Soluci&Atilde;&sup3;n Tarea 1

# Soluci&Atilde;&sup3;n Tarea 1 - F(5ica I Tarea i E...

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F(5ica I Tarea i Erler{)- tvb~ ZO\J L: Un camic" parte del rc,:nso y ace/era a i mJs2._ Die z se~undOS despoes , on Qutdmovil acelera de~de cl reooso en el mismo punto con una deeteroclon de 2- m/s'2.. ~Dbnde y cuondo el automo"il Q1can- za a\ camian? / . I AutornO\li , ACOJYli~n = )(aot.o = ~ GCJtY\\on 17 (xcorflion= xouto) =? W~-.in 1= 0 rn}~ Vlooto = 0 mJs / a aarYliorl= i m 151- Oooto = 2 ro Js2. a b 'lhcctmion == 'J,camlon + -Oca m ' .6 ni: \/bccJlYl'.on =. 0 + (.t m 15% )( \05> = 10 m I~ aO rf) 10 m/s C Gl>ondo? - 10 :t [iDO + ,ob: - - "0 :t 14.1i - ~I - -J- .r- -l.::: 4.\4 X t2.= -24.14 24,\4(.5") -1 -i - -

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cP6rx:1e? 0 Xfauto -= 0+61 uto}t.- + .L aLi: z ) 2 Xfao-to = 2.. (-2 rnJs2) (24./45) z: 58Z.--l,:. Crn) =- Xfcamion \ Gomprooo y XfcorYlion == Xicamion + LVi zzrm ion )i -t l.{CJcamion)t . XfcaMlon = Sam t 10 m/~ (24. \~ + .1-( i 2 mIt) (24. 14S) 2- 2 Xf carnien "=: 582 -:r (rn) .~_.:"t' ---
L~ Un aotdu; dism,,,uye su velOCtaetJ coo aceleraaon conStOn-te de. 24 rnls a Ib vn J~ y tccoac: 50 IY) en el txcces». d) ~ Goonto ll)a~ v,ajaro antes ce n~r 0\ a \to +eta\? Pr i mero enl (olaMDS Ox Vxf l := \Jxi2. t 20x (Xf-X, ) Ox-= ''iKf'l.,- Vx/- -=. (lb mI5)2- (2.4m'~)2._ -3.2 mJ~2. 1. (Xf-X",) 2 (50 m - Q) - A,\ lpad::ir iG:1e que I\eg a a lo~ 50 IY) torda , YXf= Vx.i fdx t -t =- " x. f - VX\ - 0 - I ~ m /5 = 5 S_ '; ~ra "ecJ Q r a 1 o~ - 3.2. mJs2. aHo tota I y recorrc X f= Xi -\- .L LVXf-VXl)t 2 Xl =- 0 t J- (0 - lb m15) (55)::: 4.0 rn J ma~ p1ra , Itl;:p r 2 .' dl d\to --total: b) uoanto +iempo le tOfllOra dctencrse desde ,O~ 24 ",Is Vx.f-== v x", tOxt t = \} )(f - Vx', - 0- 24 mJ~ := 1-.5 S ax - 3.2 rills' - - -- ---- ----'"--

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3:- Uo go-Kart recorre la primera Mitad de Lf)() pisb de \00 ri) con una rap"de-z.
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