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18 Pages

### 14_pdfsam_Cap 11

Course: ME ME 211, Spring 2009
School: Cal Poly
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Word Count: 183

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Complete COSMOS: Online Solutions Manual Organization System Chapter 11, Solution 15. Given: Separate variables Integrate using dv = kt2 dt v = 10 m/s when t = 0 and v = 10 m/s when t = 2 s. a= dv = kt 2 dt 10 22 10 dv = 0 kt dt v 10 10 = 13 kt 3 1 t 0 [(10) (10)] = 3 k ( 2 )3 0 (a) (b) Solving for k, Equations of motion. k= ( 3)( 20 ) 8 k = 7.5 m/s 4 Using upper limit of v at t, v v 10 1 =...

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Complete COSMOS: Online Solutions Manual Organization System Chapter 11, Solution 15. Given: Separate variables Integrate using dv = kt2 dt v = 10 m/s when t = 0 and v = 10 m/s when t = 2 s. a= dv = kt 2 dt 10 22 10 dv = 0 kt dt v 10 10 = 13 kt 3 1 t 0 [(10) (10)] = 3 k ( 2 )3 0 (a) (b) Solving for k, Equations of motion. k= ( 3)( 20 ) 8 k = 7.5 m/s 4 Using upper limit of v at t, v v 10 1 = kt 3 3 t 0 1 v + 10 = ( 7.5 ) t 3 3 v = 10 + 2.5 t 3 m/s Then, dx = v = 10 + t 2.5 3 dt dx = 10 + 2.5 t 3 dt Separate variables and integrate using x = 0 when t = 2 s. ( x dx 0 = 10 + 2.5 t 3 dt t 2 t 2 ( ) ) x 0 = 10 t + 0.625 t 4 4 = 10 t + 0.0625 t ( 10 )( 2 ) + ( 0.625 )( 2 ) 4 = 10 t + 0.625 t 4 [ 10] x = 10 10t + 0.625t 4 m Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.
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Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 3.Position:Velocity: x = 5t 4 4t 3 + 3t 2 ft v= a= dx = 20t 3 12t 2 + 3 ft/s dt dv = 60t 2 24t ft/s 2 dtAcceleration:When t = 2 s,x = ( 5 )( 2 ) ( 4 )( 2 ) ( 3)( 2 ) 2
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 195.Differentiate the expressions for r and with respect to time.( ) = 2 ( 2t + 4e ) rad,r = 6 4 2e t ft, 2t&amp; r = 12e t ft/s,&amp; = 12e t ft/s 2 r&amp; = 2 2 8e 2t rad/s&amp; r
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 18.Note that a is a given function of xUseUsing the limits andv dv = a dx = 600 x 1 + kx 2 dx = 600 x + 600kx3 dx v = 7.5 ft/s v = 15 ft/s when x = 0, when x = 0.45 ft,
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemSolve for x and v. x = 0.25 v= Evaluate at t = 0.2 s. 1 cos10t 201 sin10t 2 1 cos ( (10 )( 0.2 ) ) 20x = 0.271 m v = 0.455 m/sx = 0.25 v=1 sin ( (10 )( 0.2 ) ) 2Vector Mechanics for Engine
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 17.a is a function of x:a = 100 ( 0.25 x ) m/s 2Use v dv = a dx = 100 ( 0.25 x ) dx with limits v = 0 when x = 0.2 m 0 v dv = 0.2100 ( 0.25 x ) dxv x12 1 2 v 0 = (100
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 16.Note that a is a given function of x. a = 40 160 x = 160 ( 0.25 x )(a) Note that v is maximum when a = 0, or x = 0.25 m Use v dv = a dx = 160 ( 0.25 x ) dx with the lim
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 14.Given: Separate variables and integrate.v t 2 0 dv = a dt = 0 ( 9 3t ) dt = 9a = 9 3t 2v 0 = 9 t t3 (a) When v is zero. t (9 t 2 ) = 0 t = 0 and t = 3 s (2 roots) (b)
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 13.Determine velocity.v t t 0.15 dv = 2 a dt = 2 0.15 dtv ( 0.15 ) = 0.15t ( 0.15 )( 2 ) v = 0.15t 0.45 m/sAt t = 5 s,When v = 0, For 0 t 3.00 s, For 3.00 t 5 s, Determ
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 12.Given: At t = 0,v = 400 mm/s; a = kt mm/s 2 at t = 1 s, where k is a constant. x = 500 mmv = 370 mm/s, 1v t t 2 400 dv = 0 a dt = 0 kt dt = 2 ktv 400 =12 kt 2orv
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 11.Given:a = 3.24sin kt 4.32 cos kt ft/s 2 ,x0 = 0.48 ft,tk = 3 rad/sv0 = 1.08 ft/st tv v0 = 0 a dt = 3.24 0 sin kt dt 4.32 0 cos kt dt v 1.08 = = 3.24 cos kt kt 0
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 10.Given:a = 5.4sin kt ft/s 2 ,t tv0 = 1.8 ft/s, x0 = 0, 5.4 cos kt ktk = 3 rad/sv v0 = 0 a dt = 5.4 0 sin kt dt = v 1.8 = Velocity:05.4 ( cos kt 1) = 1.8cos kt 1.8
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 9.a = 3e 0.2t 0 dv = 0 a dtv 0=t 3e 0.2t dt 0vt3 = e 0.2t 0.2t0v = 15 e 0.2t 1 = 15 1 e 0.2t At t = 0.5 s,()()v = 1.427 ft/sv = 15 1 e 0.1()t 0 dx = 0 v
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 8.x = t 2 ( t 2 ) ft3v= (a) Positions at v = 0.dx 2 = 2t 3 ( t 2 ) ft/s dt2t 3 ( t 2 ) = 3t 2 + 14t 12 = 02t=14 (14) 2 (4)( 3)(12) (2)( 3)t1 = 1.1315 s and t2 = 3.5
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 7.Given:Differentiate twice. x = t 3 6t 2 + 9t + 5 v= a= (a) When velocity is zero. dx = 3t 2 12t + 9 dt dv = 6t 12 dtv=03t 2 12t + 9 = 3 ( t 1)( t 3) = 0 t = 1 s and t
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 6.Position: Where Letx = 50sin k1t k2t 2 mm k1 = 1 rad/s d = (1 t ) rad/s dt x = 50sin mm v= and k2 = 0.5 rad/s 2() = k1t k2t 2 = t 0.5t 2 radand d 2 = 1 rad/s 2 dt 2
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 4.Position:Velocity: x = 6t 4 + 8t 3 14t 2 10t + 16 in.v=a=dx = 24t 3 + 24t 2 28t 10 in./s dtdv = 72t 2 + 48t 28 in./s 2 dtAcceleration:When t = 3 s,x = ( 6 )( 3) +
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 2.x = t3 (t 2) m2v= a= (a) Time at a = 0.dx = 3t 2 2 ( t 2 ) m/s dt dv = 6t 2 m/s 2 dt0 = 6t0 2 = 0 t0 = 1 3t0 = 0.333 s(b)Corresponding position and velocity.1 1 x
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 42. Place the origin at A when t = 0. Motion of A: ( x A )0 = 0, ( v A )0 = 15 km/h = 4.1667 m/s, a A = 0.6 m/s 2v A = ( v A )0 + a At = 4.1667 + 0.6t x A = ( x A )0 + ( v
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 41.Place origin at 0. Motion of auto.( x A )0 = 0, ( vA )0 = 0,a A = 0.75 m/s 2x A = ( x A )0 + ( v A )0 t + x A = 0.375t 2 m Motion of bus.1 1 a At 2 = 0 + 0 + ( 0.75
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 40.Constant acceleration.Then,Choose t = 0 at end of powered flight. y1 = 27.5 m and a = g = 9.81 m/s 2 t = 16 s. 12 1 at = y1 + v1t gt 2 2 21 2(a) When y reaches the g
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 39.(a) During the acceleration phase x = x0 + v0t + 12 at 2Using x0 = 0, and v0 = 0, and solving for a gives a= 2x t2Noting that x = 130 m when t = 25 s, a=( 2 )(130 ) (
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 38.Constant acceleration. x0 = 0 v = v0 + at x = x0 + v0t + a= x = x0 + v0t + v= 540 = v= Then, from (3), Substituting into (1) and (2), 1 v0 2 v v0 t 12 at 2 (1) (2)Solvi
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 37.Constant acceleration. v0 = v A = 0, v = v0 + at = at x = x0 + v0t + At point B, (a) Solving (2) for a, 1212 at = at 2 2 and t = 30 s a = 6 ft/s 2 vB = 180 ft/s x0 = x A
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 36.10 km/h = 2.7778 m/s (a) Distance traveled during start test. a= dv dt 100 km/h = 27.7778 m/s 0 a dt = v0 dva= v v0 ttvat = v v0 a=27.7778 2.7778 = 3.04878 m/s 2 8
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 35.10 km/h = 2.7778 m/s (a) Acceleration during start test. a= dv dt8.2 27.7778 0 a dt = 2.7778 v dt100 km/h = 27.7778 m/s8.2 a = 27.7778 2.7778 (b) Deceleration during
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 34.(a )t dx = v = v0 1 sin dt T Integrating, using x = x0 = 0 when t = 0,t x t t 0 dx = 0 v dt = 0 v0 1 sin T dt xx 0vT t = v0t + 0 cos T t 0x = v0t +v0TcostT
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization System(b) Acceleration:a=dv = v n cos ( nt + ) dtLet v be maximum at t = t2 when a = 0.Then,cos ( nt2 + ) = 0From equation (3), the corresponding value of x isx = x0 + vncos = x0 +v x0 n v 1
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 33.(a) Given: v = v sin ( nt + )At t = 0,v = v0 = v sin orsin =v0 v(1)Let x be maximum at t = t1 when v = 0.Then, Using sin ( nt1 + ) = 0 and or cos ( nt1 + ) = 1(
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 32.The acceleration is given by a = 32.2 y 1 + 20.9 106 2vdv = ady = 32.2dy y 1 + 20.9 106 2Integrate, using the conditions v = v0 at y = 0 and v = 0 at y = ymax . Al
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 31.The acceleration is given by v dv gR 2 =a= 2 dr r gR 2dr r2Then,v dv = Integrating, using the conditions v = 0 at r = , and v = vesc at r = R0 2 vesc v dv = gR R r 2
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 30.Given: v = 7.5 (1 0.04 x )0.3with units km and km/h(a) Distance at t = 1 hr. Using dx = v dt , we get dt = dx dx = v 7.5(1 0.04 x)0.3Integrating, using t = 0 when x
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 29.x as a function of v. v = 1 e0.00057 x 154 v e 0.00057 x = 1 154 v 2 0.00057 x = ln 1 154 v 2 x = 1754.4 ln 1 154 a as a function of x. v 2 = 23716 1 e0.00057 a=v (1)2
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 25.v dv = a dx = k vdx, 1 dx = v1/2dv kx v 3/2 x0 dx = k v0 vdv = 3k vx0 = 0,v0 = 25 ft/s12v v0x x0 =2 3/2 v0 v3/2 3k()orx=2 2 3/2 3/2 3/2 ( 25) v = 3k 125 v 3
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 24.Given: a =v dv = kv2 dxSeparate variables and integrate using v = 9 m/s when x = 0.v x 9 v = k 0 dxdvln Calculate k using v = 7 m/s when x = 13 m. ln Solve for x. (a
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 23.Given: or a=v dv = 0.4v dxdv = 0.4 dxSeparate variables and integrate using v = 75 mm/s when x = 0. 75 dv = 0.4 0(a) Distance traveled when v = 0 0 75 = 0.4 xvxv
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 22.a=vvdv = 6.8 e0.00057 x dxx 0.00057 x 0 v dv = 0 6.8 edxx 0v2 6.8 e0.00057 x 0= 2 0.00057= 11930 1 e0.00057 x()When v = 30 m/s.( 30 )22= 11930 1 e0.00057 x
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 21.Note that a is a function of x.a = k 1 e x()Use v dv = a dx = k 1 e x dx with the limits v = 9 m/s when x = 3 m, and v = 0 when x = 0.0 0 x 9 v dv = 3 k (1 e ) dx(
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization System(b) Velocity when total distance traveled is 3 m.The particle will have traveled total distance d = 3 m when d xmax = xmax x or 3 2 = 2 x or x = 1 m.7 4 Using v = 12 x , which applies when x i
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 20.Note that a is a given function of x.7 a = 12 x 28 = 12 x m/s 2 3 7 Use v dv = a dx = 12 x dx with the limits v = 8 m/s when x = 0. 3 v v dv 87 = 12 dx 3 x x 0 v2
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemUsedx = v dtordt =dx dx = 2 v 40 x + 0.52()t=040dt = tdx x + 0.522Use limitx=0when40 0 dt = 0xdx 1 x = tan 1 2 0.5 0.5 x + 0.5240t = 2.0 tan 1 ( 2 x ) 2 x = tan ( 20t ) v= or
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 19.Note that a is a given function of x.Usev dv = a dx = 800 x + 3200 x3 dx()Using the limit v = 10 ft/s when x = 0,v x 3 10 v dv = 0 ( 800 x + 3200 x ) dxv 2 (10 )
Academy of Art University - CHM - 111
Chapter 1 Worksheet1. Convert: 425 m to (A) cmName: _(B) km(C) ft2. Convert: 12.73 mL to (A) cm3(B) L(C) quarts3. Sig fig calculations (all numbers are measurements) : (A) 34.7/8.23 + 1.4/0.457 =(B) 1.3 x 104 + 7.88 x 105 =4. If 1 euro is worth
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Special Problem Set ECE 585 Spring 10 SP1: For a given processor, CPU instruction access is 1 cycle; memory (data) accesses are 15 cycles(per word) and the distribution of instructions is as follows: Data transfers (integers) 10% Data transfers (FP) 5% Ar
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