objectives 1. Describe motion in terms of frame of reference, displacement, distance, vector, and scalar. 2. Understand relationship between displacement and distance, vectors and scalars. 3. Compare distance traveled to displacement. 4. Draw diagrams showing displacement and/or distance from a point of origin. Description of motion depends on the frame of reference • A frame of reference can be thought of as any spot you are doing your measurement from as long as it is not accelerating. The reference point is arbitrary, but once chosen, it must be used throughout the problem. Displacement vs. Distance • DISTANCE – The complete length of the path traveled by a moving object • DISPLACEMENT – the length of the straight-line path from a moving object’s origin to its final position Scalar vs. Vector • SCALAR – A measured quantity that has NO DIRECTION – Examples • Distance, Time, Mass, Volume • VECTOR – A measured quantity that includes DIRECTION – SIGN SHOWS DIRECTION – Example • Displacement Sign Conventions Positive Negative NORTH EAST RIGHT SOUTH WEST LEFT A bird fliesAA5ball cat meters rolls runsnorth, 58 meters meters then north. west. 7 meters south Distance = 512 8m mm Displacement = -2 +5m -8 m Example • A man drives his car 3 miles north, then 4 miles east. 4 mi East 3 mi North Distance 7 mi Displacement 5 mi Northeast What distance did he travel? What is his displacement from his point of origin? Example • Three men leave the same house on foot. The first man walks 30 feet north, then 40 feet west. The second man walks 90 feet south, then 88 feet north. The third man walks 10 feet east, then 50 feet west. • Which man has traveled the greatest distance? The second man • Who is farthest from the house? The first man • Who is closest to the house? The second man Distance vs. Displacement xi xf • There could be many distances between xf and xi many be many, distance depends on the path. • There is only one displacement between xf and xi. displacement refers to shortest distance between the xf and xi and direction from xi to xf Displacement = change in position = final position – initial position ∆x = xf ∆ denotes change - xi example • Use the diagram to determine the resulting displacement and the distance traveled by the skier during these three minutes. Distance (m) Displacement (m) AB 180 +180 BC 140 -140 CD 100 +100 TOTAL 420 +140 Example • Consider the motion depicted in the diagram below. A physics teacher walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North. Distance: 12m Displacement: 0 Distance vs. Time Graphs During what time interval was the object NOT MOVING? Distance vs. Time 18 2 – 3 seconds The interval on the graph where 14 the distance remains constant! 12 Distance (m) 16 10 8 6 4 2 0 0 1 2 3 4 Time (s) 5 6 7 Displacement vs. Time Graphs During During At what what what time distance time interval(s) interval(s) fromwas thewas the origin object thedoes object tothe the NOT object left MOVING? of stop? the origin? Displacement vs. Time 6 5 Displacement (m) 4 3 2 The object’s final position is at +1 meter (1 meter to the When the displacement is of the origin) the object has a 1 – 2 and negative, 4 – 5right seconds position to the left that of the origin Constant displacement means the object doesn’t move 1 0 -1 0 1 2 3 4 -2 -3 -4 Time (s) 5 6 7 Objectives • Know: – - Definitions of velocity and speed. – - Equation for average velocity/speed. • Understand: – - Relationship between speed, distance and time. – - Relationship between vectors and scalars. • Be able to: – - Use velocity/speed equation to find unknown. – - Draw and interpret d/t and v/t graphs. Homework – castle leaning – must show work on a separate sheet of paper Velocity vs. Speed • VELOCITY – change in DISPLACEMENT occurring over time – Includes both MAGNITUDE and DIRECTION • VECTOR • The direction of the velocity vector is simply the same as the direction that an object is moving. • SPEED – change in DISTANCE occurring over time – Inclues ONLY MAGNITUDE • SCALAR As an object moves, it often undergoes changes in speed Types of speed and velocity • • • • Initial speed Final speed Average speed Instantaneous speed • • • • Initial velocity Final velocity Average velocity Instantaneous velocity Calculate Average Speed and Average Velocity • The average speed during the course of a motion is often computed using the following formula: Does NOT include DIRECTION! • In contrast, the average velocity is often computed using this formula ∆x x f - xi vavg = ∆t = t - t f i = d t Average Velocity d v t What does this remind you of? SLOPE OF AWhat GRAPH! is happening in this Position vs. Time Position Position graph? INCREASING SLOPE CONSTANT CONSTANT POSITIVE ZERO SLOPESLOPE Time Moving with Motionless INCREASING CONSTANT Object positive velocity velocity Example • Sally gets up one morning and decides to take a three mile walk. She completes the first mile in 8.3 minutes, the second mile in 8.9 minutes, and the third mile in 9.2 minutes. – What is her average during her vavg = d / speed t walk? v = 3 mi / (8.3 min + 8.9 min + 9.2 min) avg vavg = 0.11 mi / min Example • Tom gets on his bike at 12:00 in the afternoon and begins riding west. At 12:30 he has ridden 8 miles. – What was his average velocity during his ride? vavg = d / t vavg = 8 mi / 30 min vavg = 0.27 mi / min WEST Example – finding displacement • During a race on level ground, Andre runs with an average velocity of 6.02 m/s to the east. What displacement does Andre cover in 137 s? ∆x (∆t ) vavg = ∆t (∆t ) ∆x = vavg (∆t ) = (6.02 m/s)(137 s) = 825 m Answer: 825 m East example • What is the coach's average speed and average velocity? average speed = (35 + 20 + 40) yd / 10min = 9.5 yd/min average velocity = (-35 + 20 - 40) yd / 10 min = - 5.5 yd/min Instantaneous Speed • Instantaneous Speed - the speed at any given instant in time. Average Speed versus Instantaneous Speed Average Speed - the average of all instantaneous speeds; found simply by a distance/time ratio. During your trip, there may have been times that you were stopped and other times that your speedometer was reading 50 miles per hour. Yet, on average, you were moving with a speed of In conclusion • Speed and velocity are kinematics quantities that have distinctly different definitions. Speed, scalar being a _______quantity, is the rate at which an distance object covers ___________. The average speed is the _____________ (a scalar quantity) per distance time ratio. Speed is ignorant of direction. On the vector other hand, velocity is a _________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. Using v-t Graphs Velocity (m/s) Velocity vs. Time 35 30 25 20 15 10 5 0 -5 0 -10 -15 -20 What can we DO with a v-t graph? Find distance traveled 1 2 3 4 5 6 7 8 9 10 11 Time (s) Area under under the graph to find How doArea you use thethe v-tgraph graph DISTANCE AVERAGE DISPLACEMENT? TRAVELED? VELOCITY? Area AreaononTOP TOPand = POSITIVE BOTTOM Area both onconsidered BOTTOMPOSITIVE = NEGATIVE Find displacement Find average velocity objective 1. Construct and interpret graphs of position vs. time and velocity vs. time. Graphical Interpretation of Velocity: position vs. time graph • The Meaning of Shape for a p-t Graph Straight line Constant Velocity (positive) Curved line – Changing Velocity (increasing) As the slope goes, so goes the velocity Slow, Positive, Constant Velocity Fast, Positive, Constant Velocity Fast, Negative, Constant Velocity Slow, Negative Constant Velocity Check Your Understanding • Use the principle of slope to describe the motion of the objects depicted by the two plots below. Slope is positive, increasing slope is negative, increasing velocity is positive, increasing (faster) velocity is negative, increasing (faster) The slope of the line on a position-time graph is equal to the velocity of the object Slope = constant = +10 m/s Velocity = constant = +10 m/s Slope is changing – increasing in positive direction velocity is changing – increasing in positive direction Velocity information on P-t graph • • • • Initial velocity Final velocity Average velocity Instantaneous velocity • • • • +10 m/s +10 m/s +10 m/s +10 m/s •Initial velocity •Final velocity •Average velocity •Instantaneous velocity • • • • 0 m/s ~+22 m/s +10 m/s Slope of the tangent line As the slope goes, so goes the velocity Slope is negative, increasing (steeper) Slope is negative, decreasing (flatter) velocity is negative, increasing (faster) velocity is negative, decreasing (slower) example Describe the velocity of the object between 0-5 s and between 5-10 s. The velocity is 5 m/s between 0-5 seconds The velocity is zero between 5-10 seconds Practice – determine average velocity 7.3 m/s objectives 1. Describe motion in terms of changing velocity. 2. Compare graphical representations of accelerated and nonaccelerated motions. 3. Apply kinematics equations to calculate distance, time, or velocity under conditions of constant acceleration. Acceleration • Acceleration change in VELOCITY occuring over TIME Measured in m/s2 Vector Acceleration = (change in velocity) / time ∆v vf - vi aavg = ∆t = t - t f i Questions: 1. “If an object has a large velocity, does it necessarily have a large acceleration? 2. If an object has a large acceleration, does it necessarily have a large velocity?” • Anytime an object's velocity is changing, the object is said to be accelerating; it has an acceleration. • What are the three ways to accelerate your car? Gas pedal, break, steering wheel The Direction of the Acceleration Vector • The direction of the acceleration vector depends on whether the object is speeding up or slowing down • If an object is speeding up, then its acceleration is in the same direction of its motion. – moving in the + direction, the acceleration + direction – moving in the - direction, the acceleration - direction • If an object is slowing down, then its acceleration is in the opposite direction of its motion. – moving in the + direction, the acceleration - direction – moving in • The direction of velocity and acceleration do not have to be the same!!! Negative Positive Velocity Velocity Negative Positive Acceleration Acceleration Speeding Speeding Slowing up up in indown +- direction direction Eventually speeds up in + – direction! +v -v +v -v +a -a -a +a The car will… The car will… The car will… The car will… Equations aavg v t “Acceleration is a rate of change in velocity” “The slope of a v-t graph tells what the ACCELERATION IS DOING!” v f vi at “An object’s velocity at any point in time can be found by considering: - its starting velocity - its acceleration - the amount of time over which it accelerates” Example – average acceleration • What is the acceleration of an amusement park ride that falls from rest to a speed of 28 m/s in 3.0 s? ∆v vf - vi aavg = ∆t = t - t f i aavg = (28 m/s – 0) / 3.0 s = 9.3 m/s/s down Example – finding time of accelertion • 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? ∆v aavg = ∆t ∆v aavg ∆t ∆t = ∆v = (0.0 m/s – 9.0 m/s) / -1.8 m/s2 = 5.0 s aavg Example • Two cars start at the same point . Car A starts with a velocity of -5 meters per second while Car B starts with a velocity of +3 meters per second. At the end of 15 seconds, Car A has a velocity of +25 vf = vi + at meters per second. vf = +3 m/s + (2 m/s2)(15 s) vf =acceleration? +33 m/s – What is Car A’s vf = vi + at 25 m/s = -5 m/s + a(15 s) – If Car B has the acceleration, what is its 2 a =same +2 m/s speed at 15 seconds? The Meaning of Constant Acceleration The velocity is changing by a constant amount - in each second of time. Constant motion vs. accelerated motion ticker tape constant velocity: displacement is changing by a constant amount - in each second of time. constant acceleration: displacement is increasing in each second of time. Velocity is changing by a constant amount - in each second of time Describing Motion with Velocity vs. Time Graphs The slope value of any straight line on a velocity-time graph is the acceleration of the object Velocity: Constant Slope = 0 Acceleration = 0 Velocity: increasing Slope: +, constant Acceleration: +, constant Speeding up or slowing down? • Speeding up means that the magnitude (or numerical value) of the velocity is getting large. Example: describe velocity and acceleration for each v-t diagram Velocity is positive. increasing, speeding up Velocity is positive decreasing, slowing down Velocity is negative decreasing, slowing down Velocity is negative, increasing, speeding up Acceleration is constant, positive Acceleration is constant, negative Acceleration is constant, positive Acceleration is constant, negative Example 1 Determine acceleration From 0 s to 4 s: 0 m/s2 From 4 s to 8 s: 2 m/s2 • Example 2 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. +40 m/s/s 1. t = 0 - 1 second +20 m/s/s 2. t = 1 - 4 second 3. t = 4 - 12 second -20 m/s/s objectives • Understand Relationships for all kinematics equations. • Understand Relationship between slope of kinematics graphs and quantities they represent. • Use any kinematics equation to find unknown. • Draw and interpret all kinematics graphs. Homework – castle learning vavg = vf + v i 2 vf + vi 2 ∆x vavg = ∆t ∆x = ∆t ∆x = ½ (∆t)(vf + vi) The displacement equals to the area under the velocity vs. time graph Determining the Area on a v-t Graph For velocity versus time graphs, the area bound by the line and the axes represents the displacement. The shaded area is representative of the displacement area = base x height area = ½ base x height area = ½ base x ( height1 + height2) 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). 120 m 90 m example • Determine the displacement of the object during the first second (Practice A) and during the first 3 seconds (Practice B). 5m 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). 25 m 40 m Recap ∆x vf + v i • Average velocity: v = ∆t = 2 • Acceleration: ∆v vf - vi a = ∆t = t - t f i • Final velocity: vf = vi + a∆t • Displacement: ∆x = ½ (∆t)(vf + vi) • Displacement: ∆x = vi∆t + ½ a∆t2 Another equation for final velocity • Final velocity: vf = vi + a∆t • Displacement: ∆x = ½ (∆t)(vf + vi) • Final velocity: vf2= vi2 + 2a∆x Constant Non-Zero Acceleration 1 2 d vi t at 2 “Distance or Displacement” Equation v v 2ad “Timeless or Shortcut Equation” 2 f 2 i DOES NOT INCLUDE TIME! In one dimensional motion: ∆x = d Example • A bicyclist accelerates from 5.0 m/s to a velocity of 16 m/s in 8 s. Assuming uniform acceleration, what displacement does the bicyclist travel during this time interval? ∆x = ½ (vf + vi)(∆t) 84 m Example • A racing car reaches a speed of 42 m/s. It then begins a uniform negative acceleration, using its parachute and breaking system, and comes to rest 5.5 s later. Find how far the car moves before stopping. 120 m Example • A train starts from rest and leaves Greenburg station and travels for 500. meters with an acceleration of 1.20 meters per second2. – What is the train’s vf2 = final vi2 + velocity? 2ad vf2 = 0 + 2(1.20 m/s2)(500. m) vf = 34.6 m/s – How long does it take the train to reach its final velocity? vf = vi + at 34.6 m/s = 0 + (1.20 m/s2) t t = 28.8 s Example • A driver traveling at 85. miles per hour sees a police car hiding in the trees 2.00 miles ahead. He applies his brakes, decelerating at -500. miles per hour2. 2 =mph, – If the speed limit isv 55 v 2 + will 2adhe get a ticket? vf2 = (85. f mph)2 i + 2(-500. mph2)(2.00 mi) vf = 72.3 mph *YES* – What would his acceleration need to be to not get a vf2 = vi2 + 2ad ticket? (55 mph)2 = (85. mph)2 + 2a(2.00 mi) a = -1050 mph2 Objective • Apply kinematics equations to calculate distance, time, or velocity under conditions of constant acceleration. example • A barge moving with a speed of 1.00 m/s increases speed uniformly, so that in 30.0 s it has traveled 60.2 m. What is the magnitude of the barge’s acceleration? a 6.71 102 m/s2 example • A person pushing a stroller starts from rest, uniformly accelerating at a rate of 0.500 m/s2. What is the velocity of the stroller after it has traveled 4.75 m? vf = +2.18 m/s example • An aircraft has a landing speed of 302 km/h. The landing area of an aircraft carrier is 195 m long. What is the minimum uniform accelerating required for a safe landing? a = -18 m/s2 example • A plane starting at rest at one end of a runway undergoes a uniform acceleration of 4.8 m/s2 for 15 s before takeoff. What is its speed at takeoff? How long must the runway be for the plane to be able to take off? vf = 72 m/s ∆x = 540 m Objectives 1. Relate the motion of a freely falling body to motion with constant acceleration. 2. Calculate displacement, velocity, and time at various points in the motion of a freely falling object. 3. Compare the motion of different objects in free fall. Two important motion characteristics that are true of free-falling objects • Free-falling objects do not encounter air resistance. • All free-falling objects (on Earth) accelerate downwards at a constant rate. “g” - The “Magic” Number • “g” means ACCELERATION DUE TO GRAVITY • Each planet, star, moon, or other large object has its own value for “g” Examples • “g” is 1.62 m/s2 on the Moon • “g” is 26 m/s2 on Jupiter • “g” is 9.81 m/s2 on Earth If the feather and elephant experiment were performed on the moon, would they still fall at the same rate, just like on Earth? a = -g = -9.81m/s2 for all free fall objects. (the ‘-’ sign means “downward”) • • • • Dropped from rest Throwing downward Throwing upward Throwing side ways • Since free fall motion has constant acceleration, we can apply all kinematics equations on free fall motion. Free fall – motion with constant acceleration a = -g To solve free fall motion problems, we can use kinematics equations with constant acceleration. a = -g ∆x = ∆y Velocity of a dropped object • An object falls from rest. What is its velocity at the end of one second? Two = -9.81seconds? m/s2 (‘-’ means “downward”) v f viseconds? at a = -gThree t=0 vi = 0 m/s t=1s v = -9.81 m/s t=2s v = -19.62 m/s t=3s v = -29.43 m/s Displacement of a dropped object • An object falls from rest. How far has it fallen at the end of one1 second? Two d seconds? vi t at 2 seconds? Three 2 t=0 d=0m t=1s d = -4.905 m t=2s d = -19.62 m t=3s d = -44.145 m Velocity and distance of a free falling object dropped from rest + Throwing Downward av=i =-9.81 -10 m/s 2 • An object is thrown downward from the top of a 175 meter building with an initial speed of 10 m/s. – What acceleration does it experience? • -9.81 m/s2 – What is the object’s initial velocity? • -10 m/s “a” and “vi” in SAME DIRECTION Building Example (cont.) • What is the object’s velocity as it hits the ground? • • • • vi = -10 m/s d = -175 m a = -9.81 m/s2 vf = ? vf2 = vi2 + 2ad vf2 = 0 + 2(-9.81 m/s2)(-175 m) vf = -58.6 m/s • How long does it take the object to hit the vf = vi + at ground? -58.6 m/s = -10 m/s + (-9.81 m/s2) t • t=? t = 4.95 s Throwing Upward t=2s v = +30.38 m/s t=1s v = +40.19 m/s t=0s vi = +50 m/s v f vi at • A cannon fires a shot directly upward with an initial velocity of 50 m/s. – What acceleration does the cannonball experience? • -9.81 m/s2 – What is the cannonball’s initial velocity? • +50 m/s “a” and “vi” in OPPOSITE DIRECTION Cannon (cont.) • What is the object’s velocity as it reaches the top of its flight? 0 m/s (All objects momentarily STOP at the top of their flight) • How long does it take the cannonball to reach the top of its flight? vf = vi + at 0 = +50 m/s + (-9.81 m/s2) t t = 5.1 s • What is the maximum height of the cannonball? d = v t + ½ at2 i d = (+50 m/s)(5.1 s) + ½ (-9.81 m/s2)(5.1 s)2 d = +127.4 m Question • If a ball thrown into the air, and caught at the same point you it was released. Is the speed of the ball equal to, more than, or less than the initial speed when it is caught? Acceleration is the rate at which an object changes its velocity. • To accelerate at -10 m/s/s means to change the velocity by -10 m/s each second. • If the velocity and time for a free-falling object being tossed upward from a position with speed of 20 m/s were tabulated, then one would note the pattern. Time (s) Velocity (m/s) 0 20 1 10 2 0 3 -10 4 -20 Question • If a ball thrown into the air, and caught at the same point you it was released. Is the speed of the ball equal to, more than, or less than the initial speed when it is caught? given: vi, a = -9.81 m/s2; ∆y = 0 m; Find: vf vf2 = vi2 + 2a∆y vf2 = vi2 + 0 vi vf vf2 = vi2 The ball has the same velocity, but in the opposite direction Class work • A ball is thrown straight up into the air at an initial velocity of 39.24 m/s. Fill in the table showing the ball’s position, velocity, and acceleration each second for the first 4.00 s of its motion. t (s) Y (m) v (m/s) a (m/s2) 0 0 19.62 -9.81 1 2 3 4 Free Fall by Graphs of a dropped object A position versus time graph for a free-falling object velocity increase in the negative direction – slope is neg. & increasing A velocity versus time graph for a free-falling object velocity increases in the negative direction at constant rate – slope is neg. & constant Free Fall Graphs of upward object velocity Upward: velocity is big, positive, decreasing, slope is constant (a = -9.8 m/s/s). Top, velocity is zero. Slope remains the same (acceleration is still -9.8 m/s/s) position time Downward: velocity increases in negative direction at the same constant rate of -9.81 m/s/s (slope remains the same), reaches the same speed as it started upward. Upward: displacement increases, slope is positive, decreasing (velocity is positive, decreasing) Top: slope = 0 (its velocity is zero) time Downward: displacement decreases, its slope increases in negative direction (velocity is negative, increasing)