chapter 2 - If you'd like for your Physics professor to be...

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Unformatted text preview: If you'd like for your Physics professor to be in a good mood... - You might choose to vote for his daughter Aylia Colwell to be cast in the web series "Whatever" at: - https:/lwww.facebook.com/questions! 1 79330425473271! 0 AND at: - http://www.whatevertheseries.com/nola.php 0 You might even choose to share this with your friends. 0 Voting ends this weekend. Just saying... August 25 Finish off chapter 1- vector components and some more examples. Start chapter 2. A demonstration. First clicker questions Reminders: — Read the syllabus (available online) and sign and return your clicker agreements. - Register your iCIickers when you can (you still get points. even if you have not yet registered your clicker). — Homework 1 is available on WebAssign. due one week from tonight. 1.6 VectorAddmon and Subtraction Finishing up vectOrs... When a morons hummer! by ~1. the magnitude olthe . vector remems me same. but the dream of the vectons reversed. Subtracting a vector is just adding the negative of that vector. The negative of a vector is just the vector pointing in the opposite direction. 1.7 The Components of e Vector Finish Start i and;l are the x and y components of the vectori‘. They are 'components' because they are perpendicular. There is one component for each dimension of the problem. 1.7 The Components of e Vector It IS often easier to work with the scalar components rather than the vector components. A. and A, are the scalar components of the vector A. We use 'unit vectors" in the x and y directions with length 1 so that A: A! +A,y The 'hat“ denotes unit vectors. 1.7 The Components of e Vector Vector and Vector Component Summary - Any vector can be expressed as the sum of two perpendicular vectors, called the components of that vector. 0 For two-dimensional problems. we can have those components be along the x and y axes of our coordinate system. 0 This reduces the vector to a right triangle, which we can handle mathematically with our trigonometry rules and the Pythagorean theorem. a. 1.7 The Components on Vector Example A displacement vect0r has a magnitude of 175 m and pomts at an angle of 50.0 degrees relative to the x axis. Find the x and y components of this vector. sine - Jgfr y - rsinB - (175 mxsin 50.0" )- 134 m ~‘ c058 - .1',.-"r x - rcosB - (I75 mXcos 50.0“)- I l2 m i" = (112m)i + (134m)? 1.8 Addmon of Vector: by Mans 01 Components 0 It 1.8 Addmon of Vector: by Mans 01 Components 0 .1 ,‘ C: Ali+Aj+Bli+Bj (3:91, +B,)i+(A, +B:,)y C, - .4‘ + B, (1 - .4, + B, Chapter 2 Kinematics in One Dimension Kinematics deals with the concepts that are needed to describe motion. Dynamics deals with the effect that forces have on motion. Together, kinematics and dynamics form the branch of physics known as Mechanics. 2.1 Displacement Displacement is a vector mth Origin dimensions of length. in, l {’0 Displacement *- Ai fio_—> i — i0 = initial position = final position i A}? = i — i0 = displacement 2.1 Displacement 2.1 Displacement io=—2.0mi 52:50:11}? # Ai=7.0mi Ai=i—i0 =5.0mi—(—2.0)mi=7.0mi 2.2 Speed and Velocity Average speed is the distance traveled divided by the time required to cover the distance. ‘ Distance Average speed - ‘—, Elapsed time SI units for speed: meters per second (m/s) These quantities are all scalars. 2.2 Speed and Velocity Example 1 Distance Run by a Jogger How far does a jogger run in 1.5 hours (5400 s) it his average speed is 2.22 mls? Average 5 I Distance Elapsed time Distance - (Average speedXElapsed time) x (2.22 m/s)(5400 s)- 12000 m 2.2 Speed and Velocity Average velocity is the displacement dwided by the elapsed time. Because displacement is a vector. average velocity is also a vector. Displacement Average velocity - . Elapsed time x—x0 Aii v=—=— (-10 At 2.2 Speed and Velocity Example 2 The World‘s Feeteet Jet-Engine Cu Andy Green in the car rhmsISSC set a world record 01 341.1mls m1997. To establish such a record. the diver nukes two am through the course. one in each direction. to nullify m effects From the date. determine the average veloaty lor each run. 2.2 Speed and Velocity q=£=fl=+3395mxs - A: 4.7405 2.2 Speed and Velocity The Instantaneous velocity indicates how fast the car moves and the direction of motion at each instant of time. 2.3 AMMUOII The notion of moderation emerges when a change in velocity us combmed with the time during which the change occurs. k} 91 >7 1'41 2.3 Acceleration Example 3 Aoceleranon and Increaan Vetooty Determne the average acceleration of the plane. v.40mls=0km-‘h G=2wknuh rJ-Os r-29s i=v— 2 ___ 260km/h—0km/h=+9.0kmm t—tz. 293—05 5 Nottoemcmhwdwrmmitsmtuve. 2.3 Acceleration Mecca now the Games wavelet: (1an the second second :5 large' 31th (1.513009 W aumg the first sound Th5 Is cue no a hgner average weed curing the seams second that's no 549ch time I sad second ssoom on w; shoe 2.3 Acceleration ExmvphJAczmzbnandDeauashqa§ &.r ’1- “7—...— 2.4 Eounuom or Kinematics fer Commit Acceleration .. _. AV v=_—_ a=—='— It is customary to dispense with the use of boldface symbols overdrawn with arrows for the displacement. velocity. and acceleration vectors. We will. however. continue to convey the directions with a plus 0r minus sign. 2.4 Eounuom or Kinematics fer Commit Acceleration Five kinematic variables: 1. displacement. x 2. acceleration (constant). a 3. final velocity (at time t). v 4. initial velocity. v0 5. elapsed time. t Our goal is to find general expressions that connect these variables so that we can describe motion as a function of time. 2.4 Equauons o! Khomatlcs tor Constant Accolwatton Let the object be at the origin when the clock starts. x =0 (“=0 U 2.4 Equauons o! Khomatlcs tor Constant Accolwatton 2.4 Equauons o! Khomatlcs tor Constant Accolwatton v = v” + a! x = g-(vn + v)! = “v” + v” + at )1 x - vnt + %at‘ 2.4 69000003 0! Kmamotla lb! Canton: Aocdontlon 1,-0‘ _._;_.‘ :nl'll .’ l ‘. t". K/ l-d’Oms'. c:,-9¢Drv,l I ‘6 x = vat + -2Lal2 - (6.0 m/stm s)+ § (2.0 m/s2 18.0 s)’ =+110m 2.4 690.000: of Kmanufla lb! Constant Accounaon Example 6 Catapulting a Jet Fund its displacement. v0 - Omxs a - +3 I mfs2 .r - ?? v - +62 mfs 2.‘ Equallqu o! Khmer!“ tar Cerium Acceleration v2 - v2 (62 mis), - (0 s)2 0 x=-———= V 5 20 26]m;"s‘ ) 2.‘ Equallqu o! Khmer!“ tar Cerium Acceleration Equations of Kinematk: for Constant Acceleration v = v" + at x = av" + v)! v2 - v: + Zax _ I 2 x- vut+3at 2.5 Applicant»: 0! the Equarlom o! Kinematics Reasoning Strategy 1. Make a drawing. 2. Decide which directions are to be caled positive (+) and negative H. 3. Write down the values that are given for any of the five kinematic variables. 4. Verify that the information contains values for at least three 0! the five kinematic variables. Select the appropriate equation. 5. When the motion is divided into segments. remember that the final velocity of one segment is the iniial velocity for the next. 6. Keep in mind that there may be two possible answers to a kinematim problem. 2.5 Applications of the Equations 0! Kinematics Example 8 An Accelerating Spacecraft A spacecraft is traveling with a velocity of 93250 mls. Suddenty the retrorockets are fifed. and the spacecraft begins to slow down with an acceleration whose magnitude is 10.0 rn/s’. What is the velocity of the spacecraft when the displacement of the cralt is +215 km. relative to the point where the retmrockets began firing? +215000m -1o.o m/s? +3250 m/s I 2.5 Applications of the Equations 0! Kinematics 2.5 Applications 01th. Equations 0! Kinematics -l +21sooo m 40.0 rnls’ +3250 this I v- - + 2ax = v = JV: + 2:.- v - 4 3250mm + 2110.0m/s2 izisooo m) - 12500 m/s umme Intheabsencaotairresistance. itislomdthatalbodies at the same location above the Earth tall verticaly with the same acceleration. If the distance of the fall is smdl compared to the tadius at the Earth. then the acceleration remains essentially constant throughout the descent. This idealized motion is caled free-fat! and the acceleration of a freely fall'ng body is caled the axeleration due to gravity. g=9.80m/s2 or 32.2fi/s2 umme Example 10 A Fall'ng Stone A stone is dropped from the top ota tfl building. Alter 3.00s of free fall. what is the displacement yet the stone? 300$ 1-0-0 MI I \ I ,. .. 2.6 may Falling Baum ." = V"! + #1:: = (Om/5X3.00-‘1)+#4180111“2 13.00 ' -44.1 m 2.6 may Falling Baum Example 12 Haw High Does it Go? The referee tosses the coin up with an initial speed of 5.00rw's In the absence if air resisiance. how high does the coin go above its point of release? 2.6 may Falling Baum "II ? -9.80 m/s’ 0 mls +5.00 mls 261%.:er 800“! "ll ? -9.80 m/s? 0 mls +5.00 rnls ‘! 1 2 2 ‘)- — ‘7‘ v - V” + 20y = ;:«- V - _" ' 20 a 1 v‘ -vo (0m -(5.00m I 7 ‘ -8 m 20 2(— 9.80m s‘ j 261%.:er 8m Conceptual Example 14 Acceleration Versus Velocity There are three parts to the motion of the coin. On the way up. the coin has a vector velocity that is directed upward and has decreasing magnitude. At the top of its path. the coin momentarily has zero velocity. On the way down. the coin has dommard-pohting velocity with an ‘ncreasing magnitude. In the absence 0! air resistance. does the acceleration of the coin. like the velocity. change from one part to another? 2.7 Graphical Analysis of Why and Acceleration +16 9 ._. N Position .1 (m) L & 1"=25 17WAMW NWWAWM 17WAMW NWWAWM Tangent line \ on .0 o Ax: +26 m~\ m P o E )1 C 3400 m 0 CL N .° 0 O 0 5.0 10.0 15.0 20.0 25.0 Time I (s) 17WAMW NWWAWM +36 4' N b Velocity v (tn/S) I. N ...
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This note was uploaded on 09/21/2011 for the course PHY 1500 taught by Professor Staff during the Spring '09 term at University of Central Florida.

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chapter 2 - If you'd like for your Physics professor to be...

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