Solutions for Exam 1 - Spring 2007

Solutions for Exam 1 - Spring 2007 - W 7 February 2007 CODE...

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Unformatted text preview: W 7 February 2007 CODE # 0 1:00 p.111. — 1:50 pm. PHYSICS 114 First Hour Exam INSTRUCTIONS: Answer each question. All questions carry equal weight. You should calculate your answers to three (3) significant figures. Your answer may not agree exactly with one of the possible answers, but your answer, if correct should agree with one of the possible answers to at least two (2) significant figures. Be sure to enter the code number, found in the upper right corner of this page, in the appropriate box in your scantron answer sheet. All information supplied by you on your scantron sheet M be left justified, that is it must be entered beginning at the left hand end of the entry box. l. A body moves along the x-axis starting from x0 = 7.00 m and travels out to x = 62.0 m, before turning around and traveling back to x = —8.00 In. What is the total distance traveled by this body? ' tam/tam Fae—W 2 Q1‘ {7%l—xq 111‘:,0&M 0 lo _- 3.09m 11:61.0’“ ramm— 1.00m) +(6l.bw~[—t'00n?§) —. 590m How ». mow : 11M (2%) whole trip taking 627 s. Detegmine the aviwage velocityefoz'tgiifisntlrip. 5mg:- 7Q: D“ W:1 ‘1 1‘6"“? —0 -: ’ ’M’ w W . M: W W W 5W5 pmb W7 at. .0. MW :64: magzwrxm’m MM; A03 L ilk/2W surgzo’m 2%:me “Mg/p m 659 0 2w 3. Runner A travels at a steady speed of 3.12 m/s. while runner B travels at a gfily speed of 3.45 m/s. Both runners start from the same point and run along the same road, but runner B starts 315 s after runner A. Runner B catches up to runner A after {LL : _ 9%? Bhas been runnin for: ' QBWImufibwAmfiw. m 3 [A W /7V W X24413} W WM /AF' 074 : M1 M/S M 3W Wm thQ‘mm/s b7 Bu; {Int/EVE m 66 W ~.. 0 W WM :wtggfifl «a Vg/UjUACt Halsopwfit : {a K a (j ,Lfig _, :PhgxlllLI-sp Page 1 of3 M '. W J; , In 5 9/3“; 3 we mm 2. Ajogger runs 2.85 km due North, then turns around and runs 6.45 km due Sou , the Zmififi 5 W 7 February 2007 CODE # 0 1:00 pm. — 1:50 pm. 4. Starting from rest, a light plane moves down a runway with a constant acceleration of 6.25 m/sz. Determine the time taken for the plane to reach its takeoff velocity of 38.0 m/s. 6L5 army/1C Wham: a, bi? ow O wawrm‘ 3.5“; —0 , . or , “ €= 32th ‘f—p—fii WQLWMs/VW 5. A race car starts from rest and moves with a constant acceleration of 9.55 m/sz. Determine the distance traveled when it has attained a velocity of 75.0 m/s. -7 u 1» . W UdLb U1)“ +16LL1~JCJ McoIJflLa—A )Q :3 letdown) ‘ U ‘W *3 1‘10 : 0‘0 urwoml w 70 , .i 76,0: My : ’l’llt'SDn/c W (ll/“WOW won/1w: ME/Mm MW 3L 3, 1: HS‘MZXQW I 6. A train traveling with an initial velocity of 75.0 m/s slows at a constant rate to 25.0 (3 _. Lips m/s in 27.0 5. Determine the distance traveled by the train duringthis time. U: 0 a m m mt m 5m 17 v :jZflM/S U’ ~ Eiglly :FS'I i 0 r 1;! )6 W “ L " 1 1,, I60 30 W )Wm M :3 UM: 500148 :mmfl m a a - (maul; x 2m 7. (m M‘- 1~9YWM 3:04 . vL L O. W cm mm (atom 0:11“ +0;th 6L WWW U =W+Zk ‘13 7.“ plane traveling with an initial velocityoof 3.25 X 10 m/s slows at a constant rate to I 1.20 X 102 m/s during which it travels 5.27 X 104 m. Determine the acceleration of this plane as it slowed with res ect to the direction of travel of the plane. 1 0L. u”: rob Mum—x» » H r Runny = vL-W :3 cm ‘26} :20} o »———————'+’>% , ~ 10 :0 1: s-z? xm‘tm 3, 0L: l-w Mow/s) ~ Q13 Xlfl‘M/ L: _0.g{)<,Tm[5L Vo= MWOLMZS «MW/m WWW-03 : _ 0.356 3» MW at m m" a W» J 8. A ball is dropped from a height of 65.2 m above ground level. etermine t e vaccity {/0 2 6; 2m of the ballwhen it tn'kes the'groundH I A a \ ’ Wat/(me whammy/641mm V0-0 am (r: W" M 6% #01 L L L .. . : , / r) U : O VLKQ.g0M/J (0 6mm) ’2 M‘ Q :5 NW si arm/s WWW) W9 j if; -354411‘5 : 35'4‘M(S Phsxl 14-sp07—Exam 1_cO, Page 2 of 3 W 7 February 2007 CODE # 0 1:00 pm. — 1:50 pm. 9. A ball is thrown with an initial velocity of 22.7 m/s vertically upwards from ground level. Determine the time taken by the ball to reach its maximum height above the m‘flmm m {Wm MW ‘3 . : (j‘o\ :3 D: KARI/lifts — “(fill/r /c 2) “My t: 114m]: =3 t2 AWL : mm 4.31: aw 10. Two vectors of magnitude 4.27 u and 6.92 11 are added. Irrespective of their directions, the only answer for the magnitude of the resultant vector, which cannot ‘\ JIM/MM (9 / 3+7» 41‘ WW film m“; M V212,; _ £6,th 5W lamb. flab/w. WW6 ‘\ v} 4.1:.“ 73 V» WWWWWMsm \, / Sinai mm i®+vfl ~. dumb-4m : [[.lq:,\_x t 11. Vector A has compon s, A)( = -3.55 u and Ay = 4.92 1.1. Vector B hascoT'np'onefits, Bx=3.45uand By=-2.18 u., Determine the direction ofthe vector, é=2A+§f (“Myriam Aggy“) -256? m 66 6:1 3:211:41M‘i'ui‘u, u i as"? ~ (£12? 1.2;“ : ng'om) : -6WS'ZZ" 16;»: mic] 32% W Whit/Axum m. cm m cm a? 1603—6‘t-S2L0 —. «lamb = u? m m +mmm @w l . Vector A has components, Ax = 2.55 11 and Ay : -4.l7 u. Vector l3 has components, Bx = 3.85 u and By = -3.51 u. Determine the magnitude of the vector, (3 = 2A — 4B . cut/ii Jim :m-m ~0tM-Sm : «10.3% ad. M Ma; = Neutral—Mama) : Watt ’1 C. (E; +64; :lQoaoQL +Q¥OMF . (mm Phsxl 14-sp07—Exam 1_cO, Page 3 of3 ...
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This note was uploaded on 04/09/2008 for the course PHSX 114 taught by Professor Davis during the Spring '08 term at Kansas.

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Solutions for Exam 1 - Spring 2007 - W 7 February 2007 CODE...

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