Program Explanation Given information: The physical parameters is given as, m 1 = m 2 = c 2 = c = k = 1 c 1 = 0 The initial conditions are, x 1 ( 0 ) = x 2 ( 0 ) = 0 and x ′ 1 ( 0 ) = x ′ 2 ( 0 ) = v 0 Explanation: The equation of motion of cars is defined as, m 1 x ″ 1 = k ( x 2 − x 1 ) − c 1 x ′ 1 − c ( x ′ 1 − x ′ 2 ) m 2 x ″ 2 = k ( x 1 − x 2 ) − c 2 x ′ 2 − c ( x ′ 2 − x ′ 1 ) The characteristic equation for the above equation of motion is shown below. A = [ 0 0 1 0 0 0 0 1 − k k − ( c 1 + c 2 ) c k − k c − ( c 1 + c 2 ) ] Substitute the value of physical parameters to obtain the characteristic equation as, A = [ 0 0 1 0 0 0 0 1 − 1 1 − 1 1 1 − 1 1 − 2 ] The characteristics equation of the coefficient matrix is, [ 0 − λ 0 1 0 0 0 − λ 0 1 − 1 1 − 1 − λ 1 1 − 1 1 − 2 − λ ] = 0 λ ( λ + 1 ) 3 = 0 λ = 0 , − 1 , − 1 and − 1 The Eigen value − 1 has multiplicity 3 and eigenvalue 0 has multiplicity of 0. The Eigen vector corresponding to eigenvalue − 1 and 0 is obtained as, v 0 ( λ = 0 ) = [ 1 1 0 0 ] T v 1 ( λ = − 1 ) = [ 0 1 0 − 1 ] T The third eigenvector is obtained from the solution of the equation ( A + I ) 3 v 3 = 0 as, v 3 = [ 1 1 0 − 1 ] T The four linearly independent vector solutions for the given system is written as, x 1 ( t ) = v 0 e 0 t x 2 ( t ) = v 1 e − t x 3 ( t ) = ( v 1 t + v 2 ) e − t x 4 ( t ) = ( 1 2 v 1 t 2 + v 2 t + v 3 ) e − t Further simplified vector are written as, x 1 ( t ) = [ 1 1 0 0 ] T x 2 ( t ) = [ 0 1 0 − 1 ] T e − t x 3 ( t ) = [ 1 t −

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ISBN: 8220100802454

Author: Calvis

Publisher: PEARSON

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A naïve solution for the above-stated problem is to evaluate all of the possible combinations of different pieces of varying size to find the maximum revenue-generating combination. Although this method seems the easiest and quick to practice, it has exp…

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Chapter 5.5, Problem 36P

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Program Description:Purpose of problem is to obtain the position function x1(t) and x2(t) of the railway cars and determine the distance travel by cars before stop.

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Chapter 5.5, Problem 35P

Chapter 5.6, Problem 1P

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Chapter 5 Solutions

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Ch. 5.1 - Let A=[2347] and B=[3451]. Find (a) 2A+3B; (b)...Ch. 5.1 - Prob. 2P Ch. 5.1 - Find AB and BA given A=[203415] and B=[137032].Ch. 5.1 - Prob. 4P Ch. 5.1 - Prob. 5P Ch. 5.1 - Prob. 6P Ch. 5.1 - Prob. 7P Ch. 5.1 - Prob. 8P Ch. 5.1 - Prob. 9P Ch. 5.1 - Prob. 10P

Ch. 5.1 - Prob. 11P Ch. 5.1 - Prob. 12P Ch. 5.1 - Prob. 13P Ch. 5.1 - Prob. 14P Ch. 5.1 - Prob. 15P Ch. 5.1 - Prob. 16P Ch. 5.1 - Prob. 17P Ch. 5.1 - Prob. 18P Ch. 5.1 - Prob. 19P Ch. 5.1 - Prob. 20P Ch. 5.1 - Prob. 21P Ch. 5.1 - Prob. 22P Ch. 5.1 - Prob. 23P Ch. 5.1 - Prob. 24P Ch. 5.1 - Prob. 25P Ch. 5.1 - Prob. 26P Ch. 5.1 - Prob. 27P Ch. 5.1 - Prob. 28P Ch. 5.1 - Prob. 29P Ch. 5.1 - Prob. 30P Ch. 5.1 - Prob. 31P Ch. 5.1 - Prob. 32P Ch. 5.1 - Prob. 33P Ch. 5.1 - Prob. 34P Ch. 5.1 - Prob. 35P Ch. 5.1 - Prob. 36P Ch. 5.1 - Prob. 37P Ch. 5.1 - Prob. 38P Ch. 5.1 - Prob. 39P Ch. 5.1 - Prob. 40P Ch. 5.1 - Prob. 41P Ch. 5.1 - Prob. 42P Ch. 5.1 - Prob. 43P Ch. 5.1 - Prob. 44P Ch. 5.1 - Prob. 45P Ch. 5.2 - Prob. 1P Ch. 5.2 - Prob. 2P Ch. 5.2 - Prob. 3P Ch. 5.2 - Prob. 4P Ch. 5.2 - Prob. 5P Ch. 5.2 - Prob. 6P Ch. 5.2 - Prob. 7P Ch. 5.2 - Prob. 8P Ch. 5.2 - Prob. 9P Ch. 5.2 - Prob. 10P Ch. 5.2 - Prob. 11P Ch. 5.2 - Prob. 12P Ch. 5.2 - Prob. 13P Ch. 5.2 - Prob. 14P Ch. 5.2 - Prob. 15P Ch. 5.2 - Prob. 16P Ch. 5.2 - Prob. 17P Ch. 5.2 - Prob. 18P Ch. 5.2 - Prob. 19P Ch. 5.2 - Prob. 20P Ch. 5.2 - Prob. 21P Ch. 5.2 - Prob. 22P Ch. 5.2 - Prob. 23P Ch. 5.2 - Prob. 24P Ch. 5.2 - Prob. 25P Ch. 5.2 - Prob. 26P Ch. 5.2 - Prob. 27P Ch. 5.2 - Prob. 28P Ch. 5.2 - Prob. 29P Ch. 5.2 - Prob. 30P Ch. 5.2 - Prob. 31P Ch. 5.2 - Prob. 32P Ch. 5.2 - Prob. 33P Ch. 5.2 - Prob. 34P Ch. 5.2 - Prob. 35P Ch. 5.2 - Prob. 36P Ch. 5.2 - Prob. 37P Ch. 5.2 - Prob. 38P Ch. 5.2 - Prob. 39P Ch. 5.2 - Prob. 40P Ch. 5.2 - Prob. 41P Ch. 5.2 - Prob. 42P Ch. 5.2 - Prob. 43P Ch. 5.2 - Prob. 44P Ch. 5.2 - Prob. 45P Ch. 5.2 - Prob. 46P Ch. 5.2 - Prob. 47P Ch. 5.2 - Prob. 48P Ch. 5.2 - Prob. 49P Ch. 5.2 - Prob. 50P Ch. 5.3 - Prob. 1P Ch. 5.3 - Prob. 2P Ch. 5.3 - Prob. 3P Ch. 5.3 - Prob. 4P Ch. 5.3 - Prob. 5P Ch. 5.3 - Prob. 6P Ch. 5.3 - Prob. 7P Ch. 5.3 - Prob. 8P Ch. 5.3 - Prob. 9P Ch. 5.3 - Prob. 10P Ch. 5.3 - Prob. 11P Ch. 5.3 - Prob. 12P Ch. 5.3 - Prob. 13P Ch. 5.3 - Prob. 14P Ch. 5.3 - Prob. 15P Ch. 5.3 - Prob. 16P Ch. 5.3 - Prob. 17P Ch. 5.3 - Prob. 18P Ch. 5.3 - Prob. 19P Ch. 5.3 - Prob. 20P Ch. 5.3 - Prob. 21P Ch. 5.3 - Prob. 22P Ch. 5.3 - Prob. 23P Ch. 5.3 - Prob. 24P Ch. 5.3 - Prob. 25P Ch. 5.3 - Prob. 26P Ch. 5.3 - Prob. 27P Ch. 5.3 - Prob. 28P Ch. 5.3 - Prob. 29P Ch. 5.3 - Prob. 30P Ch. 5.3 - Prob. 31P Ch. 5.3 - Prob. 32P Ch. 5.3 - Prob. 33P Ch. 5.3 - Verify Eq. 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Program Description: Purpose of problem is to obtain the position function x 1 ( t ) and x 2 ( t ) of the railway cars and determine the distance travel by cars before stop.

Solution Summary: The author explains that the equation of motion of railway cars is defined as, lm_1x

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Program Description:Purpose of problem is to obtain the position function x1(t) and x2(t) of the railway cars and determine the distance travel by cars before stop.

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Chapter 5 Solutions