - Typical acceleration units is (m/s)/s or m/s2 The Direction of the Acceleration Vector • The direction of the acceleration vector depends on two things: – whether the object is speeding up or slowing down – whether the object is moving in the + or - direction • The general RULE OF THUMB is: If an object is slowing down, then its acceleration is in the opposite direction of its motion. • This RULE OF THUMB can be applied to determine whether the sign of the acceleration of an object is positive or negative. • Consider the two data tables below. Object is speeding up Object is slowing in the positive down in the negative direction, its direction, its acceleration is in the acceleration is in the same direction of its opposite direction of its motion – positive. motion – positive. • Consider the two data tables below. Object is speeding up Object is slowing down in the negative in the positive direction, direction, its its acceleration is in the acceleration is in the opposite direction of its same direction of its motion – negative. motion – negative. Check Your Understanding • Use the equation for acceleration to determine the acceleration for the following two motions. Use a = (v - v ) / t and pick any Use a = (v -v ) / t and pick any two points. f i two points.f i a = (0 m/s - 8 m/s) / (4 s) a = (8 m/s - 0 m/s) / (4 s) a = (-8 m/s) / (4 s) a = (8 m/s) / (4 s) a = -2 m/s/s a = 2 m/s/s Calculating average acceleration • A shuttle bus slows to a stop with an average acceleration of -1.8 m/s2. How long does it take the bus to slow from 9.0 m/s to 0.0 m/s? Given: aavg = -1.8 m/s2 ; vi = 9.0 m/s; vf = 0 m/s unknown: t = ? a = ∆v / t -1.8 m/s2 = (0 – 9.0 m/s) / t t = (0 – 9.0 m/s) / (-1.8 m/s2) = 5.0 s example • Text book – page 49 – practice 2B - #5 Given: aavg = 4.7 x 10-3 m/s2 ; t = 5.0 min = 5 x 60 s = 300 s unknown: a. ∆v = ? b. If vi = 1.7 m/s vf = ? a. a = ∆v / t 4.7 x 10-3 m/s2 = ∆v / 300 s ∆v = (4.7 x 10-3 m/s2) x 300 s = 1.4 m/s b. ∆v = vf - vi 1.4 m/s = vf – 1.7 m/s vf = 1.4 m/s + 1.7 m/s = 3.1 m/s 9/14 do now • Practice packet: Acceleration - page 7 Homework castle learning: acceleration1 Objective – review acceleration • The direction of acceleration • Practice packet: Acceleration - page 8 • Text book – page 50 - Conceptual challenge: 1, 2, 3 1. No; the ball could change direction at the very top. 2. Slowing down 3. Jennifer’s acceleration could be positive if she is moving in the negative direction. 9/15 do now • Observe when a ball is thrown upward, describe 1. Its direction of velocity, give the reason for your description 2. Its direction of acceleration, give the reason for your description Lesson 2 Objective : Describing Motion with Diagrams 1. Ticker Tape Diagrams 2. Vector Diagrams Ticker Tape Diagrams • A common way of analyzing the motion of objects in physics labs is to perform a ticker tape analysis. 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 - say every 0.10 second. 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 representation of the object's motion. • The distance between dots on a ticker tape represents the object's position change during that time interval. • The analysis of a ticker tape diagram will also reveal if the object is moving with a constant velocity or accelerating. Vector Diagrams • Vector diagrams are diagrams which depict the direction and relative magnitude of a vector quantity by a vector arrow. Vector diagrams can be used to describe the velocity of a moving object during its motion. • In a vector diagram, the magnitude of a vector quantity is represented by the size of the vector arrow. • Vector diagrams can be used to represent any vector quantity. In future studies, vector diagrams will be used to represent a variety of physical quantities such as acceleration, force, and momentum. It will become a very important representation of an object's motion as we proceed further in our studies of the physics of motion. • This animation depicts some information about the car's motion. The velocity and acceleration of the car are depicted by vector arrows. The direction of these arrows are representative of the direction of the velocity and acceleration vectors. Lab 2: determine your speed • Purpose: question: How can I find my speed while I am in motion? • Material: spark timer, spark recording tape, ruler • Procedure: (write down what you did to answer the question) • Data: (set up a data table) • conclusion: • (save your data tape) A complete lab write-up includes a Title, a Purpose, the material, a Procedure, a Data section, and a Conclusion. The Data section should include collected data for several trials; column headings should be labeled and units shown. One sample calculation should be shown. Speed values for all the trials (except those which are obvious ) should be averaged; should be indicated as such in the Data section (a is a wholly scientific means of doing so). The Conclusion should respond to the question raised in the Purpose of the lab. 9/15 do now • Observe when a ball is thrown upward, describe 1. Its direction of velocity, give the reason for your description 2. Its direction of acceleration, give the reason for your description 9/16 Do now - practice • Practice packet – – Page 12 #5, #6 Homework Hand out – due tomorrow Quiz tomorrow – you may have your homework for reference only. I have hall duty today – no post session. Brief review • What are the words we learned to describe motion? Distance, displacement, speed, velocity, acceleration Scalar quantity: Distance, speed Vector quantity: Displacement, velocity, acceleration Ticker tape diagram: Vector diagram: Lesson 3 objective : Describing Motion with Position vs. Time Graphs 1. The Meaning of Shape for a p-t Graph 2. The Meaning of Slope for a p-t Graph 3. Determining the Slope on a p-t Graph The Meaning of Shape for a p-t Graph • To begin, consider a car moving with a constant, rightward (+) velocity - say of +10 m/s. If the position-time data for such a car were graphed, then the resulting graph would look like this: Note that a motion described as a constant, positive velocity results in a line of constant and positive slope when plotted as a position-time graph. • Now consider a car moving with a rightward (+), changing velocity - that is, a car that is moving rightward but speeding up or accelerating. If the position-time data for such a car were graphed, then the resulting graph would look like this: Note that a motion described as a changing, positive velocity results in a line of changing and positive slope when plotted as a position-time graph. • The position vs. time graphs for the two types of motion - constant velocity and changing velocity (acceleration) - are depicted as follows. Constant Velocity Positive Velocity Positive Velocity Changing Velocity (acceleration) The Importance of Slope • The slope of the line on a position-time graph reveals useful information about the velocity of the object. It is often said, "As the slope goes, so goes the velocity." Whatever characteristics the velocity has, the slope will exhibit the same (and vice versa). – If the velocity is constant, then the slope is constant (i.e., a straight line). – If the velocity is changing, then the slope is changing (i.e., a curved line). – If the velocity is positive, then the slope is positive (i.e., moving upwards and to the right). • "As the slope goes, so goes the velocity." Constant, Positive, Constant, Positive, Slow Velocity Fast Velocity Constant, Positive, Constant, Positive, smaller slope bigger slope Constant, negative Constant, negative Slow Velocity Fast Velocity Constant, negative, Constant, negative, smaller slope bigger slope Negative (-) Velocity Leftward (-) Velocity Slow to Fast Fast to Slow Negative slope, Negative slope, increasing decreasing Check Your Understanding • Use the principle of slope to describe the motion of the objects depicted by the two plots below. In your description, be sure to include such information as the direction of the velocity vector (i.e., positive or negative), whether there is a constant velocity or an acceleration, and whether the object is moving slow, fast, from slow to fast or from fast to slow. Be complete in your description. positive slope, increasing Negative slope, increasing Positive velocity slow to fast. negative velocity slow to fast. The Meaning of Slope for a p-t Graph • The shape of the line on the graph (straight, curving, steeply sloped, mildly sloped, etc.) is descriptive of the object's motion. The slope of a position vs. time graph reveals information about an object's velocity. • For example, a small, negative, constant slope means a slow, negative, constant velocity; a changing slope (curved line) means a changing velocity. • Consider a car moving with a constant velocity of +10 m/s for 5 seconds. The diagram below depicts such a motion. Note that during the first 5 seconds, the line on the graph slopes up 10 m for every 1 second along the horizontal (time) axis. That is, the slope of the line is +10 meter/1 second. In this case, the slope of the line (10 m/s) is equal to the velocity of the car. • Consider a car moving at a constant velocity of +5 m/s for 5 seconds, abruptly stopping, and then remaining at rest (v = 0 m/s) for 5 seconds. For the first five seconds the line on the graph slopes up 5 meters for every 1 second along the horizontal (time) axis. the line on the position vs. time graph has a slope of +5 meters/1 second for the first five seconds. Thus, the slope of the line on the graph equals the velocity of the car. During the last 5 seconds (5 to 10 seconds), the line slopes up 0 meters. That is, the slope of the line is 0 m/s - the same as the velocity during this time interval. an important principle • The slope of the line on a position-time graph is equal to the velocity of the object Determining the Slope on a p-t Graph • The slope of the line on a position versus time graph is equal to the velocity of the object. – If the object is moving with a velocity of +4 m/s, then the slope of the line will be +4 m/s. – If the object is moving with a velocity of -8 m/s, then the slope of the line will be -8 m/s. – If the object has a velocity of 0 m/s, then the slope of the line will be 0 m/s. • Let's begin by considering the position versus time graph To calculate the slope we must use the slope equation. 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) • To determine the slope for the graph: 9/17 do now • Describe the velocity in the following graphs using words: positive or negative, constant or increasing or decreasing or zero: A B C p p p t t t Positive, constant Positive, increasing zero objectives • Using Position-time graph to describe motion • Quiz • Finish and hand in all your homework – including Castlelearning, school website, handout Example – calculate slope • Slope = (y2 – y1) / (x2 – x1) • Slope = (26.0 m – 2.0 m) / (0.0 s – 8.0 s) = -3.0 m/s Check Your Understanding • Determine the velocity of the object as portrayed by the graph below. • Slope = (y2 – y1) / (x2 – x1) • Slope = (25 m – 5 m) / (5 s – 0 s) = 4 m/s Class work • Describing Motion with Position-Time Graphs • Text book – page 47 #3 • Review for the quiz 9/17 do now • Describe the velocity in the following graphs using words: positive, constant, increasing, decreasing, zero: A B C p p p t t t Positive, constant Positive, increasing zero objective • Quiz • Finish and hand in all your homework 9/20 do now • Use a new sheet of paper p As the slope goes, so goes the velocity t Describe the velocity of the object as portrayed by the graph using words such as (positive or negative), (constant or zero or increasing or decreasing). Lesson 4 objective: Describing Motion with Velocity vs. Time Graphs: 1. The Meaning of Shape for a v-t Graph 2. The Meaning of Slope for a v-t Graph 3. Relating the Shape to the Motion 4. Determining the Slope on a v-t Graph 5. Determining the Area on a v-t Graph The Meaning of Shape for a v-t Graph • Consider a car moving with a constant, positive velocity - say of +10 m/s. A car moving with a constant velocity is a car with zero acceleration. Note that a motion described as a constant, positive velocity results in a line of zero slope (a horizontal line has zero slope) when plotted as a velocity-time graph. Furthermore, only positive velocity values are plotted, corresponding to a motion with positive velocity. • Now consider a car moving with a positive, changing velocity - that is, a car that is moving rightward but speeding up or accelerating. Since the car is moving in the positive direction and speeding up, the car is said to have a positive acceleration. Note that a motion described as a changing, positive velocity results in a sloped line when plotted as a velocity-time graph. The slope of the line is positive, corresponding to the positive acceleration. Furthermore, only positive velocity values are plotted, corresponding to a motion with positive velocity. • The velocity vs. time graphs for the two types of motion - constant velocity and changing velocity (acceleration) - can be summarized as follows. Positive Velocity Positive Velocity Zero Acceleration Positive Acceleration The Importance of Slope • The slope of the line on a velocity-time graph reveals useful information about the acceleration of the object. – If the acceleration is zero, then the slope is zero (i.e., a horizontal line). – If the acceleration is positive, then the slope is positive (i.e., an upward sloping line). – If the acceleration is negative, then the slope is negative (i.e., a downward sloping line). • The slope of a velocity-time graph reveals information about an object's acceleration. But how can one tell whether the object is moving in the positive direction (i.e., positive velocity) or in the negative direction (i.e., negative velocity)? And how can one tell if the object is speeding up or slowing down? • The answers to these questions hinge on one's ability to read a graph. A positive velocity means the A negative velocity means the object is moving in the object is moving in the positive direction; negative direction; The slope of a velocity-time graph reveals information about an object's acceleration. But how can one tell if the object is speeding up or slowing down? Speeding up means that Slowing down means that the magnitude (or the magnitude (or numerical value) of the numerical value) of the velocity is getting large. velocity is getting smaller Check Your Understanding • Consider the graph at the right. The object whose motion is represented by this graph is ... (include all that are true): 1. moving in the positive direction. 2. moving with a constant velocity. 3. moving with a negative velocity. 4. slowing down. 5. changing directions. 6. speeding up. 7. moving with a positive acceleration. 8. moving with a constant acceleration. The Meaning of Slope for a v-t Graph • The shape of a velocity versus time graph reveals pertinent information about an object's acceleration. – If the acceleration is zero, then the velocity-time graph is a horizontal line (i.e., the slope is zero). – If the acceleration is positive, then the line is an upward sloping line (i.e., the slope is positive). – If the acceleration is negative, then the velocity-time graph is a downward sloping line (i.e., the slope is negative). – If the acceleration is great, then the line slopes up steeply (i.e., the slope is great). • The shape of the line on the graph (horizontal, sloped, steeply sloped, mildly sloped, etc.) is descriptive of the object's motion. • Consider a car moving with a constant velocity of +10 m/s. A car moving with a constant velocity has an acceleration of 0 m/s/s. Time (s) Velocity (m/s) 0 10 1 10 2 10 3 10 4 10 5 10 Note that the line on the graph is horizontal. That is the slope of the line is 0 m/s/s. In this case, it is obvious that the slope of the line (0 m/s/s) is the same as the acceleration (0 m/s/s) of the car. • Consider a car moving with a constant velocity of +10 m/s. A car moving with a constant velocity has an acceleration of 0 m/s/s. Time (s) Velocity (m/s) 0 10 1 10 2 10 3 10 4 10 5 10 Note that the line on the graph is horizontal. That is the slope of the line is 0 m/s/s. In this case, it is obvious that the slope of the line (0 m/s/s) is the same as the acceleration (0 m/s/s) of the car. • Consider a car moving with a constant velocity of +10 m/s. A car moving with a constant velocity has an acceleration of 0 m/s/s. Time (s) Velocity (m/s) 0 0 1 10 2 20 3 30 4 40 5 50 Note that the line on the graph is diagonal - that is, it has a slope. The slope of the line can be calculated as 10 m/s/s. The slope of the line (10 m/s/s) is the same as the acceleration (10 m/s/s) of the car. • Consider the motion of a car which first Time (s) Velocity (m/s) travels with a constant velocity (a=0 m/s/s) 0 2 of 2 m/s for four seconds and then accelerates at a rate of +2 m/s/s for four 1 2 seconds. That is, in the first four seconds, 2 2 the car is not changing its velocity (the 3 2 velocity remains at 2 m/s) and then the car increases its velocity by 2 m/s per second 4 2 over the next four seconds. 5 4 6 6 7 8 8 10 Velocity (m/s) The slope of the line on a velocity-time graph is equal to the acceleration of the object Time (s) Check Your Understanding • The velocity-time graph for a two-stage rocket is shown below. Use the graph and your understanding of slope calculations to determine the acceleration of the rocket during the listed time intervals. 1. t = 0 - 1 second 40 m/s2 2. t = 1 - 4 second 20 m/s2 3. t = 4 - 12 second -20 m/s2 Relating the Shape to the Motion • The shape of a velocity vs. time graph reveals pertinent information about an object's acceleration. – If the acceleration is zero, then the velocity-time graph is a horizontal line - having a slope of zero. – If the acceleration is positive, then the line is an upward sloping line - having a positive slope. – If the acceleration is negative, then the velocity-time graph is a downward sloping line - having a negative slope. – If the acceleration is great, then the line slopes up steeply - having a large slope. • The shape of the line on the graph (horizontal, sloped, steeply sloped, mildly sloped, etc.) is descriptive of the object's motion. 9/21 do now Use words like constant, positive, zero, negative, increasing, decreasing to describe velocity and acceleration of the object in graph A and graph B A B position t Velocity: Constant, positive Velocity: Constant, positive Acceleration: Zero (zero slope) Acceleration: Zero objectives 1. Relating the Shape to the Motion 2. Determining the Slope on a v-t Graph 3. Determining the Area on a v-t Graph Use words like constant, positive, zero, negative, increasing, decreasing to describe velocity and acceleration of the object. Constant, negative velocity, Zero acceleration (zero slope) Use words like constant, positive, zero, negative, increasing, decreasing to describe velocity and acceleration of the object. increasing, positive velocity, Constant, positive acceleration (constant, positive slope) Use words like constant, positive, zero, negative, increasing, decreasing to describe velocity and acceleration of the object. positive velocity, slowing down Acceleration is in the direction opposite of its motion – negative, constant (negative, constant slope) is Use words like constant, positive, zero, negative, slowing down, speeding up to describe velocity and acceleration of the object. negative velocity, slowing down Positive, constant acceleration (positive, constant slope) Use words like constant, positive, zero, negative, increasing, decreasing to describe velocity and acceleration of the object. negative velocity, speeding up. Negative, constant acceleration. (negative, constant slope) Check Your Understanding • Describe the motion depicted by the following velocity-time graphs. In your descriptions, use words such as positive, negative, speeding up or slowing down, constant, zero to describe velocity and acceleration during the various time intervals (e.g., intervals A, B, and C). A: positive, constant velocity, zero acceleration B: positive velocity, slowing down, constant negative acceleration C: positive, constant velocity, zero acceleration A: positive, decreasing velocity, negative, constant acceleration B: zero velocity, zero acceleration C: negative velocity, speeding up, negative constant acceleration A: positive, constant velocity, zero acceleration B: positive velocity, slowing down, negative, constant acceleration C: negative velocity, speeding up, negative constant acceleration Determining the Slope on a v-t Graph • The slope of the line on a velocity versus time graph is equal to the acceleration of the object. – If the object is moving with an acceleration of +4 m/s/s (i.e., changing its velocity by 4 m/s per second), then the slope of the line will be +4 m/s/s. – If the object is moving with an acceleration of -8 m/s/s, then the slope of the line will be -8 m/s/s. – If the object has a velocity of 0 m/s, then the slope of the line will be 0 m/s. • Let's begin by considering the velocity versus time graph To calculate the slope we must use the slope equation. 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) • To determine the slope for the graph: Check Your Understanding • Consider the velocity-time graph below. Determine the acceleration (i.e., slope) of the object as portrayed by the graph. The acceleration (i.e., slope) is 4 m/s/s. Determining the Area on a v-t Graph • A plot of velocity-time can be used to determine the acceleration of an object (the slope). • A plot of velocity versus time can also be used to determine the displacement of an object. • For velocity versus time graphs, the area bound by the line and the axes represents the displacement. 9/21 do now Use words like constant, positive, zero, negative, increasing, decreasing to describe velocity and acceleration of the object in graph A and graph B A B position t Velocity: Constant, positive Velocity: Constant, positive Acceleration: Zero (zero slope) Acceleration: Zero objective Determining the Area on a v-t Graph Determining the Area on a v-t Graph • A plot of velocity-time can be used to determine the acceleration of an object (the slope). • A plot of velocity versus time can also be used to determine the displacement of an object. • For velocity versus time graphs, the area bound by the line and the axes represents the displacement. • The diagram below shows three different velocity-time graphs; the shaded regions between the line and the time-axis represents the displacement during the stated time interval. 180 m 80 m 125 m – 20 m = 105 m example • Determine the displacement (i.e., the area) of the object during the first 4 seconds (Practice A) and from 3 to 6 seconds (Practice B). The area of under the line The area of under the line represents the displacement. represents the displacement. A = (4 s) • (30 m/s) = 120 m A = (3 s) • (30 m/s) = 90 m That is, the object was displaced That is, the object was displaced 120 m during the first 4 seconds of 90 m during the interval of time motion between 3 and 6 seconds. example • Determine the displacement of the object during the first second (Practice A) and during the first 3 seconds (Practice B). The area of under the line The area of under the line represents the displacement. represents the displacement. Area = ½ • b • h Area = ½ • b • h Area = ½ • (1 s) • (10 m/s) = 5 m Area = ½ • (3 s) • (30 m/s) = 45 m example • Determine the displacement of the object during the time interval from 2 to 3 seconds (Practice A) and during the first 2 seconds (Practice B). The area of a trapezoid represents The area of a trapezoid represents the displacement the displacement Area = ½ • b1 • h1 – ½ b2 • h2 (b1 = 3 Area = ½ • b1 • h1 – ½ b2 • h2 (b1 = 3 s, s, b2 = 2s, h1 = 30 m/s, h2 = 20 m/s.) b2 = 2s, h1 = 30 m/s, h2 = 10 m/s.) Area = 45 m – 20 m = 25 m Area = 45 m – 5 m = 40 m That is, the object was displaced 25 That is, the object was displaced 40 m m during the time interval from 2 to 3 during the time interval from 1 to 2 seconds. seconds. 9/22 Do now A. What does the slope in position-time graph represent? B. What does the slope in velocity-time graph represent? C. What does the area bound by the line and the axes in velocity-time graph represent? 9/23 do now • Describe the object’s velocity in the diagram A and B. choose one word for each blank. p v t t A B Velocity: ____________ (positive or Velocity: ____________ (positive or negative), _____________(constant or negative), _____________(constant or zero or increasing or decreasing) zero or increasing or decreasing) objectives • Position-time graph lab • Review graphs Homework – castle learning: Lab 3 : Position-Time Graphs Lab Direction: Open up logger pro program – open file folder on top – click physics with vernier (physics with computer) – click 01a Question: How can the following types of motion be described with a position-time graph? (moving in the positive direction versus moving in the negative direction; moving fast versus moving slow; moving with a constant speed versus moving with a gradually changing speed; speeding up versus slowing down; etc.) Purpose: To contrast the shape and slope of the position-time graphs for the following types of motion: – moving in the + direction versus moving in the - direction – moving fast versus moving slow – a constant speed motion versus a gradually changing speed – a speeding up motion versus a slowing down motion (use incline) – combinations of the above (use incline) Material: – Computer, Logger Pro program, motion sensor, computer interface, Report requirement: A complete lab write-up includes Title, Purpose, Materials, Data section, and Conclusion/Discussion. – The Data section should include one graph for each contrasting set of two motions; axes should be labeled; labels or color coding or some other method should be used to distinguish between the two motions. – The Conclusion/Discussion section should provide a thorough discussion of the differences in the position-time graphs for the variety of motions under study. Class work 1. Describing Motion with Velocity-Time Graphs 2. Describing Motion Graphically 3. Interpreting Velocity-Time Graphs 4. Graphing Summary 5. Kinematics Graphing - Mathematical Analysis 6. Text book – page 59 - #5, 6 9/24 do now • The graph below shows the velocity of a race car moving along a straight line as a function of time. What is the magnitude of the displacement of the car from t = 3.0 seconds to t = 4.0 seconds? objectives • Graph quiz • Practice packet – finish pages up to page 23. 9/27 do now • The velocity-time graph represents the motion of a 3- kilogram cart along a straight line. The cart starts at t = 0 and initially moves north. What is the magnitude of the displacement of the cart 1. during t = 0 and t = 5 seconds? 2. during t = 5 and t = 7 seconds? 3. During t = 0 and t = 7 seconds? • The velocity-time graph represents the motion of a 3- kilogram cart along a straight line. The cart starts at t = 0 and initially moves north. What is the acceleration of the cart 1. during t = 0 and t = 3 seconds? 2. during t = 3 and t = 7 seconds? • The velocity-time graph represents the motion of a 3- kilogram cart along a straight line. The cart starts at t = 0 and initially moves north. What is the velocity of the cart 1. At t = 3 seconds? 2. during t = 6 seconds? • The graph represents the relationship between the displacement of an object and its time travel along a straight line. a. What is the magnitude of the object's total displacement during t = 0 to t = 8.0 seconds? b. What is the average speed of the object during t = 0 to t = 4.0 seconds? c. What is the magnitude of acceleration of the object during t = 0 to t = 2.0 seconds? 9/27 do now • A locomotive starts at rest and accelerates at 0.12 meters per second squared to a speed of 2.4 meters per second in 20. seconds. Describe the locomotive’s acceleration and velocity. 1. Acceleration: _______________(constant, increasing, decreasing, zero) 2. velocity: _______________(constant, increasing, decreasing, zero) objectives • Free fall homework • Castle learning – free fall Lesson 5 objective : relate the motion of a Free Falling body to motion with constant Acceleration 1. Introduction to Free Fall 2. The Acceleration of Gravity 3. Representing Free Fall by Graphs 4. How Fast? and How Far? 5. The Big Misconception Homework: castle learning Introduction to Free Fall • A free-falling object is an object which is falling under the sole influence of gravity. Any object which is being acted upon only be the force of gravity is said to be in a state of free fall. • There are two important motion characteristics which 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 9.81 m/s/s. The dot diagram at the right depicts the acceleration of a free-falling object. The fact that the distance which the object travels every interval of time is increasing is a sure sign that the ball is speeding up as it falls downward • A free-falling object is an object which is falling under the sole influence of gravity. A free-falling object has an acceleration of 9.81 m/s/s, downward (on Earth). • It is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. • The symbol g is used to represent acceleration of gravity. • g = 9.81 m/s/s (m/s2), downward • Acceleration is the rate at which an object changes its velocity. 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 Time (s) Velocity (m/s) for a free-falling object being dropped from a position of rest were 0 0 tabulated, then one would note the 1 - 9.8 following pattern. 2 - 19.6 3 - 29.4 4 - 39.2 5 - 49.0 For an object falling freely from rest, it’s speed increases by 9.81 m/s every second. Representing Free Fall by Graphs • We can describe the motion of objects through the use of graphs - position versus time and velocity vs. time graphs. • A free falling object is accelerating. Position vs. time graph Position (m) Position (m) or Time (s) Time (s) Downward is set to be Downward is set to be positive. negative. Representing Free Fall by Graphs • A free falling object is accelerating. Velocity vs. time graph velocity (m/s) Velocity (m/s) Time (s) or Time (s) Downward is set to be Downward is set to be positive. negative. How Fast? and How Far? • Free-falling objects are in a state of acceleration. Specifically, they are accelerating at a rate of 9.81 m/s/s. • The velocity of a free-falling object is changing by 9.8 m/s every second. • If dropped from a position of rest, the object will be traveling – At 1 s – v = 9.81 m/s – At 2 s – v = 19,6 m/s – At 3 s – v = 29.4 m/s. • Thus, the velocity of a free-falling object which has been dropped from a position of rest is dependent upon the time for which it has fallen. The formula for determining the velocity of a falling object after a time of t seconds is • vf = g * t where g is the acceleration of gravity. • g = 9.81 m/s/s vf = g * t • where g is the acceleration of gravity. The value for g on Earth is 9.8 m/s/s. • The above equation can be used to calculate the velocity of the object after any given amount of time when dropped from rest. • Example Calculations: – At t = 6 s vf = (9.81 m/s2) * (6 s) = 58.8 m/s – At t = 8 s vf = (9.81 m/s2) * (8 s) = 78.4 m/s • The distance which a free-falling object has fallen from a position of rest is also dependent upon the time of fall. This distance can be computed by use of the formula; d = ½ ·g·t2 where g is the acceleration of gravity (9.81 m/s/s on Earth). • Example • t=1s d = ½ ·(9.81 m/s2)·(1 s)2 = 4.9 m • t=2s d = ½ ·(9.81 m/s2)·(2 s)2 = 19.6 m • t=5s d = ½ ·(9.81 m/s2)·(5 s)2 = 122.5 m • Try to remember these distances: • 1s–5m • 2 s – 20 m • 3 s – 45 m • 4 s – 80 m • 5 s – 123 m The Big Misconception Acceleration is • The acceleration of a free- the same at all falling object (on earth) is 9.81 points. m/s/s. This value (known as the acceleration of gravity) is the same for all free-falling objects regardless of how long they have been falling, or whether they were initially dropped from rest or thrown up into the air. The Big Misconception • “Doesn't a more massive object accelerate at a greater rate than a less massive object?" The answer is absolutely not! That is, absolutely not if we are considering the specific type of falling motion known as free-fall. Free-fall is the motion of objects which 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. practice • Free fall work sheet • Text book – page 65 - #1, 3, 4, 6a, b. 9/28/10 do now • Sketch a graph best represents the relationship between mass and acceleration due to gravity for objects near the surface of the Earth. Homework: castlelearning: kinematics 1, kinematics 2 Lesson 6 objective : Describing Motion with Equations 1. The Kinematics Equations 2. Kinematics Equations and Problem-Solving 3. Kinematics Equations and Free Fall 4. Sample Problems and Solutions 5. Kinematics Equations and Graphs • The kinematics equations are a set of four equations which can be utilized to predict unknown information about an object's motion if other information is known. d = vit + ½ a·t2 vf = vi + a·t vf2 = vi2 + 2a·d v +v d = vavg·t = i f ·t 2 d: displacement t: time a: acceleration vi: initial velocity vf: final velocity Kinematics Equations and Problem- Solving • We are using the equations to determine unknown information about an object's motion. The strategy of finding the unknown involves the following steps: 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 which 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 A • 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. Given: unknown: vi = +30.0 m/s vf = 0 m/s a = - 8.00 m/s2 d = ?? vf2 = vf2 + 2·a·d (0 m/s)2 = (30 m/s)2 + 2(- 8.00 m/s2)·d d = 56.3 m The value seems reasonable. It takes a car about 4 seconds to come to a stop. Its average speed is 15 m/s, so the distance should be close to 60 m. Example B • Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6.00 m/s2 for a time of 4.10 seconds. Determine the displacement of Ben's car during this time period Diagram: Given: unknown: vi = 0 m/s t = 4.10 s d = ?? a = 6.00 m/s2 d = vit + ½ ·a·t2 d = (0 m/s)(4.10 s) + ½ (6.00 m/s2)·(4.10 s)2 The value seems reasonable. Its speed at the end d = 50.4 m of 4.10 s is about 24 m/s, so its average is about 12 m/s, its distance should be around 48 m. Kinematics Equations and Free Fall • A free-falling object is an object which is falling under the sole influence of gravity. Such an object will experience a downward acceleration of 9.81 m/s/s. • Whether the object is falling downward or rising upward towards its peak, if it is under the sole influence of gravity, then its acceleration value is 9.8 m/s/s. • Like any moving object, the motion of an object in free fall can be described by four kinematics equations. d = vit + ½ a·t2 vf = vi + a·t vf2 = vi2 + 2a·d vi+vf d = vavg·t = ·t 2 • There are a few conceptual characteristics of free fall motion: 1. An object in free fall experiences an acceleration of -9.81 m/s/s. (The “-” sign indicates a downward acceleration.) 2. If an object is merely dropped (as opposed to being thrown) 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. This value can be used as one of the motion parameters in the kinematic equations; for example, the final velocity (vf) after traveling to the peak would be assigned a value of 0 m/s. 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 which 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. Example A • Luke Autbeloe drops a pile of roof shingles from the top of a roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground. Diagram: Given: unknown: vi = 0.0 m/s d = -8.52 m t = ?? a = - 9.81 m/s2 d = vit + ½ ·a·t2 -8.52 m = (0 m/s)(t) + ½ (-9.81 m/s2)·t2 t = 1.32 s The value seems reasonable enough. It falls about 5 m at the end of 1 s, about 20 m at the end of 2 s. Example B • Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height. Diagram: Given: unknown: vi = 26.2 m/s vf = 0 m/s d = ?? a = -9.81 m/s2 vf2 = vf2 + 2·a·d (0 m/s)2 = (26.2 m/s)2 + 2(- 9.81 m/s2)·d d = 35 m The value seems reasonable. It takes the vase less than 3 seconds to reach the top. The distance should be less than 45 meters and more than 20 m.. example • A spark timer is used to record the position of a lab cart accelerating uniformly from rest. Each 0.10 second, the timer marks a dot on a recording tape to indicate the position of the cart at that instant, as shown. • The linear measurement between t = 0 second to t = 0.30 is 5.4 cm. • Calculate the magnitude of the acceleration of the cart during that time interval. Given: t = 0.30 s, d = 5.4 cm, vi = 0 Find: a = ? d = vit + ½ at2 a = 120 cm/s2 9/29 do now • Which combination of graphs above best describes free-fall motion? [Neglect air resistance.] 1 D kinematics review • Lesson 1 : Describing Motion with • Lesson 4 : Describing Motion with Words Velocity vs. Time Graphs Introduction to the Language of Kinematics The Meaning of Shape for a v-t Graph Scalars and Vectors The Meaning of Slope for a v-t Graph Distance and Displacement Relating the Shape to the Motion Speed and Velocity Determining the Slope on a v-t Graph Acceleration Determining the Area on a v-t Graph • Lesson 2 : Describing Motion with • Lesson 5 : Free Fall and the Diagrams Acceleration of Gravity Introduction to Diagrams Introduction to Free Fall Ticker Tape Diagrams The Acceleration of Gravity Vector Diagrams Representing Free Fall by Graphs • Lesson 3 : Describing Motion with How Fast? and How Far? Position vs. Time Graphs The Big Misconception The Meaning of Shape for a p-t Graph • Lesson 6 : Describing Motion with The Meaning of Slope for a p-t Graph Equations Determining the Slope on a p-t Graph The Kinematics Equations Kinematics Equations and Problem-Solving Kinematics Equations and Free Fall Sample Problems and Solutions Kinematics Equations and Graphs 9/30 do now • A roller coaster, traveling with an initial speed of 15 meters per second, decelerates uniformly at -7.0 meters per second2 to a full stop. Approximately how far does the roller coaster travel during its deceleration? [show work] vi vf vavg a t d 15 m/s 0 m/s -7 m/s2 ? vf2 = vf2 + 2·a·d 0 = (15 m/s)2 + 2 (-7 m/s2)·d -(15 m/s)2 = 2 (-7 m/s2)·d d = 16 m example • A 1.0-kilogram ball is dropped from the roof of a building 40. meters tall. What is the approximate time of fall? [Neglect air resistance.] [show work] vi vf vavg a t d 0 m/s -9.81 m/s2 ? -40 m d = vit + ½ ·a·t2 t = 2.86 s objectives • Review • Test is tomorrow – do your homework – castle learning assignments! Homework note book • Text book – P65 - #1, 2, 4, 5 – P53 – #2 (18.8 m); #4 (no; the plane needs 1 km to land.) – P55 - #3 (-7.5 m/s; 19 m); #4 (2.5 s; 32 m) – P58 - #3 (16 m/s, 7 s) – P64 – #4 (3.7 m; 0.76 s) objectives • Lab – determine the acceleration • Review • Test is tomorrow – do your homework – castle learning assignments! 10/1 do now • A ball is thrown straight downward with a speed of 0.50 meter per second from a height of 4.0 meters. What is the speed of the ball 0.70 second after it is released? [Neglect friction.] vi vf vavg a t d -0.50 m/s ? -9.81 m/s2 0.7 s -4.0 m vf = vi + a·t vf = -7.4m/s Speed is 7.4m/s • The velocity-time graph represents the motion of a 3- kilogram cart along a straight line. The cart starts at t = 0 and initially moves north. What is the magnitude of the displacement of the cart between t = 3 and t = 7 seconds? [Show work] 0m example • The graph represents the relationship between the displacement of an object and its time travel along a straight line. 1. What is the magnitude of the object's total displacement during t = 0 seconds and t = 8.0 seconds? 0m 2. What is the average speed of the object during t = 0 seconds and t = 4.0 seconds? [show work] 2 m/s 3. What is the magnitude of acceleration of the object during t = 0 seconds and t = 2.0 seconds? 0m objectives • Lab • Chapter 1 test • Homework note book is due Lab 4 : determine acceleration Question: How can you find the acceleration of a moving cart on an incline? Purpose: To determine the acceleration of a moving cart on an incline. Material: spark timer, data recording tape, ruler, incline, tapes, graph paper Report requirement: A complete lab write-up includes Title, Purpose, Materials, Data section, and Conclusion/Discussion. The Data section should include – Data table: • collected data for three trials; • column headings should be labeled and units shown. • displacement values for all the trials (except those which are bvious ) should be averaged; • One sample calculation of acceleration should be shown. • Final average of the acceleration should be shown. – Displacement vs. time graph: • Title of the graph • Axises of displacement and time with appropriat scale and unit • Plot the data points on the graph • Draw the best fit line – Velocity vs. time graph: • Title of the graph • Axises of velocity and time with appropriate scale and unit • Find velocity at a time instant by using kinematics equations v(t) = 2 · V average (0 – t) • Plot the data points on the graph • Draw the best fit line • Determine the acceleration of the object • The Conclusion should respond to the question raised in the Purpose of the lab. – Procedure – Compare acceleration determined in velocity time graph with the acceleration value determined in data table. Discuss your result. 2. Time of Free-Fall A long string of metal washers is constructed such that the first one is 30 cm from the bottom end, the second is 130 cm from the bottom end, and the third is 2.74 m from the bottom end. When the string is released, the washers will sound out at constant time intervals. 1. Units of Measure A meter stick, yard stick, 1 kilogram mass and 1 pound weight are laid out on a demonstration table for inspection and comparison of relative size. Mechanics Runner's Speed • 5. Take a stopwatch and a meter stick to a running track or sidewalk. Use the meter stick to measure 20 m. Mark the distance with pieces of masking tape. Measure the time it takes to walk 20 m. Calculate your average speed. Measure the time it takes you to jog, or run, the same distance. What is your average speed? Materials: stopwatch; Meter stick; Masking tape.