Test1 - TEST 1 (PHYS 14101-3) 29 Oeteber EDGE, Number ef...

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Unformatted text preview: TEST 1 (PHYS 14101-3) 29 Oeteber EDGE, Number ef Prebleme 6, Duretien 1.5 h Name: Student Number: thlem 1- {3 marks} An nhjeet stems Frem peint. A at time L1 = D. Al. time 111 = E e, 1.5- = 3 5, t3 = 4 e '11: peeeee thrflugh the 1.1011115 13? C, and D reepeetively. Determine [a] the tern] dlepleeement {Hem A be D], (b) average veleeity {I'mm A In £1}, and {e} average speed {I'm-m A be E} ef the ebjeet. 533 I I III“.- III-HI III-II I I I I I I Solution. Peeltien vech are ree=rmn=—fii—1 riiel=r{4}=fii+ij- Displaeelnem is . . fir = TUE} -‘ I‘|{E_,.1J = m' Amer 43 velueity is _ _ 3g fir _ 111 —_] v“? = .fli 4 — fl = fiflli—j} mfe- Average speed W131 Etruj “neg = ‘Flt affi+ # 3.? wife- I J:- | 1:: eel—r ennu- =1 + Problem 2. {3 marks} The positienei' an ehjeet mating in an rep plane is given by :- [ti = [311!) — t?) i- 2.Ut_i._ with r in meters and t in setsintls. In unit—vester netatien, calculate {a} pesitien r [It], {1:} instantaneeus veleeitp v {t}. {e} instantaneeus aeselet‘atien a [th fer t = 5.1:} 3. Calculate {ti} displacement fir, {e} average Vela-tit}- vary. and {f} average aeeeieratien aim,Sir fer the first five seeends ei' mntien t = 5.0 s. Calculate [g] diatanee after five semnds ei' :tnntitinl (it) magnitude ef t'eieeity. {i} speed. and {1} angle between the positive t‘liIEL'titin at the 1-: axis and a line tangent tn the particles path at t = 5.0 s Seiutien. Instantaneeus Witheitj‘ is are} T swat: 41’]. tit-anti. V {t} at _ at 1 + a't J {—Eflti — 2m] mfs. Instantaneaus aeeeleratien is _ ave] _ dfi-2.flt]__ti(-2.fl]l. am“ a‘ a: L at] = —2.m mfsz. At time t = 5 s we ehtain {it} r{5} = [fifli — lilflj] m. {h} v [5} = {—iflfli - Elij} mfs. is} ata} = elfli me? Hi rm} = 3110i m, rffi} = {tifli — lflflj} m1 fin- : HIE} erfifl} = {—Efifli - lflflj} m. {3} ii. —25 {i' it] D' vet-g = I: = ‘51_—fl I J I [-5fli— mfs. if} _ flit-r {t} _ VIIEr} — viii] _ -1El.fli — Eflj +2.3] 3”“ ‘ at _ a—tl _ a = —2 Eli net's: is} iris tantt = Ifiirl = #253 -'- ll]'2 2 W a 26.9 111 [hi tartan = the? + s2 = M a me mfs {i} 1.! [5] = ]1.:' as 10.2 mf's {H t. as -i —_m t _ E = ens—l {—0.93} E awash-DEE} 2 163.5”. Prat-[em 3. [1 mark} Figure shows the pasitiamwrsua—time graph of a particle. rum-lug alflng the x-amlfi- Draw the particle‘s velucity w; it} and acceleratiun HIM graphs. Pruhlem :1. {2 nmrlcs'j Figure chews the velccity graph 1J1 {t} cf a particle mew-leg elcng the :c—exis. The particle alerts from :r: [{l} -.: #1 m. at E = fl. Draw {a} peejticn :t: {6] and {b} acceleration {1:13} greplm fer the particle. Prehlem .5. [1 mark} A jet plane is erujeing el; Sflfl wife when suddenly the pilet turns the up to full threttie. After trawling -1.EI km, the jet is Ineeiug with a speed at" :me mfe. {a} What is the Jet e eeeeleretien. stemming it me be e. eeuetent acceleratien‘;’ [th How Iflng dues it take te eeeelerete the Jet? Selutiun, Equetiene ef metien are 1:: {e = e: {e} +ert. Then we put 11' {fl} = fl, 1: {I} : 40% m, U311.” = 3m mfs, Hi. [i] = 4m mfe and ehtein from the 335$!“ '01. equatitme dflfl = Sflfl + 41:15, :2 met: = seet+ at} . a First e uetien e'vee { } e e1 m t = . '51:: After eubstitutien in the eeeend one we obtain lflfl '12 eeue = ene— + r," , e; - er , dzeTe :1: e2 er 4 = E flu: frem here eI = fi m 3.3 mfefl 4 lib] t— 1m— -—mfl —E:=I]ee _ e: 35ft! _ T Pruhlem E- [1 mark} A stem is threw: vertically rue-were: with an initial speed 13f MU} = EDWIer fmm :he reef er a building, I: = 25 m ubeve the greund- Take 9 2 1e wage? '.. Hew high does it gnu with respect In the gmmld Iva-ml? 2. What is the speed ef the erene at impact [en the ground Ina-ell? Selutien. Equetiene ef mmiun are 1'1: '3} em = y{fl:-+ep{fl}t+ U1..- + Ere-t1 yet-3 —-2 . II: eur case Ellie system of equati-ene reduefi m the fellewing eyetem eye} = SU—lflt. em = 25+3flt—5E. {EL} ‘Ir‘r'e put e'y [t’] = [I and: mlu'e the first equatim1 fl=30wlflf 7% t“ =3e. After euhetituhien et' thie result 1:1 tn the eeeeud equet'ten we ehtein yu’J E H=Ee+3flwj 4(3): :71: m. {11] We put 3: III”) 2 I] and melee the second equatien u = 25 + em” — 5:”. GI t"*-fit“—e=e =- t”=3=~.r’fi Peeielve eelutien ghfi t" = 3 '1' Ell—4 m 5.? 3. Then 1.1!, 3"] = 30 — llltl'it'“I “a: 3E] —1fl[fi.T} m —37 mfs. ...
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This note was uploaded on 01/10/2011 for the course PHYS 1410B taught by Professor Terekidi during the Fall '10 term at York University.

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Test1 - TEST 1 (PHYS 14101-3) 29 Oeteber EDGE, Number ef...

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