Explanation Given: The given acceleration vector is and initial conditions a → ( t ) = − 32 k ^ ; v → ( 0 ) = 3 i ^ − 2 j ^ + k ^ , r → ( 0 ) = 5 j ^ + 2 k ^ Formula used: The integrations formula is: v ( t ) = ∫ a → ( t ) d t r → ( t ) = ∫ v → ( t ) d t Calculation: The velocity vector is v ( t ) = ∫ a → ( t ) d t = − ∫ 32 k ^ . d t = − 32 t k ^ + c → Where c → = c 1 i ^ + c 2 j ^ + c 3 k ^ Applying condition v → ( 0 ) = 3 i ^ - 2 j ^ + k ^ we get v → ( 0 ) = c 1 i ^ + c 2 j ^ + c 3 k ^ = 3 i ^ − 2 j ^ + k ^ ⇒ c 1 = 3 , c 2 = − 2 , c 3 = 1 So the velocity at time t is v → ( t ) = 3 i ^ − 2 j ^ + ( 1 − 32 t ) k ^ By integrating we get, r → ( t ) = ∫ v → ( t ) d t = 3 t

10th Edition

ISBN: 9781285057095

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 12.3, Problem 22E

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To calculate: The velocity and position vector at t=0, position vector at t=2 for given acceleration vector a→(t)=−32k^.

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Chapter 12.3, Problem 21E

Chapter 12.3, Problem 23E

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The velocity and position vector at t = 0 , position vector at t = 2 for given acceleration vector a → ( t ) = − 32 k ^ .

Solution Summary: The author explains how to calculate the velocity and position vector at t=0 — the acceleration vector is 3stackreli-2

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To calculate: The velocity and position vector at t=0, position vector at t=2 for given acceleration vector a→(t)=−32k^.

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Chapter 12 Solutions