PHYS 1441 – Section 002 Lecture #6Slide 22D Coordinate SystemsComponents and Unit VectorsUnit VectorsExamples of Vector Operations2D Displacement2D Average VelocitySlide 92D Average AccelerationDisplacement, Velocity, and Acceleration in 2-dimKinematic Quantities in 1D and 2DA Motion in 2 DimensionMotion in horizontal direction (x)Motion in vertical direction (y)Slide 16Kinematic Equations in 2-DimEx. A Moving SpacecraftHow do we solve this problem?Ex. continuedFirst, the motion in x-direciton…Now, the motion in y-direction…The final velocity…If we visualize the motion…What is a Projectile Motion?Monday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu1PHYS 1441 – Section 002Lecture #6Monday, Sept. 27, 2010Dr. Jaehoon Yu•Components of the 2D Vector•Understanding the 2 Dimensional Motion•2D Kinematic Equations of Motion•Projectile Motion•Maximum Range and HeightToday’s homework is homework #4, due 10pm, Tuesday, Oc. 5!!Monday, Sept. 20, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu32D Coordinate Systems•They make it easy and consistent to express locations or positions•Two commonly used systems, depending on convenience, are–Cartesian (Rectangular) Coordinate System•Coordinates are expressed in (x,y)–Polar Coordinate System •Coordinates are expressed in distance from the origin ® and the angle measured from the x-axis, θ(rθ)•Vectors become a lot easier to express and computeO (0,0)(x1,y1)r1θ11x =How are Cartesian and Polar coordinates related?1r =y1x1+x+y1y = tanθ1= x12+y12( )1y r1 r1=(r1θ) 1x θ1=tan−1y1x1⎛⎝⎜⎞⎠⎟ cosθ1 sinθ1Monday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu4Components and Unit VectorsCoordinate systems are useful in expressing vectors in their components(Ax,Ay)AAθAyAxxyxA = Aur=}Components(+,+)(-,+)(-,-) (+,-)yA =} Magnitude Aurcosθ( )2+ Aursinθ( )2 = Aur2cos2θ +sin2θ( ) = Aur Aurcosθ Aursinθ Aur= Aur= Ax2+Ay2Monday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu5Unit Vectors•Unit vectors are the ones that tells us the directions of the components•Dimensionless •Magnitudes these vectors are exactly 1•Unit vectors are usually expressed in i, j, k or ir, jr, kr Aur=So a vector AA can be expressed as yA = Aurcosθ + Aursinθ xA +irirjrjrMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu6Examples of Vector OperationsFind the resultant vector which is the sum of AA=(2.0ii+2.0jj) and B B =(2.0ii-4.0jj) Cur=Find the resultant displacement of three consecutive displacements: dd11=(15ii+30j j +12kk)cm, dd22=(23ii+14j j -5.0kk)cm, and dd33=(-13ii+15jj)cm Dur= Cur= Dur=θ = 2.0ir+2.0jr( )+ 2.0ir−4.0jr( ) = 2.0+2.0( )ir tan−1CyCx= tan−1−2.04.0=−27o = 15+23−13( )ir25i=r dur1+dur2+dur3= 15ir+30jr+12kr( )+ 23ir+14jr−5.0kr( )+ −13ir+15jr( ) 25( )2+ 59( )2+ 7.0( )2= 65(cm) 4.0( )2+ −2.0( )2 = 16+4.0 = 20 =4.5(m) + 2.0−4.0( )jr =4.0ir −2.0jrm( ) Aur+Bur= + 30+14+15( )jr +12−5.0( )kr +59jr +7.0kr(cm)MagnitudeMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu7 initial positiono=rr final position=rr Displacement: 2D Displacement Δrr = rr −rroMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu8 rv =Average velocity is the displacement divided by the elapsed time. rr −rro = t −to ΔrrΔt2D Average VelocityMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu9The instantaneous velocity indicates how fast the car moves and the direction of motion at each instant of time. rv = limΔt→ 0ΔrrΔtMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu10 ra= rvo rv Δrv2D Average Acceleration rv−rvo = t −to ΔrvΔtMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu11Displacement, Velocity, and Acceleration in 2-dim•Displacement: Δrr=•Average Velocity: vr≡•Instantaneous Velocity: vr≡•Average Acceleration ar≡•Instantaneous Acceleration: ar≡How is each of these quantities defined in 1-D? rrf−rri ΔrrΔt= rrf−rritf−ti limΔt→ 0ΔrrΔt ΔvrΔt= vrf−vritf−ti limΔt→ 0ΔvrΔtMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu12Kinematic Quantities in 1D and 2DQuantities 1 Dimension 2 DimensionDisplacementAverage VelocityInst. VelocityAverage Acc.Inst. Acc. Δx= Δrr=vx=ΔxΔt=xf−xitf−ti vr≡ΔrrΔt=rrf−rritf−ti vr≡limΔt→ 0ΔrrΔt vx≡limΔt→ 0ΔxΔt ar≡ΔvrΔt=vrf−vritf−tiax≡ΔvxΔt=vxf−vxitf−ti ar≡limΔt→ 0ΔvrΔt ax≡limΔt→ 0ΔvxΔtWhat is the difference between 1D and 2D quantities? xf−xi rrf−rriMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu13This is a motion that could be viewed as two motions combined into one. (superposition…)A Motion in 2 DimensionMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu14Motion in horizontal direction (x) x =12vxo+vx( )t x =vxot +12axt2 vx=vxo+axt vx2=vxo2+2axxMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu15Motion in vertical direction (y) y =12vyo+vy( )t y =vyot +12ayt2 vy=vyo+ayt vy2=vyo2+2ayyMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu16Imagine you add the two 1 dimensional motions on the left. It would make up a one 2 dimensional motion on the right.A Motion in 2 DimensionMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu17x-component x =12vxo+vx( )t Δx=vxot +12axt2 vx=vxo+axt vx2=vxo2+2axx y =12vyo+vy( )t Δy =vyot +12ayt2 vy=vyo+ayt vy2=vyo2+2ayyKinematic Equations in 2-Dimy-componentMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu18In the x direction, the spacecraft in zero-gravity zone has an initial velocity component of +22 m/s and an acceleration of +24 m/s2. In the y direction, the analogous quantities are +14 m/s and an acceleration of +12 m/s2. Find (a) x and vx, (b) y and vy, and (c) the final velocity of the spacecraft at time 7.0 s.Ex. A Moving SpacecraftMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu19How do we solve this problem?1. Visualize the problem Draw a picture!2. Decide which directions are to be called positive (+) and negative (-).3. Write down the values that are given for any of the five kinematic variables associated with each direction.4. Verify that the information contains values for at least three of the kinematic variables. Do this for x and y separately. Select the appropriate equation.5. When the motion is
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