HW2_solutions

HW2_solutions - MAT“ g; a; Somme: m5: ms Homework 2 A - ....

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Unformatted text preview: MAT“ g; a; Somme: m5: ms Homework 2 A - . Afififi Dynamics dx dv dv . . If x = x(t) , then Va) : __ and 510‘) = — = 12—. You must use one of the acceleratlon expressrons to dt dt dx obtain v and then use 11(1) = 6:1;— to obtain x(t). t 1. If a = 3t ~ 2 and at time t = 0, x = x0 and v = v0. Find v and x usino= definite integration. 2. If a = -2v2 and v(x°) = v0 and x(to=0) = x0, then using indefinite integration find v. Hint: It will be v(x). Also, find the integral (you do not have to integrate it) that yields x(t). 3. Ifa = 2 cos t and at time t = 1 sec, x = 0 while at time t = 10 sec, x = 4. Find v. (Make sure your calculator is in radian mode) 4. If a = -2x and v = v0 and x = x0 at time t = 0’ sec. Find v. Hint: It will be v(x). Also, find the integral (you do not have to integrate it) that yields x(t). 5. Given a. x(t)=sin(l+t2) Find in), 550). b. no) 2 ln(2t) Find m), 560). c. W): COS: Find 5cm. 1 +t d. x(t) 2 4% + 2t —5 Find kg), 550). 6. Given the following figure where 0 S x S 1 , find a. the location of the area centroid, b. the area moment of inertia around the x—axis, c. the area moment of inertia around the y—axis, and d. the polar area moment of inertia. 1.0 y=«/; ' \4 0.2 0.0 . 0.0 0.2 0.4 0.6 0.8 1.0 1.2 E I 01‘7 63') f-A/M’FV‘AH K,, (K / [Hus/“‘2 pa“ 1 ___E__ Ci:ng a=3t~2 )(1Xo Mal/€710 at t=0 v E Wits} W< mm a i v ’C : m 0‘;an 4‘? a“ = 035V i l E E ¥ % g i g E E i E i E E s g E § 1 § E E E E i g 4 § i i i l! E E E 1 £ % i é E E s I E i i E E E i E E z i i i I; E i :i 1 xo + u, t H. 3 E: «W W0 + a /.. a )0 x , e A, % «la .7 .W. a” +u 2 Z = _ v a“ 8 . s C 6 X _ flu. M b o x) D : 7 ‘= I e U v7». VnD vsA 0' : 1M : E .e vm X OEIIIL V U /.. «a , 7 2 K 2 / V9 .U_ U W : e .2 3a 0 a 0 a? v / v > a __ v M .. \sl 7. o ) U N. «a 4.,A U a 2 = e V x m Id; 1 WM U Cw QESEHEV v fix V . . V tr; 5w”). E ’7 7 I E “R66, 3 0" i E: Gwen a=2ws£ CA" 4.“: Is ¥¢D i» E 9 E ‘t= Ios )xz‘f i i E V g Fudo 110:) E {Q 5 1% £1]:— QGOS’L‘ 3? Std-7]: QCOSEJIf'F c, ZE at g got Kfi’g g v '1 QSI‘A‘C ‘* C) E mm) NLSo Mmfim 437 r. 1) =27 gdx = $110“? *CL E v ! J E N at 5 i = (420.951erle + CZ ; % I i (15‘ . E n ”“ 0:- —2c051+c,m+ CZ we a: far g I 2) M1 L1: "QCosl'O+ 0,10 +6z xz‘t at ‘t’rlox‘ xv 2)-.) "\=')ws|01-2aasl+ Cf) i cl: Jc;(L|+Z&astD—ZC.ASI) :— 0J3? E i E g 59 E 'U" Qant-POJBB E i E E ‘ I E E 1 g % E i E E g i i s i 5 i , 4%) (gig-‘W/Wmm 2t" z N '6 '1) 6" a J: b m t :‘E‘ G \u, k): 5%- N “’1 K 9‘. ow, Fail 2. UK?) =—2x :9 12/14" Jdoufi. INT“. Bur ng , I Suowao DEFudfis 2. vi: 002+ Qfxz-X ) :— U1+Q[x2“xl) [25.3me 052056 +% 51m 5% X=Ko Uz’VVO E E 6%): z + l/@Z+2[kd’—x2) )g Dal-e 2(on—X7') Fla/0 n1 lflfl‘ML 7745/22 . .537 (ff/fR/WN/M? 'a) X = 3m (14‘) 91x a £05 in t‘))/-?é) m' = COS [/ftz) djx : Quart?) — at m make i l i L E 61%" : 2w5[/+62)*‘lt2 Jm’lez) E E varé A ' b) X: haw) 4x: ,I_ .291 jflf‘znzn' i df 2k T ;t["‘*”"“’t r E ‘ 42X: —_/_ dtz 7' gc) X, cost g’x: ‘J‘ml" .- cost [2%) Mtz M “t; (H1591 ‘ My“ = {Ht/’2); ‘ Locoiham a! m @eflmlcj 5) area $mwenj- a? Mama x-a‘e'ff C) [I l[ l\ (gfiaxif 00> P04“ Ma (workout-OF \unhc. fr?» (MW/w .. ‘ ,.a“...“mm...”_...,.,__,._..W.V...__...A.r. wmwaqfl..-“VW...mw.~.......,m-,.......~_~..‘.,.. H=fj d; dx a i r EQTNS ,4 = d3? — I“ WN‘E/Lt .V (3) 3(— A = fix dX E}? Lacuna/V of a g A; m M} W Jxx dx 5"} rm,W-‘me..qw_wwmnr‘fimmx-«wmwww ..,.w....wmwmm..“WWW” .mwmvhwwww. NW." "mm. w.W....,.........—.m.vw._--ww»...w‘mmnnwaflvm ,F.,,..wm.—_...VW.,-<u... .. 1”. “WW WWWWWM. 65> K WWW/i7 ' a??? / i a s y x i, X a N e\ >7! l N \>< v K Y H D; \ >< $ I b\x W \a K \K ...
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HW2_solutions - MAT“ g; a; Somme: m5: ms Homework 2 A - ....

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