Program Explanation Given information: The acceleration a ( t ) = 1 t + 4 and the initial position x 0 = 1 and the initial velocity v 0 = − 1 . 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 1 t + 4 . 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 ) = ∫ 1 t + 4 d t ........ (1) Now take u for t + 4 and then differentiate with respect to x as, u = t + 4 d u d t = 1 d u = d t Nowsubstitute u for t + 4 and d t for d u in equation (1) and integrate the equation on both the sides. ∫ 1 t + 4 d t = ∫ ( u ) − 1 2 d u Now evaluate the above integral as, v ( t ) = 2 ( u ) 1 2 + C 1 ........ (2) Now substitute t + 4 for u in equation (2). v ( t ) = 2 ( t + 4 ) 1 2 + C 1 ........ (3) Now put the initial condition v 0 = − 1 in equation (3) to obtain the value of constant C . − 1 = 2 ( 0 + 4 ) 1 2 + C 1 − 1 = 4 + C 1 C 1 = − 5 Now substitute − 5 for C 1 in equation (3) to obtain particular velocity equation

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ISBN: 9780135904206

Author: Edwards

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 1.2, Problem 16P

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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 is v0=v(0) .

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Chapter 1.2, Problem 15P

Chapter 1.2, Problem 17P

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Chapter 1 Solutions

DIFFERENTIAL EQUATIONS+..-MYLAB ACCESS

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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 is 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 is v0=v(0) .

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Chapter 1 Solutions