Program Explanation Given information: The acceleration a ( t ) = 3 t and the initial position x 0 = 0 and the initial velocity v 0 = 5 . Concept used: The differential equation d y d x = f ( x ) g ( y ) can be calculated with the help of separating the variables as, ∫ d y g ( y ) = ∫ f ( x ) d x Calculation: The constant acceleration function can be expressed as, d v d t = a ( t ) The function a ( t ) is 3 t . The function v ( t ) represents the velocity function. Now integrate the equation on both the sides to obtain the general velocity function v ( t ) . v ( t ) = ∫ ( 3 t ) d t v ( t ) = 3 t 2 2 + C 1 Now put the initial condition v 0 = 5 in equation (3) to obtain the value of constant C . 5 = 3 ( 0 ) 2 2 + C 1 C 1 = 5 Now substitute 5 for C 1 in equation (3) to obtain particular velocity equation

5th Edition

ISBN: 9780321974235

Author: Calvis

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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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…

Question

Chapter 1.2, Problem 13P

Program Plan Intro

To find: The position function x(t) of a moving particlehas the acceleration a(t) and the initial position x0=x(0) and the initial velocity v0=v(0) .

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Chapter 1.2, Problem 12P

Chapter 1.2, Problem 14P

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Chapter 1 Solutions

EBK DIFFERENTIAL EQUATIONS

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The position function x ( t ) of a moving particlehas the acceleration a ( t ) and the initial position x 0 = x ( 0 ) and the initial velocity v 0 = v ( 0 ) .

Solution Summary: The author explains how the position function x(t) of a moving particle has the acceleration and the initial position.

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To find: The position function x(t) of a moving particlehas the acceleration a(t) and the initial position x0=x(0) and the initial velocity v0=v(0) .

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Chapter 1 Solutions