Monday, Feb. 9, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu1PHYS 1441 – Section 004Lecture #6Monday, Feb. 9, 2004Dr. Jaehoon Yu• Chapter three: Motion in two dimension– Vector and Scalar– Properties of vectors– Vector operations– Components and unit vectors– 2D Kinematic Equations– Projectile MotionMonday, Feb. 9, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu2Announcements• There will be a quiz this Wednesday, Feb. 11– Will cover•Sections A5 –A9• Chapter 2• Everyone is registered to homework system: Good job!!!• E-mail distribution list (phys1441-004-spring04)– 48 of you subscribed as of 11am this morning– Test message was sent last night• Please reply to ME for verificationMonday, Feb. 9, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu3Kinematic Equations of Motion on a Straight Line Under Constant Acceleration()tavtv xxixf +=221tatvxxxxiif += +()tvvtvxx xixfixf+==−2121()ifxf xxavv xxi−+= 222Velocity as a function of timeDisplacement as a function of velocities and timeDisplacement as a function of time, velocity, and accelerationVelocity as a function of Displacement and accelerationYou may use different forms of Kinetic equations, depending on the information given to you for specific physical problems!!Monday, Feb. 9, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu4Vector and ScalarVector quantities have both magnitude (size) and directionScalar quantities have magnitude onlyCan be completely specified with a value and its unitForce, gravitational pull, momentumNormally denoted in BOLD BOLD letters, FF, or a letter with arrow on topFTheir sizes or magnitudes are denoted with normal letters, F, or absolute values:For FEnergy, heat, mass, weightNormally denoted in normal letters, EBoth have units!!!Monday, Feb. 9, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu5Properties of Vectors• Two vectors are the same if their and the are the same, no matter where they are on a coordinate system.xyAABBEEDDCCFFWhich ones are the same vectors?A=B=E=DA=B=E=DWhy aren’t the others?C:C: The same magnitude but opposite direction: C=C=--A:A:A negative vectorF:F: The same direction but different magnitude sizes directionsMonday, Feb. 9, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu6Vector Operations• Addition: – Triangular Method: One can add vectors by connecting the head of one vector to the tail of the other (head-to-tail)– Parallelogram method: Connect the tails of the two vectors and extend– Addition is commutative: Changing order of operation does not affect the results A+B=B+AA+B=B+A, A+B+C+D+E=E+C+A+B+DA+B+C+D+E=E+C+A+B+DAABBAABB=AABBA+BA+B• Subtraction: – The same as adding a negative vector:A A -B = A B = A + (-BB)AA--BBSince subtraction is the equivalent to adding a negative vector, subtraction is also commutative!!!• Multiplication by a scalar is increasing the magnitude A, BA, B=2A A AAB=2AB=2AAB 2=A+BA+BA+BA+BAA--BBORMonday, Feb. 9, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu7Example of Vector AdditionA car travels 20.0km due north followed by 35.0km in a direction 60.0owest of north. Find the magnitude and direction of resultant displacement.NE60οθr()()22sincosθθBBAr ++=20AABB60cos60sintan1BAB+=−θFind other ways to solve this problem…()θθθcos2sincos2222ABBA +++=θcos222ABBA ++=()()60cos0.350.2020.350.2022××++=)(2.482325 km==60cos0.350.2060sin0.35tan1+=−N W wrt to9.385.373.30tan1 ο==−BcosθBsinθMonday, Feb. 9, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu8Components and Unit VectorsCoordinate systems are useful in expressing vectors in their components22yxAAA +=(Ax,Ay)AAθAyAxxyθcosAAx=A=JG}Components(+,+)(-,+)(-,-) (+,-)• Unit vectors are dimensionless vectors whose magnitude are exactly 1• Unit vectors are usually expressed in i, j, k or • Vectors can be expressed using components and unit vectorskji , ,A=JGSo the above vector AA can be written asθsinAAy=}Magnitude()()22cos sinAAθθ+JG JG()222cos sinAθθ=+JGA=JGxyAiAj+=GGcos sinAiAjθθ+JGGJGGMonday, Feb. 9, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu9Examples of Vector OperationsFind the resultant vector which is the sum of AA=(2.0ii+2.0jj) and B B =(2.0ii-4.0jj)C =JGFind the resultant displacement of three consecutive displacements: dd11=(15ii+30j j +12kk)cm, dd22=(23ii+14j j -5.0kk)cm, and dd33=(-13ii+15jj)cmD=JGC =JGD =JGθ=()()2.0 2.0 2.0 4.0ij ij++−GGGG()2.0 2.0 i=+G1tanyxCC−=12.0tan 274.0−−=−D()15 23 13 i=+−G25i=G123ddd++=JGJGJG()()()15 30 12 23 14 5.0 13 15ijk ijk ij++ ++− +−+GGG GG G GG() () ( )22 225 59 7.0++=65( )cm()( )224.0 2.0+−16 4.0 20 4.5( )m=+= =()2.0 4.0 j+−G4.0i=G()2.0jm−GAB+=JGJG()30 14 15 j+++G()12 5.0 k+−G59 j+G7.0 ( )kcm+GMagnitudeMonday, Feb. 9, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu10Displacement, Velocity, and Acceleration in 2-dim• Displacement:ifrrr −=∆• Average Velocity:ififttrrtrv−−=∆∆≡• Instantaneous Velocity:dtrdtrvt=∆∆≡→∆ 0lim• Average Accelerationififttvvtva−−=∆∆≡• Instantaneous Acceleration:220limdtrddtrddtddtvdtvat=⎟⎟⎠⎞⎜⎜⎝⎛==∆∆≡→∆How is each of these quantities defined in 1-D?Monday, Feb. 9, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu112-dim Motion Under Constant Acceleration• Position vectors in x-y plane:ir=Gfr=G• Velocity vectors in x-y plane:iv=Gfv=Gxfv=• How are the position vectors written in acceleration vectors?fr =Gfx=fv=Gyfv=fy=()iixi y j=+GGVelocity vectors in terms of acceleration vectoriixiyj+GGffxi y j+GGxi yivi v j+GGxf yfvi v j+GGxixvat+yi yvat+()()xi x yi yvativatj+++ =GGivat+GG212ixi xxvt at++212iyi yyvt at++fxiGfyj+=G212ixi xxvt ati⎛⎞++⎜⎟⎝⎠G212iyi yyvt atj⎛⎞+++⎜⎟⎝⎠G()xi yivi v j t++GG()212xyai a j t++GGir=JGivt+G212at+GMonday, Feb. 9, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu12Example for 2-D Kinematic EquationsA particle starts at origin when t=0 with an initial velocity vv=(20ii-15jj)m/s. The particle moves in the xy plane with ax=4.0m/s2. Determine the components of velocity vector at any time, t.xfvCompute the velocity and speed of the particle at t=5.0 s.5tv=G()()22xyspeed v v v== +G()vtGyfvxiv=xat+20=()4.0 /tms+yiv=yat+15=− 0t+()15 /ms=−Velocity vector()xvti=G()yvtj+G()20 4.0ti=+G15 ( / )jm s−G,5xtvi==G,5ytvj=+G()20 4.0 5.0 i=+×G15j−G()40 15 /ijms=−GG() ( )2240 15 43 /ms=+−=Monday, Feb. 9, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu13Example for 2-D Kinematic Eq. Cnt’dθ=Determine the x and y components
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