1Chapter 4 – 2D and 3D MotionI. DefinitionsII. Projectile motionIII. Uniform circular motionIV. Relative motionPosition vector: extends from the origin of a coordinate system to the particle.)1.4(ˆˆˆkzjyixr ++=)2.4(ˆ)(ˆ)(ˆ)(12121212kzzjyyixxrrr −+−+−=−=∆I. DefinitionsAverage velocity:)3.4(ˆˆˆktzjtyitxtrvavg∆∆+∆∆+∆∆=∆∆=Displacement vector: represents a particle’s position change during a certaintime interval.2Instantaneous velocity:)4.4(ˆˆˆˆˆˆkdtdzjdtdyidtdxdtrdkvjvivvzyx++==++=-The direction of the instantaneous velocity of a particle is always tangent to the particle’s path at the particle’s positionInstantaneous acceleration:Average acceleration:)5.4(12tvtvvaavg∆∆=∆−=)6.4(ˆˆˆˆˆˆkdtdvjdtdvidtdvdtvdkajaiaazyxzyx++==++=II. Projectile motionMotion of a particle launched with initial velocity, v0and free fall accelerationg.- Horizontal motion: ax=0  vx=v0x= constant- Vertical motion: ay= -g = constantRange (R): horizontal distance traveled by a projectile before returning to launch height.)7.4()cos(0000tvtvxxxθ==−)8.4(21)sin(21200200gttvgttvyyy−=−=−θ)9.4(sin00gtvvy−=θ)10.4()(2)sin(02002yygvvy−−=θThe horizontal and vertical motions are independent from each other.3- Trajectory: projectile’s path.)11.4()cos(2)(tancos21cossincos)8.4()7.4(20020200000000θθθθθθvgxxyvxgvxvyvxt−=→−=→=→+000== yx- Horizontal range: R = x-x0; y-y0=0.)12.4(2sincossin2cos21tancos21cos)sin(21)sin(0cos)cos(020200002202020000002000000θθθθθθθθθθθgvvgRvRgRvRgvRvgttvvRttvR==→−=−=−==→=(Maximum for a launch angle of 45º )Overall assumption: the air through which the projectile moves has no effect on its motion  friction neglected.122: A third baseman wishes to throw to first base, 127 feet distant. His best throwing speed is 85 mi/h. (a) If he throws the ball horizontally 3 ft above the ground, how far from first base will it hit the ground? (b) From the same initial height, at what upward angle must the third baseman throw the ball if the first baseman is to catch it 3 ft above the ground? (c) What will be the time of flight in that case?( )mfootmfeetsmmimshhmi91.01305.03/38116093600185=⋅=⋅⋅xyv0h=3ftB3B1xmax0 xB1=38.7mVertical movementHorizontal movementsttmgttvyyy43.09.491.00212200=→−=−−=−34.16)43.0)(/38(380max00maxBfrommssmtxtvxxx===−=−The ball will hit ground at 22.3 m from B1xv0h=3ftB3B1y38.7m6.72sin5.013.0cossin144463.1899.4sin38cos387.381cos387.38cos7.389.4sin38sin9.42100000200=→=→=→===→===→==→−==−θθθθθθθθθθstvtmvtvtvgttvyyxyyθ4N7: In Galileo’s Two New Sciences, the author states that “for elevations (angles of projection) which exceed or fall short of 45º by equal amounts, the ranges are equal…” Prove this statement.δθθδθθθ−=+==45454521( )[ ]( )( )[ ]( )δθδθδθδθ290sin452sin'290sin452sin20202020−=−=+=+=gvgvRgvgvRxv0x=R=R’?yθ=45º02sin:max020=→= hatdgvRRangeθbabababababasincoscossin)sin(sincoscossin)sin(−=−+=+[ ][ ])2cos()2sin(90cos)2cos(90sin')2cos()2sin(90cos)2cos(90sin20202020δθδθδθδθδθδθgvgvRgvgvR=−==+=III. Uniform circular motionMotion around a circle at constant speed.)13.4(2rva =)14.4(2vrTπ=θθθφθθθθθθtancossintansincosˆsinˆcosˆˆˆˆˆˆˆ)cos(ˆ)sin(ˆˆ22222222===→=+=+=−+−=+−=+−==⋅+⋅−=+−=+=xyyxxyppppyxaaradiusalongdirectedarvrvaaajrvirvjvrvivrvjdtdxrvidtdyrvdtvdajrxviryvjvivjvivv- Period of revolution:- Acceleration: centripetalvyvx-Velocity: tangent to circle in the direction of motion.Magnitude of velocity and acceleration constant. Direction varies continuously.554. A cat rides a merry-go-round while turning with uniform circular motion. At time t1= 2s, the cat’s velocity is: v1= (3m/s)i+(4m/s)j, measured on an horizontal xy coordinate system. At time t2=5s its velocity is:v2= (-3m/s)i+(-4m/s)j. What are (a) the magnitude of the cat’s centripetal acceleration and (b) the cat’s average acceleration during the time interval t2-t1?v1v2xyIn 3s the velocity is reversed  the cat reaches the oppositeside of the circlemrsmrsvrTsmv77.4/532/54322=→=→==+=ππ2222/23.577.4/25smmsmrvac===22212/33.3ˆ)/67.2(ˆ)/2(3ˆ)/8(ˆ)/6(smajsmismsjsmismtvvaavgavg=−−=−−=∆−=IV. Relative motion1DParticle’s velocity depends on reference frame)15.4(BAPBPAvvv+=Frame moves at constant velocity)16.4()()()(PBPABAPBPAaavdtdvdtdvdtd=→+=0Observers on different frames of reference measure the same accelerationfor a moving particle if their relative velocity is constant.62005.00)(flightflightylaunchlandflightsxbglaunchlandgttvyytvvxxx−==−−=∆=−jvivvvjvivvivvysxgyxrelssˆˆ)(ˆˆˆ00000,0+−=+=−=75. A sled moves in the negative x direction at speed vswhile a ball of ice is shot from the sled with a velocity v0= v0xi+ v0yj relative to the sled. When the ball lands, its horizontal displacement ∆xbgrelative to the ground (from its launch position to its landing position) is measured. The figure gives ∆xbgas a function of vs. Assume it lands at approximately its launch height. What are the values of (a) v0xand (b) v0y?The ball’s displacement ∆xbsrelative to the sled can also be measured. Assume that the sled’s velocity is not changed when the ball is shot. What is ∆xbswhen vsis (c) 5m/s and (d) 15m/s?Launch velocity relativeto groundDisplacements relative to groundsyyxysxbgyflightflightyvgvgvvgvvvxgvtgttv00000020222)(25.00−=−=∆=→−=ab∆xbg= a+ b vssmvgsmgvsmvxsmvsmmgvxxsbgyy/10)/10(220/100/6.19/208020000=→⋅−=→=→=∆=→−=−smvsgvtyflight/226.1910422200=+===Displacements relative to the sledRelative to the sled, the displacement does not dependon the sled’s speed Answer (c)= Answer (d)mssmtvxflightxbs404)/10(0=⋅==∆(iii) A dog wishes to cross a river to a point directly opposite as shown. It can swim at 2m/s in still water and the river is flowing at 1m/s. At what angle with respect to the line joining the starting and finishing points should it start