1、Motion in Two and Three Dimensions Dr.Huaqiu Deng E-mail:Position and DisplacementPosition vectorRectangular coordinates),(kjizyxzyxr5m)2m,(-3m,km)5(jm)2(im)3(r)kji(-)kji(11122212zyxzyxrrrk)-(j)-(i)-(121212zzyyxxrSample Problem k(3m)i(12m)12rrrkm)8(jm)2(im)9(km)5(jm)2(im)3(21rrDisplacement interval
2、timeyin velocit changeon accelerati averagetvtvva12avgkjikjizyxzyxaaadtdvdtdvdtdvdtvdaAverage and Instantaneous Acceleration instantaneous acceleration:kjikjizyxzyxaaadtdvdtdvdtdvdtvdakjizyxrk)-(j)-(i)-(121212zzyyxxrkjikjizyxvvvdtdzdtdydtdxdtrdv62.0)2.76.0(tdtddtdvaxx44.0)1.944.0(tdtdayj)(0.44m/si)m
3、/s 62.0(22a222m/s 76.0yxaaao135tanxyaaSample ProblemProblem:A rabbit runs across a parking lot on which a set of coordinate axes has,strangely enough,been drawn.The coordinates(meters)of the rabbits position as functions of time t(seconds)are given by x=-0.31t2+7.2t+28,y=0.22t2-9.1t+30.find the velo
4、city at time t=15s.Solution:ji000yxvvv000000sincosvvvvyxj gaFeatures:the horizontal motion and the vertical motion are independent of each other Projectile MotionProjectile Motion Analyzed Horizontal Motion:velocity remains constanttvtvxxx)cos(0000The Vertical Motion:as free-fall motion000022100sin
5、vvgtvvgttvyyyyyy2002000)cos(2)(tan 0,0vgxxyyxThe Equation of PathProjectile Motion Analyzed tvxxR)cos(000The Horizontal Range 0)sin(221000gttvyy0202singvR maximum 4500RSample Problemo148tanhxm5.555)cos(00tvx500)sin(22100gttvyProblem:In the Figure,a rescue plane flies at 198 km/h(=55.0 m/s)and consta
6、nt height h=500 m toward a point directly over a victim,where a rescue capsule is to land.(a)What should be the angle?of the pilots line of sight to the victim when the capsule release is made?(b)As the capsule reaches the water,what is its velocity in unit-vector notation and in magnitude-angle not
7、ation?Solution:(a)m500,0 m/s,5500hvst1.10Sample Problemm/s 55cos00vvxm/s 99sin00gtvvyjm/s)99(im/s)55(v019.60tanxyvv(b)m/s 11322yxvvvSample Problemoo002063 and 27 2singvRm 690)452sin(o20maxgvRProblem:The figure shows a pirate ship 560m from a fort defending a harbor entrance.A defense cannon,located
8、at sea level,fires balls at initial speed v0=82 m/s.(a)At what angle 0 from the horizontal must a ball be fired to hit the ship?(b)What is the maximum range of the cannonballs?Solution:(a)(b)Even constant speed,the velocity changes in direction.The velocity is always directed tangent to the circle i
9、n the direction of motion.rva2vrT2Uniform Circular MotionThe acceleration is always directed radially inward.Centripetal accelerationvelocity depends on the reference frame BAPBPAxxxBAPBPAvvvBAPBPAaaaFrame A on the groundFrame B on moving carObject P on another carRelative Motion in One Dimensionkm/
10、h 52km/h 78 PBBAPBPAvvvvSample ProblemProblem:In the figure,suppose that Barbaras velocity relative to Alex is a constant vBA=52 km/h and car P is moving in the negative direction of the x axis.(a)If Alex measures a constant vPA=-78 km/h for car P,what velocity vPB will Barbara measure?(b)If car P b
11、rakes to a stop relative to Alex(and thus relative to the ground)in time t=10 s at constant acceleration,what is its acceleration aPA relative to Alex?(c)What is the acceleration aPB of car P relative to Barbara during the braking?Solution:(a)km/h 130 PBv20m/s 2.2s 10km/h)78(0tvvaPA20m/s 2.2s 10km/h
12、)130(km/h 52tvvaPBSample Problem(b)(c)BAPBPArrrBAPBPAvvvBAPBPAaaaRelative Motion in Two Dimensions Sample ProblemWGPWPGvvvProblem:In the figure,a plane moves due east while the pilot points the plane somewhat south of east,toward a steady wind that blows to the northeast.The plane has velocity vPW r
13、elative to the wind,with an airspeed(speed relative to the wind)of 215 km/h,directed at angle south of east.The wind has velocity vWG relative to the ground with speed 65.0 km/h,directed 20.0 east of north.What is the magnitude of the velocity vPG of the plane relative to the ground,and what is?Solution:20cos65sin2150oo5.16km/h 22820sin655.16cos215ooPGvHomework#10 of P78#20#27#56#69#73
