Program Explanation Given information: The acceleration a ( t ) = 50 sin 5 t and the initial position x 0 = 8 and the initial velocity v 0 = − 10 . 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 The integration of cos x is sin x . Calculation: The constant acceleration function can be expressed as, d v d t = a ( t ) The function a ( t ) is 50 sin 5 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 ) = ∫ ( 50 sin 5 t ) d t ........ (1) Now take u for 5 t and then differentiate with respect to x as, u = 5 t d u d t = 5 d u 5 = d t Nowsubstitute u for 5 t and d t for d u in equation (1) and integrate the equation on both the sides. ∫ ( 50 sin 5 t ) d t = ∫ ( 50 sin u ) d u 5 = 10 cos u + C 1 Therefore, the velocity can be expressed as, v ( t ) = 10 cos u + C 1

5th Edition

ISBN: 9780321974235

Author: Calvis

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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Chapter 1.2, Problem 18P

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 17P

Chapter 1.2, Problem 19P

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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