Ch12_Sec1-5_Ex-11-15-22-41-45--57

# Ch12_Sec1-5_Ex-11-15-22-41-45--57 - 12-11 A article is...

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Unformatted text preview: 12-11. A article is moving along a straight line such that its posi ion is deﬁned by s = (10:2 + 20) mm, where t is in scco ds. Determine (a) the displacement of the particle du ng the time interval from r= 1 s to t= S s, (b) the we age velocity of the particle during this time interval,'an (c) the acceleration when t = 1 s. K; .r =10t1 +20 (n) sh.=10<1>1+2o=30mm sis, =10(5)‘+2o=270mm As=270-30-240mm A... (b) At=5-l =43 A: 240 vn'aEBTSmm’S A". d2: (c) ”Edam/:3 (forum) A... w gums-1» ration (175‘ xes about 3 l5. Tegts reveal that a 1' al driver can react to a before begim, ‘3 avoid a collision. It s for a driver .ig 0.1% aicohol in his stem to do the same. If sun» was are :zaveling on a raight road at 30 mph (4- ~"» and their cars can accelerate air 2 ft/sz, detern: the shurtest stoppinv: distance d for each from t .nomcn: they see tha pedestrians. Moral: If you mu 'ink,piuase don‘tdriv. - Stopping Dahlia: : For nonar. vet. the car mom tdistnncx - . m = «(0.15) = 33.0 ﬁbefore he 0' xacts and decelcmas the car . .~;.ping distancectnbeob ' usng ‘6wilhso=d’=33.0ﬁand l = 132 h before he v“ ..+2a.(s—:o) a; ~4’+2(—2)(.d- 33.0) d = 517 ft Ans or she rr: ‘5 and decelerm d1: car. '11:: mopping distance For aim-ME: car mm .dism of d’ = ox= 44(3) unbeobuinedgus' ‘ ‘ (1+): Eq. —6widuov=d’= 132 fund o=0. ”2 =D|zi+2ac(5-5)) oz =44’+2(—2 4—132) d = 2; fl Ans 12-2 A particle is moving along a straight line such that " s a celeration is deﬁned as a = (-20) m/sz. where v is i m ters per second. If v = 20 m/s when s = 0 and t= 0:1 de ermine the particle’s position, velocit , and acceleration as functions of time. ‘ article travels along a curve deﬁned by the u-‘ﬁonﬁi ([3 - 3:2 + 2t) m, where t is in seconds Draw “Ina ‘ 1H and a—t graphs for the particle for 0 5 IS 3 s. 1:11.577 5. and 1:0.4226 s, sI..‘..m = ~o.385 m Il..lo.4zzs = 0.385 m bile traveling along a - 7. The v—ngraph ofacarw . :2: is shownL Dkaw the s—t and a—t graphs for the motlon. OStSS as—aﬂmhulsz A15 ssrszo a=er=£9:—2-0=0mlsz A: 20-5 20Sts30 ‘ a--£’=-9:£=-21nlsz A! 30—20 Frommev—nnphu n =5:, tz=20.r. um, =30s. .1le1=%(5 )=50m .11 =A. +A1= 0+20(20—5)=3§0m :; =Al M; + , =350+ goo-20x20) =450m Theequﬁons ' [theponions afﬁxes-tgnphm 05:55: van; dt=vdn L'dveﬂkdt: ssh: 55:52.0: in”; and: ;o"’Is'2°"' s=2o:-so 205530: 1 v=2(30-t); ¢r=vda '“dysjz'ozeo-odx: = _‘z +60:-- 450 12-45. e snowmobile moves along a straight course according the v-t graph. Construct the s-t and a-t graphs for the pa e 50-s time interval. When I = 0, s = 0. r—t awkwmpommmmmormxmuobmbymm dr; ' =—.‘Forltimeinuvnl undo I. u=;—:r-Gr) rnlr. d1: 5 b=ﬂdl ENE-4r nee)»:- Alkali“, .r=§(301)=180m l Par'mlhm‘ml 30 s<¢sso s, l «1:st: I' (bar 124: I". 30. .r = (m- rso) rn AN=503. :812(50)-180=420m d-l Gmphmmnfmwﬂoninmofﬁmetmbeobuhedby . do applymgau—.,Forﬁme'iuuvll 015t<3031nd303<1550 I. 4! _d1) 2 d 0-7;:330Amhzandaszuso, respectively. 30 E’mh, 12—45 ...
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