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Unformatted text preview: 1211. 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=27030240mm A... (b) At=5l =43 A: 240
vn'aEBTSmm’S A".
d2:
(c) ”Edam/:3 (forum) A... w gums1» 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 aimME: car mm .dism of d’ = ox= 44(3) unbeobuinedgus'
‘ ‘ (1+): Eq. —6widuov=d’= 132 fund o=0. ”2 =Dzi+2ac(55)) oz =44’+2(—2 4—132) d = 2; fl Ans 122 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:—20=0mlsz A: 205 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+ goo20x20) =450m
Theequﬁons ' [theponions afﬁxestgnphm 05:55: van; dt=vdn L'dveﬂkdt: ssh: 55:52.0: in”; and: ;o"’Is'2°"' s=2o:so 205530: 1 v=2(30t); ¢r=vda '“dysjz'ozeoodx: = _‘z +60: 450 1245. e snowmobile moves along a straight course
according the vt graph. Construct the st and at graphs
for the pa e 50s time interval. When I = 0, s = 0. r—t awkwmpommmmmormxmuobmbymm dr; '
=—.‘Forltimeinuvnl undo I. u=;—:rGr) rnlr. d1:
5 b=ﬂdl ENE4r 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 dl Gmphmmnfmwﬂoninmofﬁmetmbeobuhedby . do
applymgau—.,Forﬁme'iuuvll 015t<3031nd303<1550 I. 4!
_d1) 2 d
07;:330Amhzandaszuso, respectively. 30 E’mh, 12—45 ...
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 Fall '08
 Hudyma
 Dynamics

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