Program Explanation Given information: The acceleration a ( t ) = − 20 and the initial position x 0 = 5 and the initial velocity as v 0 = − 15 . 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 Now integrate the equation on both the sides. ∫ d v = ∫ a d t v = a t + C 1 The function v ( t ) represents the velocity. Therefore, the general velocity function v ( t ) can be expressed as, v ( t ) = a t + C 1 ......... (1) Now substitute − 20 for a and 0 for t in equation (1) to obtain the value of constant C . v ( 0 ) = − 20 ( 0 ) + C 1 ........ (2) Now put the initial condition v 0 = − 15 in equation (1) to obtain the value of constant C . − 15 = − 20 ( 0 ) + C 1 − 15 = C 1 Now substitute − 15 for C 1 and − 20 for a in equation (1) to obtain particular velocity equation

MyLab Math with Pearson eText -- 24-Month Standalone Access Card -- For Differential Equations and Boundary Value Problems: Computing and Modeling Tech Update

5th Edition

ISBN: 9780134872971

Author: Edwards, C., Penney, David, Calvis

Publisher: PEARSON

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Chapter 1.2, Problem 12P

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 as v0=v(0) .

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Chapter 1.2, Problem 11P

Chapter 1.2, Problem 13P

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Chapter 1 Solutions

MyLab Math with Pearson eText -- 24-Month Standalone Access Card -- For Differential Equations and Boundary Value Problems: Computing and Modeling Tech Update

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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 as 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 as v0=v(0) .

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Chapter 1 Solutions