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3 Pages

### hwsolB17

Course: ENEE 241, Spring 2008
School: Maryland
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Word Count: 159

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17 HW Solution ______________ (i) Using microseconds (and noting that 1 MHz = 1/(1 microsecond)) : f1 = 3 ; f2 = 2.5; gamma = 1.45 t = (0 : 0.1 : 10).' ; u1 = cos(2*pi*f1*t) ; w1 = sin(2*pi*f1*t) ; u2 = cos(2*pi*f2*t) ; w2 = sin(2*pi*f2*t) ; V = [u1 w1 u2 w2] ; c = (V'*V)\(V'*s)% s as in data17.txt Obtain: c = 2.2657 (=a1) -9.8499 (=b1) 4.8427 (=a2) -8.1701 (=b2) (ii) plot(t,s,t,V*c,'-.'),...

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17 HW Solution ______________ (i) Using microseconds (and noting that 1 MHz = 1/(1 microsecond)) : f1 = 3 ; f2 = 2.5; gamma = 1.45 t = (0 : 0.1 : 10).' ; u1 = cos(2*pi*f1*t) ; w1 = sin(2*pi*f1*t) ; u2 = cos(2*pi*f2*t) ; w2 = sin(2*pi*f2*t) ; V = [u1 w1 u2 w2] ; c = (V'*V)\(V'*s)% s as in data17.txt Obtain: c = 2.2657 (=a1) -9.8499 (=b1) 4.8427 (=a2) -8.1701 (=b2) (ii) plot(t,s,t,V*c,'-.'), ... title('Plot of s (solid) and s^{hat} (dotted)'), ... xlabel('t (microseconds)') see ; % hw17graph1.pdf (iii) Estimate of R1/R2 equals ((a1^2 + b1^2)/(a2^2 + b2^2))^(-gamma) = 0.8349 (iv) S = fft(s) ; plot(abs(S)), title('Plot of abs(fft(s))') % see hw17graph2.pdf The ratio of the two peaks is given by 450.7/411.6 = 1.0951 Note that the quantity (a1^2 + b1^2)/(a2^2 + b2^2))^(1/2) which estimates the ratio of the two amplitudes in the expression for s(t), equals 1.06. (The taller peak corresponds to f1 = 3 MHz, the shorter peak to f2 = 2.5 MHz.)
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Maryland - ENEE - 241
HW 18 Solutions_V = [-2.8867 4.0276 2.1607 0.0369 0.3649 -6.1858 0.0999 -6.0291 0.4916 4.0812 0.5427 2.7973 ](i) Rounded to four significant digits:V'*V = [ 16 -32 8-32 73 -7 8 -7 29 ](ii) Forward elimination on V'*V:m x1 x2 x3_ 16*-
Maryland - ENEE - 241
HW 19 Solution_i) V = [ v1 v2 v3 ]v1'*v2 = 3(a+bj) - 18j + 6j(a-jb)v1'*v2 = 0 for orthogonality, so Re( v1'*v2 ) = 3a + 6b = 0 -&gt; a = -2bIm( v1'*v2 ) = 3b - 18 + 6a = 0 -&gt; b = -2, a = 4The same terms are obtained in v2'*v3 and v3'*v1, so these i
Maryland - ENEE - 241
HW 20 Solutions_Let a = sqrt(3) throughout.v0 = [ 1 1 1 1 1 1 ].'v1 = [ 1 (1+j*a)/2 (-1+j*a)/2 -1 (-1-j*a)/2 (1-j*a)/2 ].'v2 = [ 1 (-1+j*a)/2 (-1-j*a)/2 1 (-1+j*a)/2 (-1-j*a)/2 ].'v3 = [ 1 -1 1 -1 1 -1 ].'v4 = [ 1 (-1-j*a)/2 (-1+j*a)/2 1 (-1-j*a)/2 (-1+j
Maryland - ENEE - 241
HW 21 SOLUTION_(i) X1 = [ 0 1 0 0 0 0 0 0 0 1 ].' The Fourier frequencies present in x1 are: w = (2*pi/10) = pi/5 and w = 9*(2*pi/10) = 9*pi/5 The corresponding Fourier sinusoids are v1[n] = exp(j*pi*n/5) and v9[n] = exp(j*9*pi*n/5) = exp(-
Maryland - ENEE - 241
HW 22 SOLUTION_(i) Since s is real-valued, S has circular conjugate symmetry, i.e.,S = [6 -5-3*j 3+2*j -7 3-2*j -5+3*j ].'(ii) s[0] + . + s[6] = S[0] = 6(iii) We compute the modulus and angle of each entry of S to obtain the amplutude and phase spe
Maryland - ENEE - 241
HW 24 Solution_s = [ a b c d e f g h ].'S = [ 4 -1+2j 5*j 3+j -6 3-j -5*j -1-2j ].'S1:R*s &lt;-&gt; R*S = [ 4 -1-2j -5*j 3-j -6 3+j +5*j -1+2j ].'s1 = P*R*s &lt;-&gt; F^(-1)*R*S = S1The entries of R*S are multiplied by those of the k = 8-1 = 7th sinusoi
Maryland - ENEE - 241
HW 25 Solution_s = [ a b c d e f g h ].'S = [ 4 -1+2j 5*j 3+j -6 3-j -5*j -1-2j ].'S1:R*s &lt;-&gt; R*S = [ 4 -1-2j -5*j 3-j -6 3+j +5*j -1+2j ].'s1 = P*R*s &lt;-&gt; F^(-2)*R*S = S1The entries of R*S are multiplied by those of the k = 8-1 = 6th sinusoi
Maryland - ENEE - 241
HW 29 Solution_The problem parameters (frequency f0, amplitude A and phase phi) are randomly generated in MATLAB: n = (0:19).' ; f0 = 0.15 + 0.25*rand(1) ; A = 10*rand(1) ; phi = pi*rand(1) ; x = A*cos(2*pi*f0*n + phi) + 0.3*A*randn(size(n) ;% (i
Maryland - ENEE - 241
S 30.1 (P 4.1)_(d[.] = delta[.], the unit impulse function)(i) x[n] = d[n+2] - d[n+1] + 5d[n] - d[n-1] + d[n-2]In other words,x[0] = 5x[1] = x[-1] = -1x[2] = x[-2] = 1x[n] = x[-n] = 0 for n&gt;2The DTFT of d[n-m] equals exp(-j*m*w), soX(exp(j*
Maryland - ENEE - 241
S 32.1 (P 4.5)_i) MATLAB code:b = [1 -3 1 1 -3 1].';H = fft(b,256);A = abs(H);q = angle(H);ii) H(exp(j*w) = 1 - 3*exp(-j*w) + exp(-j*2*w) + exp(-j*3*w) - 3exp(-j*4*w) + exp(-j*5*w)= exp(-j*5*w/2)*(exp(j*5*w/2) - 3*exp(j*3*w/2) + exp(j*w/2)
Maryland - ENEE - 241
S 33.1 (P 4.4)_(i) H(z) = 1 - z^(-1 - *z^(-2) + z^(-3)(ii) H(exp(j*omega) = 1 - exp(-j*omega) - exp(-j*2*omega) + exp(-j*3*omega)= exp(-j*3*omega/2). *(exp(j*3*omega/2) - exp(j*omega/2) - exp(-j*omega/2) + exp(-j*3*omega/2)= exp(-j*3*omega/2)*(
Maryland - ENEE - 241
S 34.1 (P 4.3)_Divide by 2*pi to express the three frequencies in cycles per sample:3/28, 9/35 and 17/48Since these are rational (i.e., integer fractions), thesignal is periodic. The period is the smallest integerL such that each frequency can be
Maryland - ENEE - 241
S 35.1 (P 4.10)_(i)bar(0:36, b1), title('Filter Impulse Response'), xlabel('Time n')The impulse response is symmetric about n=M/2=20.(ii)f = (0:999)/1000;H = fft(b1,1000);Ha = abs(H);plot(f,Ha), title('Amplitude Response'), xlabel('\Omega/
Maryland - ENEE - 241
S 36.1 (P 4.13)_(i) The response to delta[n-m] is h[n-m]. Therefore the response tox = delta[n+1] - delta[n-1]isy = h[n+1] - h[n-1]The first nonzero value of h occurs at n=0 and the last one at n=4.The first nonzero value of y occurs at n=-1 (pro
École Normale Supérieure - ACCOUNTING - ACC202
Zishan Yousuf Intermediate Accounting 3300 Take Home AssignmentGeneral JournalJuly 1, 2007 Cash Common Stock July 1, 2007 Auto Cash Accounts Payable July 3, 2007 Supplies Accounts Payable Prepaid Insurance Cash Accounts Receivable Fees Earned Accounts P
Cal Poly - ME - ME 211
COSMOS: Complete Online Solutions Manual Organization SystemChapter 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 dt10 22 10 dv = 0 kt dtv10 10=13 kt 3 1t0[(
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
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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 )
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