Go over vector and vector algebra Displacement and position in 2-D Average and instantaneous velocity in 2-D Average and instantaneous acceleration

Similar documents
Chapter 4: Motion in Two Dimensions Part-1

Physics 120 Spring 2007 Exam #1 April 20, Name

Physics 207, Lecture 3

PHY2053 Summer C 2013 Exam 1 Solutions

Physics 101 Lecture 4 Motion in 2D and 3D

Chapter 3: Vectors and Two-Dimensional Motion

( ) ( ) ( ) ( ) ( ) ( ) j ( ) A. b) Theorem

Physics 201 Lecture 2

Science Advertisement Intergovernmental Panel on Climate Change: The Physical Science Basis 2/3/2007 Physics 253

Rotations.

Physics 15 Second Hour Exam

Ch.4 Motion in 2D. Ch.4 Motion in 2D

Physics 201, Lecture 5

Uniform Circular Motion

Chapter 6 Plane Motion of Rigid Bodies

PHYS 1443 Section 001 Lecture #4

Axis. Axis. Axis. Solid cylinder (or disk) about. Hoop about. Annular cylinder (or ring) about central axis. central axis.

Chapter 2 Linear Mo on

Use 10 m/s 2 for the acceleration due to gravity.

Calculus 241, section 12.2 Limits/Continuity & 12.3 Derivatives/Integrals notes by Tim Pilachowski r r r =, with a domain of real ( )

5-1. We apply Newton s second law (specifically, Eq. 5-2). F = ma = ma sin 20.0 = 1.0 kg 2.00 m/s sin 20.0 = 0.684N. ( ) ( )

ME 141. Engineering Mechanics

Motion in Two Dimensions

Physics 201 Lecture 3. dx v x. l What was the average velocity? l Two legs with constant velocity but. l Average velocity:

Motion on a Curve and Curvature

L4:4. motion from the accelerometer. to recover the simple flutter. Later, we will work out how. readings L4:3

() t. () t r () t or v. ( t) () () ( ) = ( ) or ( ) () () () t or dv () () Section 10.4 Motion in Space: Velocity and Acceleration

EGN 3321 Final Exam Review Spring 2017

Chapter I Vector Analysis

LECTURE 5. is defined by the position vectors r, 1. and. The displacement vector (from P 1 to P 2 ) is defined through r and 1.

Class Summary. be functions and f( D) , we define the composition of f with g, denoted g f by

_ J.. C C A 551NED. - n R ' ' t i :. t ; . b c c : : I I .., I AS IEC. r '2 5? 9

Faraday s Law. To be able to find. motional emf transformer and motional emf. Motional emf

2D Motion WS. A horizontally launched projectile s initial vertical velocity is zero. Solve the following problems with this information.

Motion. ( (3 dim) ( (1 dim) dt. Equations of Motion (Constant Acceleration) Newton s Law and Weight. Magnitude of the Frictional Force

1.B Appendix to Chapter 1

CHAPTER 10: LINEAR DISCRIMINATION

Answers to test yourself questions

Field due to a collection of N discrete point charges: r is in the direction from

s = rθ Chapter 10: Rotation 10.1: What is physics?

Chapter Fifiteen. Surfaces Revisited

3 Motion with constant acceleration: Linear and projectile motion

Physics 111. Uniform circular motion. Ch 6. v = constant. v constant. Wednesday, 8-9 pm in NSC 128/119 Sunday, 6:30-8 pm in CCLIR 468

1. Kinematics of Particles

Name of the Student:

Physic 231 Lecture 4. Mi it ftd l t. Main points of today s lecture: Example: addition of velocities Trajectories of objects in 2 = =

A 1.3 m 2.5 m 2.8 m. x = m m = 8400 m. y = 4900 m 3200 m = 1700 m

Review: Transformations. Transformations - Viewing. Transformations - Modeling. world CAMERA OBJECT WORLD CSE 681 CSE 681 CSE 681 CSE 681

(b) 10 yr. (b) 13 m. 1.6 m s, m s m s (c) 13.1 s. 32. (a) 20.0 s (b) No, the minimum distance to stop = 1.00 km. 1.

f(x) dx with An integral having either an infinite limit of integration or an unbounded integrand is called improper. Here are two examples dx x x 2

Invert and multiply. Fractions express a ratio of two quantities. For example, the fraction

Circuits 24/08/2010. Question. Question. Practice Questions QV CV. Review Formula s RC R R R V IR ... Charging P IV I R ... E Pt.

EECE 260 Electrical Circuits Prof. Mark Fowler

Lecture 5. Plane Wave Reflection and Transmission

Average & instantaneous velocity and acceleration Motion with constant acceleration

MODEL SOLUTIONS TO IIT JEE ADVANCED 2014

PHYSICS 1210 Exam 1 University of Wyoming 14 February points

PHUN. Phy 521 2/10/2011. What is physics. Kinematics. Physics is. Section 2 1: Picturing Motion

Electric Potential. and Equipotentials

Motion in a Straight Line

Physics 110. Spring Exam #1. April 23, 2008

Mechanics Physics 151

( ) ( ) Physics 111. Lecture 13 (Walker: Ch ) Connected Objects Circular Motion Centripetal Acceleration Centripetal Force Sept.

1. Viscosities: μ = ρν. 2. Newton s viscosity law: 3. Infinitesimal surface force df. 4. Moment about the point o, dm

1. Consider a PSA initially at rest in the beginning of the left-hand end of a long ISS corridor. Assume xo = 0 on the left end of the ISS corridor.

when t = 2 s. Sketch the path for the first 2 seconds of motion and show the velocity and acceleration vectors for t = 2 s.(2/63)

One of the common descriptions of curvilinear motion uses path variables, which are measurements made along the tangent t and normal n to the path of

Addition & Subtraction of Polynomials

Maximum likelihood estimate of phylogeny. BIOL 495S/ CS 490B/ MATH 490B/ STAT 490B Introduction to Bioinformatics April 24, 2002

COMP 465: Data Mining More on PageRank

Homework 3 MAE 118C Problems 2, 5, 7, 10, 14, 15, 18, 23, 30, 31 from Chapter 5, Lamarsh & Baratta. The flux for a point source is:

Physics 1502: Lecture 2 Today s Agenda

I-POLYA PROCESS AND APPLICATIONS Leda D. Minkova

Rotational Kinematics. Rigid Object about a Fixed Axis Western HS AP Physics 1

Section 35 SHM and Circular Motion

1. The sphere P travels in a straight line with speed

September 20 Homework Solutions

HERMITE SERIES SOLUTIONS OF LINEAR FREDHOLM INTEGRAL EQUATIONS

Lecture 5 Single factor design and analysis

Physics 11b Lecture #2. Electric Field Electric Flux Gauss s Law

D zone schemes

Course Outline. 1. MATLAB tutorial 2. Motion of systems that can be idealized as particles

On Fractional Operational Calculus pertaining to the product of H- functions

ESS 265 Spring Quarter 2005 Kinetic Simulations

Today - Lecture 13. Today s lecture continue with rotations, torque, Note that chapters 11, 12, 13 all involve rotations

U>, and is negative. Electric Potential Energy

Lecture 3. Electrostatics

PHYS 2421 Fields and Waves

2-d Motion: Constant Acceleration

Motion in One Dimension

Chapter 4 Kinematics in Two Dimensions

Reinforcement learning

Mark Scheme (Results) January 2008

β A Constant-G m Biasing

Lecture 9-3/8/10-14 Spatial Description and Transformation

Technical Appendix for Inventory Management for an Assembly System with Product or Component Returns, DeCroix and Zipkin, Management Science 2005.

DEPARTMENT OF CIVIL AND ENVIRONMENTAL ENGINEERING FLUID MECHANICS III Solutions to Problem Sheet 3

Chapter 2. Motion along a straight line. 9/9/2015 Physics 218

Notes on the stability of dynamic systems and the use of Eigen Values.

Physics 105 Exam 2 10/31/2008 Name A

Transcription:

Mh Csquee

Go oe eco nd eco lgeb Dsplcemen nd poson n -D Aege nd nsnneous eloc n -D Aege nd nsnneous cceleon n -D Poecle moon Unfom ccle moon Rele eloc*

The componens e he legs of he gh ngle whose hpoenuse s A A A A n A A A nd Acos ) Asn ) A A A A n A A A A A o n A A O,

Knemc bles n one dmenson Poson: ) m Veloc: ) m/s Acceleon: ) m/s Knemc bles n hee dmensons ) zk Poson: ) zk Veloc: Acceleon: ) zk m/s All e ecos: he decon nd mgnudes m m/s z k

In one dmenson ) ) ) = - 3. m, ) = +. m Δ = +. m + 3. m = +4. m In wo dmensons Poson: he poson of n obec s descbed b s poson eco ) lws pons o pcle fom ogn. Dsplcemen: ) ) ) )

Aege eloc g Insnneous eloc lm d d lm g d d d d g g, d d g, s ngen o he ph n - gph;

Emple : Moon of Tule A ule ss he ogn nd moes wh he speed of = cm/s n he decon of 5 o he hozonl. ) Fnd he coodnes of ule seconds le. b) How f dd he ule wlk n seconds?

Emple : The egle descend An egle peched on ee lmb 9.5 m boe he we spos fsh swmmng ne he sufce. The egle pushes off fom he bnch nd descends owd he we. B dusng s bod n flgh, he egle mnns consn speed of 3. m/s n ngle of. deg below he hozonl. ) How long does ke fo he egle o ech he we? b) how f hs he egle eled n he hozonl decon when eches he we?

Emple : The egle descend An egle peched on ee lmb 9.5 m boe he we spos fsh swmmng ne he sufce. The egle pushes off fom he bnch nd descends owd he we. B dusng s bod n flgh, he egle mnns consn speed of 3. m/s n ngle of. deg below he hozonl. ) How long does ke fo he egle o ech he we? b) how f hs he egle eled n he hozonl decon when eches he we?

Aege cceleon g g, g g, Insnneous cceleon lm lm g d d d d d d d d The mgnude of he eloc he speed) cn chnge The decon of he eloc cn chnge, een hough he mgnude s consn Boh he mgnude nd he decon cn chnge

Poson Aege eloc Insnneous eloc Acceleon e no necessl sme decon. ) d d d d d d lm ) g g g,, d d d d d d lm ) d d d d d d d d d d d d ) nd ),, )

Moons n hee dmensons e ndependen componens Consn cceleon equons Consn cceleon equons hold n ech dmenson = begnnng of he pocess; whee nd e consn; Inl eloc nl dsplcemen ; ) )

Defne coodne ssem. Mke skech showng es, ogn. Ls known qunes. Fnd,,,, ec. Show nl condons on skech. Ls equons of moon o see whch ones o use. Tme s he sme fo nd decons. = = ), = = ), = = ), = = ). He n s pon long he decon of f s consn. ) )

Emple 3: A Humme Accelees A hummngbd s flng n such w h s nll mong ecll wh speed of 4.4 m/s nd cceleng hozonll m/s. Assumng he bd's cceleon emns consn fo he me nel of nees, fnd ) he hozonl nd ecl dsnces hough whch moes n.55 s nd b) s nd eloc componens he me =.55 s. c ) wh s he bd s eloc?

-D poblem nd defne coodne ssem: - hozonl, - ecl up +) T o pck =, = = Hozonl moon + Vecl moon Hozonl: =, consn eloc moon Vecl: = -g = -9.8 m/s, = Equons: g f ) ) Hozonl Vecl

X nd Y moons hppen ndependenl, so we cn e hem sepel Hozonl g g Vecl T o pck =, = = Hozonl moon + Vecl moon Hozonl: =, consn eloc moon Vecl: = -g = -9.8 m/s nd e conneced b me ) s pbol

-D poblem nd defne coodne ssem. Hozonl: = nd ecl: = -g. T o pck =, = =. Veloc nl condons: cn he, componens. s consn usull. cos chnges connuousl. sn Equons: Hozonl Vecl g g

Inl condons = ): =, = = cosθ nd = snθ Hozonl moon: Vecl moon: Pbol; θ = nd θ = 9? g g cos n g

Inl condons = ): =, = = cosθ nd = snθ, hen g h sn g g Hozonl Vecl R cos sn sn g g g g

Inl condons = ): =, = = cosθ nd = snθ, hen Hozonl Vecl g g g sn g g h h h g h sn g g g h g g

Complemen lues of he nl ngle esul n he sme nge The heghs wll be dffeen The mmum nge occus poecon ngle of 45 o R sn g

Poson Aege eloc Insnneous eloc Acceleon e no sme decon. ) d d d d d d lm ) g g g,, d d d d d d lm ) d d d d d d d d d d d d ) nd ),, )

If pcle moes wh consn cceleon, moon equons e Poecle moon s one pe of -D moon unde consn cceleon, whee =, = -g. f f f ) ) f f f ) ) ) f