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### hwB37

Course: ENEE 241, Spring 2008
School: Maryland
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Word Count: 429

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241 ENEE 02 HOMEWORK ASSIGNMENT 37 Due Mon 05/12 A FIR lter acts on the input sequence given by x[0 : 5] = 1 3 5 2 6 2 T ; x[n] = 0 for n &lt; 0 and n &gt; 5 to produce the output sequence given by y [0 : 11] = 2 3 0 10 23 27 61 51 54 14 2 T and y [n] = 0 for n &lt; 0 and n &gt; 11. (i) (2 pts.) What is the length of the lter coecient vector b? (As usual, it is assumed that the rst and last...

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241 ENEE 02 HOMEWORK ASSIGNMENT 37 Due Mon 05/12 A FIR lter acts on the input sequence given by x[0 : 5] = 1 3 5 2 6 2 T ; x[n] = 0 for n < 0 and n > 5 to produce the output sequence given by y [0 : 11] = 2 3 0 10 23 27 61 51 54 14 2 T and y [n] = 0 for n < 0 and n > 11. (i) (2 pts.) What is the length of the lter coecient vector b? (As usual, it is assumed that the rst and last entries in b are nonzero.) (ii) (6 pts.) Use the FFT command in MATLAB to determine the coecient vector b. (Alternatively, b can be determined by solving a triangular system of linear equations.) (iii) (6 pts.) Without performing a convolution, determine the response of the lter to the input sequence x(1) [ ] whose nonzero entries are x(1) [0 : 11] = 1 3 5 2 6 2 1 3 5 2 6 2 T (iv) (6 pts.) Without performing a convolution, determine the response of the lter to the input sequence x(2) [ ] whose nonzero entries are x(2) [0 : 9] = 2 6 10 4 11 7 5 2 6 2 T Solved Examples S 37.1 (P 4.21 in textbook). Consider the FIR lter y [n] = x[n] 2x[n 1] x[n 2] + 4x[n 3] x[n 4] 2x[n 5] + x[n 6] The response of the lter to the input : x[0 3] = (x[] = 0 otherwise) is given by y [0 : 9] = 1 1 0 7 8 13 20 1 11 4 T a0 a1 a2 a3 T (y [] = 0 otherwise); while the response to x[0 : 3] = ([] = 0 otherwise) is given by x y [0 : 9] = 1 2 2 10 8 10 18 2 9 4 T c0 c1 c2 c3 T ([] = 0 otherwise). Determine the response of the lter to y x[0 : 7] = ([] = 0 otherwise). x a0 a1 a2 a3 2c0 2c1 2c2 2c3 T S 37.2 (P 4.24 in textbook). Consider the following MATLAB code: a b A B C c = = = = = = [ 2 -1 0 -1 2 ]. ; [ 3 -2 5 1 -1]. ; fft(a,9); fft(b,9); A.*B; ifft(C); (i) Without using MATLAB, determine the vector c. (ii) Without performing a new convolution, determine the vector e obtained by the following code: a d A D E e = = = = = = [ 2 -1 0 -1 2 ]. ; [-3 2 -5 -1 1 3 -2 5 1 -1]. ; fft(a,14); fft(d,14); A.*D; ifft(E); S 37.3 (P 4.25 in textbook). The data le s7.txt contains a vector s of length L = 64. Let b= 2 1 3 1 3 1 3 1 2 T (i) Use the CONV function in MATLAB to compute the convolution c = b s. What is the length of c? (ii) Compute c using three convolutions of output length 25 = 32, followed by a summation. Your answer should be the same as for part (i).
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Maryland - ENEE - 241
HW 1 SOLUTION_(i) Plot (-1,sqrt(3) and (1,-1) on the (x,y) plane.(ii) |z1| = sqrt(-1)^2+(sqrt(3)^2) = sqrt(1+3) = 2angle(z1) = arctan(-sqrt(3) + pi = 2*pi/3 = 2.0944|z2| = sqrt(1+1) = sqrt(2) = 2.828angle(z2) = arctan(-1/1) = -pi/4(iii) z1^3 =
Maryland - ENEE - 241
HW 2 SOLUTION_(i) The polar coordinates are |u| = 1, angle(u) = pi/7 |v| = 1, angle(v) = 6*pi/7 |w| = 1, angle(u) = 2*pi/5(ii) |z-u| = |z-v| is tha perpendicular bisector of the segmentjoining u and v. This straight line is the y-axis (x = 0) (ii
Maryland - ENEE - 241
HW 3 Solution_(i) We know that the value 2.5 occurs at t=0, 0.160 and 0.180 sec andat no other intermediate times. Since a sinusoid crosses any givenlevel (except its maximum and minimum) twice in each period, it followsthat the period T equals exact
Maryland - ENEE - 241
HW 4 Solution_(i) For x[.], we have w=3*pi/4 = (3/8)*(2*pi). Thusthe fundamental period of x[.] equals N = 8.N = 8;n = 0 : 4*N-1;x = cos(3*pi*n/4 + 2*pi/3);stem(n,x)(ii) The values of w such that cos(w*n) is periodic with fundamental period N
Maryland - ENEE - 241
HW 5 Solution_(i) Clearly, A = 4. The period T equals 0.7 sec, therefore f = 10/7 Hz andW = 20*pi/7cos(phi) = 2*sqrt(3)/A = sqrt(3)/2, therefore phi = pi/6 or -pi/6. By inspection, t=0 corresponds to a point in the last quarter-cycle of the cosine wa
Maryland - ENEE - 241
Maryland - ENEE - 241
HW 7 Solution _ (i) A = [-4 3 0 2 -2 -1 1 2 2 ] A*[1 ; 0 ; 1] = [-4 ; 1 3] (ii) (a) B*x = [2; 3; 1]. B is a transformation matrix that circularly shifts the elements of the vector x. (b) B^2*x = [3; 1; 2]. Multiplying by B circularly shifts t
Maryland - ENEE - 241
HW 8 Solution_(i) The columns of A are obtained by right-multiplying A by thethree unit vectors.We have[1 ; -1 ; 0] + [2 ; 1 ; 0] = [3 ; 0 ; 0]thereforeA*[3 ; 0 ; 0] = A*[1 ; -1 ; 0] + A*[2 ; 1 ; 0]= [-4 ; -5 ; -6] + [ 1 ; -4 ; 0] = [-3 ; -9 ;
Maryland - ENEE - 241
HW 9 Solution_Note that the dimensions of P and Q can be inferred from the result,i.e., if P*A*Q is m*n, then P is m*3 and Q is 4*n.(i) P = 0 0 10 1 01 0 0Q = I(ii) P = I Q =0 0 0 10 1 0 00 0 1 01 0 0 0(iii) P = 0 0 11 0 00
Maryland - ENEE - 241
HW 10 Solution_(i) To solve U*x = b for arbitrary x = [x1 x2 x3 x4].' and b = [b1 b2 b3 b4].':x4 = b4 -2*x4 + x3 = b3 ; therefore x3 = 2*b4 + b3x2 - 2*x3 + x4 = b2 ; therefore x2 = -b4 + 2*(2*b4+b3) + b2 = b2 + 2*b3 + 3*b4x1 - 2*x2 + x3 = b1
Maryland - ENEE - 241
HW 11 Solution_Forward elimination:m x1 x2 x3 x4b_210-1-2-2401-23-1/21-3/263/235-3697-323m x1 x2 x3 x4b_210-1-20-2107-10-262363067029m x1 x2 x3 x4b_210-1-20-2107005229-20010
Maryland - ENEE - 241
HW 12 Solution_m x1 x2 x3 x4 _ 1* 4 4 4-2 2 2 0 2-1 1 1 3 3-1 1 3 1 2_ 1 4 4 4 0-6* -8-6-1/2 0-3-1-1-1/6 0-1-3-2_ 1 4 4 4 0-6 -8-6 0 0 3* 2 5/9 0 0-5/3 -1_ 1 4 4 4 0-6 -8-6 0 0 3
Maryland - ENEE - 241
HW 13 Solution_Matrix A_p m x1 x2 x3 x4_* 1* 1 1 1 -1 1 -1 1 -1 -1 1 2 4 8 -1 1 -3 6 -9_* 1 1 1 1* 0 -2* 0 -2 1/2 0 1 3 7 -2 0 -4 5 -10_* 1 1 1 1* 0 -2 0 -2* 0 0 3* 6 -5/3 0 0 5 -6_* 1 1 1 1* 0 -2 0 -2* 0 0 3 6 0 0 0
Maryland - ENEE - 241
HW 14 Solution_(i) |v1| = |v2| = sqrt(1 + 8 + 16) = 5&lt;v1,v2&gt; = 4-8+4 = 0(ii) &lt;v1,a&gt; = 3+12+12 = 27, &lt;v2,a&gt; = 12-12+3 = 3|a| = sqrt(9 + 18 + 9) = sqrt(36) = 6f1 = ( &lt;v1,a&gt; / |v1|^2 )*v1 = (27/25)*v1 Angle between a and v1 = arccos(27/(6*5) = 0.4
Maryland - ENEE - 241
HW 15 Solutions _ v1 = [ 5 3 -1 1 ].' v2 = [ 5 -1 7 3 ].' v3 = [ 1 1 -10 0 ].' s = [ 3 -1 4 4 ].' (i) We form the matrix V = [v1 v2]: V = [ 5 5 3 -1 -1 7 1 3 ] Then V'*V = [ 36 18 18 84 ] V'*s = [ 12 56 ] Solve (V'*V)*c
Maryland - ENEE - 241
HW 17 Solution_(i) Using microseconds (and noting that 1 MHz = 1/(1 microsecond) :f1 = 3 ; f2 = 2.5; gamma = 1.45t = (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] ;
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