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Quiz2Solutions

# Quiz2Solutions - QUIZ 2 OCTOBER 8 2010 1 AIRAIN STARTS FROM...

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Unformatted text preview: QUIZ 2 OCTOBER 8, 2010 1) AIRAIN STARTS FROM REST AND ACCELERATES UNIFORMLY FOR 350 5, AT WHICH POINT IT HAS ACQUIRED AVELOCI'IY 0F 35 M/S. THE TRAIN THEN MOVES AT A CONSTANT mgcm or 35 M/S FOR 400 5‘ THE TRAIN THEN DECEERAIES UNIFORMLY AT 0.05 M/Sz, UNTIL IT IS BROUGHT TO A HALT. A. DRAW THEVELOCIIY. VSTIMEGRARIIEQRTHIS MDTION-.AXESAND TICK. MARKS MUST BE LABELED. THE END POINTS OF ALL LINES MUST HAVE- THEJRCOORDINATESLABEIED- 71-0 fwwmh, e/ML ENG; mm, = Maw \ ﬂ; 0’ BSWSL—yfo‘ﬁgvfkgéﬂﬂ \$fz=IWSOS kmmmm_ﬂ_’_*~#_____.n (INS-u 5/ O M/S> (go) 3% @5b W7 3'6‘0 lHSb t,Csl B. DRAW T'HIE ACCELERATION VS TIME GRAPH FOR THIS MOTION. AXES AND TICK MARKS MUST BE LABELED. THE END POINTS OF ALL LINES MUST THEIR COORDINATES LABELED. «JP/DIM mam/won, m Aim 3mm in; m WWW . \f’ V3“ JG'V‘W WFK 1&0/ 6M 56/714404‘ 9% M/‘U’QI’; a: ; 0x )5?“ a? 3 g ‘6 § 5 ? LJ o‘\ K 00'?” T W M ' x Tum? :SSU £50 (9:0 ,. 0.0"?“ g QUIZ 2 OCTOBER 8, 2010 C WHAT IS THE ACCELERATION DURING IliEEIRSI 350—5 OE (10¢ 5my17 (“CM/L has ar’ﬁfr ww’“ a us,—t WA. Wt Mae/lawman dwwam aémma-o 5 me- D. WHAT IS THE DISTANCE TRAVELED BY THE TRAIN DURING W 0km Sgt/eff“; mm]; +0 obtains, um 19mm 1Mka am.“ WWV 0\ W000” U\$.ﬁm_¢oxm7ﬂln oggvﬂ’s MKS dismay, so W cm ka 16mm 0mm MK {.40 fw'owcolm WM LD) a“ 1m Wu. A=VLU~ %‘ J: ngmms—?50;)Lf>9‘w/s> : ii'L'LS‘Ow ~ TWA H1MMR¥WWI vhe, (mix uSt Ovo' lama/“NRC {T‘WMV‘S ‘: x .Q I , m L o \— d 2 vlt + VLCJTL>~> _‘2-oL¢ (WM/57L?“ 53r/74 ﬁg /5 X?) S) _, vﬁ:-v;"3 MA -> 07/4 WW9 + 2(«0’7’m/5LN‘U WWW~/W'n 95 0k: Illawm / 7: CCL‘C/UJOAYC M QwaLgc grad, MM, NJ ﬁlwwkhk; ’hro gym/ire, cksﬁoncc 0M Mu bio”) elm/{wk Ciacdu‘w’ﬁ 0m TA— Facts )‘ZJZ.§'O 1M W ﬁmw bow/f D. W CW" CCA-‘owu 3% a% whim/Lem “tw— QW “'U‘EQM3 ’jﬁ gull W (MS *Z‘WCS MMaX M MSG HAWK M 50mm, wow! . W“ M“ “WW2: m covébni" m,“ m MW w WV 06L K NJW-W PMwa a? 0X: ['gfm/s)(qoo5) :._ WW. ’fﬁc wig gauow aaderoépka“ ’ J 4 , ,. {farms momma“ \w 50 dc Vu'sxabsxgzmggzmwm. ? has 7’7,‘ (51 m/j N“ F. TISTHETRAIN’ SAVERAGEACCELERATIONFORTHE T500505 111511192 _ In“ (*W‘C‘xax/ “Mam-hﬁ n n’ S gym/mac, [w «AJan OW .C/ {Marat w‘ ﬁnk/me“ We, \rUOLﬁ», 5gfs from 9 he Zé‘ava WﬁMvoéouﬁwm 9H5?» => Q33: ow, QUIZ 2 OCTOBER 8, 2010 2) JOE HAS BEEN DRIVING A CAR AT 60 KMPH FOR 10 MINUTES. JOE IS NOT FULLY ALERT AND N THENEXT 2 MINUTES, HIS SPEED INCREASES AT STEADY RATE TO 80 KMPH. REALIZING THIS, JOE SLOWS THE CAR DOWN AT A STEADY RATE IN THENEXT 1 MINUTETO 6O KMPH. A. DRAW THE VELOCITY VERSUS TIME GRAPH FOR THIS MOTION. AXES AND TICK MARKS MUST BE LABELED. THEEND POINTS OF ALL LINES MUST HAVE COORDINATES LABELED. a FIN/5A,; JOE-‘5 wa¥ KM?“ fb bra/WM Qe 0 » WV \AM;)S“S uv‘ QMW M‘QS MUM- (xw—. HWM/W“) ‘ l (0' h (7/, '1’; ELM“? B. DRAW TI-LE ACCELERATION VS TIMIE GRAPH FOR THIS MOTION. AXES AND TICK MARKS MUST BE LABELED. THE END POINTS OF ALL LINES MUST HAVE THEIR COORDINATES Aglwklw gﬁ‘rjzt' [J fa‘CMcMiaH/ a,“ a'ﬁﬂqyjﬁpg} a K envy“ vﬂocﬂry vsﬂv‘mo 7mpbx. 9 12mm 0 i is B, . LO f Cwiu'l QUIZ 2 ' OCTOBER 8,2010 C. HOW FAR DIDJOE TRAVEL INTHE FIRST 11 MINUTES DESCRIBED IN THE PROBLEM? f6)", CCMUKV’A’L Maw 42m» Mbwvaﬁelw om N194)? ('0 vvxiwv\"C‘/§ 3 M = UW/wmuom) zlo low DUO 00"“ W M95” 0”“? 976‘ @mw‘ Wemi-tueﬁIWQPEF’WS Stab—o... 73;, M MD fax 4% Eff)" I! I’Vu‘M 3. 7“ ‘ L X; 2 X6 + UTE—£1 P3145th Km 3 (0 [W4 4-“ EW/Wr’m>0 “‘“341/‘1—(‘15'7‘tﬁy‘1goma §m m” ’ % % D. HOW FAR DID JOE TRAVEL IN THE LAST 2 MINUTES DESCRIBED IN THE PRQBLEM? ,7 f “(V‘k/Vlbu.) (Aow C44“ M ell/Vi“ 9; Wiwwxﬂth WAX 5;..(‘1‘35 "U01 140w- T‘Nb {AWK {rum U, H VL’WnKKa-eﬁg‘ aka U06 (aw\43_¢.ev‘t 916,11“, ’CMLW‘KHL/ "W9 diagrvwmg 9:244, dw—E—kwcg bum/tack {row H +3 )m mfng_ ’1) , . 9 \$3“ 7' vim/{L 2 7R2; r cd’ =97 Um ¢-( 1 [MM/wen) 4*(vW5—k’m/wiw") (“A”) 2>v§uu 1- IQVbW/WLW XML 7 1% + hair" =\ XML=‘(L1w’1M/wmwhw“ VLW'W/Mmf’wﬁ - ~ 1 t-Ls‘bw ' . W, I m: we? me, I «n 3~‘/7,u+”-> Km} 1m: Immmwwwﬁz WM WW) 1 M“ ‘ E. WHAT ISJOE' SAVERAGE ACCELERATION FORTHEFIRST l 1 MINUTES a WW} “:9- DESC IN PROBLEM? ' ’ ’1, E. WHAT IS DE 8 AVERAGE SPEED FOR THE 13 MINUTES DESCRIBED IN THE PROBLEM. 7:; ave/«r JWJ/7QV‘ 714/ W Wﬁm i3 WMLMLLS CS 774’“ "Mm! dam“, choﬂ ﬂax/3M 5mg. , W . Mtw yg (MD/HALLS 77 [’5 MM a f m . _ ’ {6‘9 O / 9 XCQSWIQQS/ ﬁrs/W3 (A, mQHMgbe/ S w. Af ( {imam “39'” 1° 7% ll sawmou/ WWW 4!, (gm/W a“ a 55W” ’{Viamﬁ'lc’ (1%va Cal/ta; gracing? Int, :3 A1“? l/LU" / A?» 3 ﬁx ‘9 ? A. )(l?~?“7rb“‘/r«2w”' I‘M/W“) 7 :1 a ’33 “Lg/k m = (2 WM ‘ WM L 7/“ é A,~+AL 2/. 33 law * > A3 3 X 1,14 / = wa 3) I43 7/21 (MM>(L’5’5WM;M a} W/M“) 7— 'HO‘S-bw ‘ ‘ , .MARM3 M55“ +95%” 'Pr+'(A-,+A;) HA3 +43) 3%; 1m MK 5M‘017‘CHWBM difhnco Pgrvvw 0—— l\ Ytgo (Add ‘ ’%,‘1‘(’5‘\au~ »‘ H.0% Lav; ED“; kw mww “WIS nah hm, «ﬂaw mswut. ...
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