Explanation Results used: Suppose an object moves along a line with a velocity (known) v ( t ) ,for t ≥ 0 . Then its position is found by solving the initial value problem s ( t ) = ∫ v ( t ) d t , s ( 0 ) = s 0 , where s 0 is the (known) initial position. If the acceleration of an object a ( t ) is given, then its velocity is found by solving the initial value problem v ( t ) = ∫ a ( t ) d t , v ( 0 ) = v 0 , where v 0 is the (known) initial velocity. Given: Acceleration of the moving object is a ( t ) = − 32 . Initial velocity of the moving object is v ( 0 ) = 20 . Initial position of the moving object is s ( 0 ) = 0 . Calculation: By the above results, the velocity of the object is obtained as follows. v ( t ) = ∫ a ( t ) d t = ∫ ( − 32 ) d t = − 32 ∫ d t = − 32 t + C Use the initial condition v ( 0 ) = 20 in the velocity function, 20 = − 32 × 0 + C = 0 + C 20 = C Substitute the value of C = 20 in the equation v ( t ) = − 32 t + C , v ( t ) = − 32 t + 20

Calculus, Single Variable: Early Transcendentals (3rd Edition)

3rd Edition

ISBN: 9780134766850

Author: William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz

Publisher: PEARSON

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Chapter 4.9, Problem 97E

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To find: The position function of the object which moving along a line with the acceleration function a(t)=−32 and with the initial velocity and position conditions v(0)=20 and s(0)=0 .

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Chapter 4.9, Problem 96E

Chapter 4.9, Problem 98E

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The position function of the object which moving along a line with the acceleration function a ( t ) = − 32 and with the initial velocity and position conditions v ( 0 ) = 20 and s ( 0 ) = 0 .

Solution Summary: The author explains the position function of an object moving along a line with the acceleration function a(t)=-32 and the initial velocity and position conditions v(0)=20.

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To find: The position function of the object which moving along a line with the acceleration function a(t)=−32 and with the initial velocity and position conditions v(0)=20 and s(0)=0 .

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Chapter 4 Solutions