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# 05sep08 - The stopping distance of a car going 50 mph(27...

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Unformatted text preview: The stopping distance of a car going 50 mph (27 m/s) is 53.6 in. What is the magnitude of the acceleration in units of the gravitational acceleration g (g = 9 8 m/s )7 Assume the acceleration is constant. (A) 0.23 (3)0.35 (C) 0.47 (D) 0. 69 (E) 1.08 SOLUTION: leen vn=27 m/S att=0 and given v=0 when x=53.6 in. Also, at i=0. xn=o, Since a=constant, (1) x = x0 + vat + (112)28 and (2) v = v0 + at Solve (2) to get the time when car stops: 1 = (v» vo)/a = »viJa Substitute time t into (1) x = x0 + vn(- we) + (1/2)a(—Vg/a)2 Multiply and collect terms. x = - vozla + (1/2)a(-vi.la)2 = «oz/(2a) So alg=(-vi,2)l(29x) Substitute numbers: 3/9 = (0-272)/(2x9.8x53.6) = -0._69 [accelerimnn noel A car starts from rest and accelerates at 2.0 m/sz. How far will it have gone when it reaches a speed of 30 m/s (67.1 mph)? (A) 90m (B) 110m (C) 140m (D) 160m (E) 225m Answer dislance x : xvo in. (1/1)il7,butslnze x.=u and v.43, we have xiulllat’ Newlndmenmemxulvlng i=v...t,wnitn hemmeslusl v: 3! in iev/a, Sthsltlule [=V/B lnlb t/zm’ anugei x= (mm ell/zlalv/a)‘ llal ainsiiiuiing wage! xelau m/sl “zoom/Swem immmtimumq A rock is 'H'irown Gain 4e ground Verl‘ically upward. Le‘l‘llw x-cvis be Vertical) increasing upward. A+ -l-lie. rock’s hi’ggefgoi'gf ll\$ decelem-Hon is SL‘tUlIciiJ. M l i). m iinlliist Inuit I '2: ; 1M c f L‘ : l. ”l i) = ‘ ii \ iitttlei .enl Kare ltntt b L l q . W l e ~3‘L¥ A; bascbﬂ is H? STY‘W «P owl is Why He. W 4,40 sec. led-er: How M3“- Jul #4: Lam rise? 1‘:1A=‘DM;K:1D:7M ‘ “ea,“ --~ ‘ l ”W” ls l mu, eon/S \ ens/X REW- was: CM_ J/gwt «if: hint exexduhglltt‘ Emu? geiwnt :> L”.—'-ﬂt Ewe e7 unﬁt 1 l 1 7(_: »L\t lint :‘ikt ...
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