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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