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# Practice1 - PROBLEM 11.18 Point A oscillates with an...

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PROBLEM 11.18 Point A oscillates with an acceleration a = 144(20 - x), where a and x are expressed in in.ls2 and inches, respectively. Knowing that the system starts at time 1= 0 with v = 0 and x = 19 in., determine the position and the velocity of A when 1= 0.2 s. SOLUTION Note that a is a given function of x. a = 144(20 - x)in.ls2 Use vdv = adx = 144(20 - x) with the limits v = 0 when x = 19 in. 2 v 20 - X 2jX 2 2 - 0 = -144 ( 2 ) 19 = 72 - 72(20 - x) [ or v = ±12~1- (20 - x)2 in.ls Use dx = vdl, or dl = _dx = _----;==dx=== v ±12~1-(20-x)2 with the limits I = 0 when x = 19 in. 1 dx or I = ±- r ----r====== 12 9 ~1 _ (20 _ X)2 Let u = 20 - x. u = 1 in. when x = 19 in. dx = -du U _ 1 f du _ 1 . -1 I _ 1 ( . -1 1!) I = +-.. ~ = +-sm u = +- sm u - - u 2 12 ,,1 - 12 1 12 2 u = Sin(; + 121) = cos(±12t) = cos(121) = 20 - x Then, x = 20 - cos(121) in. and v = 12sin(121) in.ls At I = 0.2 s, 121 = 2.4 rad x = 20 - cos(2.4) v = 12sin(2.4) x = 20.7 in ...... v = 8.11 in.ls .....

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PROBLEM 11.21 The acceleration of a particle is defined by the relation a :::;; -kfv, where k is a constant. Knowing that x ° and v :::;; 25 ftls at t :::;; 0, and that v :;:: 12 ftls when x :;:: 6 ft, detennine (a) the velocity of the particle when x = 8 ft, (b) the time required for the particle to come to rest. SOLUTION vdv :::;; adx = -kJ';dx, xo 0, Vo = 25 ftls 1 112 dx:;:: --v dv k F dx = -.!.. t J';dv _~v3'2J'" .Ixo ko 3k "'0 Noting that x 6 ft when v 12 ftls, 6 :;:: ~[125 -123'2J ::;; 55.62 3k k or k :::;; 9.27 ,Jmls Then, (a) When (b) x:::;; 8 ft, v3'2 :::;; 125 - 13.905x yl'2 :::;; 125 - (13.905)(8) :;:: 13.759 (ftlS)3'2 dv adt = -kFvdt Idv dt=--- k ylf2 v 5.74 ftls .... 1 [1I2J'" 2( 112 112) t :::;; -k . 2 v 1.'0:::;; k Vo - y Atrest, v:::;; 0 t = 2y~2 k = (2)(25)112 9.27 t:;:: 1.079 s ....
PROBLEM 11.27 Experimental data indicate that in a region downstream of a given

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