Explanation Given info: The mass of the object is m , the force component in the x direction at t = 0 is F x ( t ) = k 1 + k 2 y and the force component in the y direction at t = 0 is F y ( t ) = k 3 t , The mass of the object is m , the force component in the x direction at t = 0 is F x ( t ) = k 1 + k 2 y , the force component in the y direction at t = 0 is F y ( t ) = k 3 t , the position vector as a function of time in y direction is 1 m k 3 t 3 6 and the velocity vector in x direction is k 1 t m + k 2 1 m 2 k 3 t 4 24 . Explanation: The velocity vector of the given object is given as: v → ( t ) = v x ( t ) i ^ + v y ( t ) j ^ (I) v → ( t ) is the velocity vector as a function of time. v x ( t ) is the velocity vector in the x direction. v y ( t ) is the velocity vector in the y direction. Formula to calculate the force along the y direction is: F y ( t ) = m d v y ( t ) d t (II) F y ( t ) is the force along the y direction. d v y ( t ) d t is the acceleration of the object along the y direction. Substitute k 3 t for F y ( t ) in the equation (II) to find v y → ( t ) : k 3 t = m d v y ( t ) d t The integration of the above equation is: ∫ ( k 3 t ) d t = ∫ ( m d v y ( t ) d t ) d t [ k 3 t 2 2 ] = m [ v y ( t ) ] v y ( t ) = 1 m k 3 t 2 2 + C 1 (III) At t = 0 The velocity vector in y direction is: v y ( 0 ) = 1 m k 3 ( 0 ) 2 2 + C 1 C 1 = 0 Substitute 0 for C 1 in the equation (III) to find v y ( t ) : v y ( t ) = 1 m k 3 t 2 2 + 0 v y ( t ) = 1 m k 3 t 2 2 Thus the velocity vector in the y direction is 1 m k 3 t 2 2 . Formula to calculate the force along the x direction is: F x ( t ) = m d v x ( t ) d t (IV) F x ( t ) is the force along the x direction. d v x ( t ) d t is the acceleration of the object along the x direction. Substitute k 1 + k 2 y for F x ( t ) in the equation (IV): k 1 + k 2 y = m d v x ( t ) d t The integration of the above equation is: ∫ ( k 1 + k 2 y ) d x = ∫ ( m d v x ( t ) d t ) d t (V) Formula to calculate the velocity of the object in y direction is: v y ( t ) = d y d t Substitute d y d t for v y ( t ) in the equation (III) to find y : d y d t = 1 m k 3 t 2 2 The integration of the above equation is: ∫ ( d y d t ) d t = ∫ ( 1 m k 3 t 2 2 ) d t y ( t ) = 1 m k 3 t 3 6 + C 2 (VI) At t = 0 The position vector in y direction is: y ( 0 ) = 1 m k 3 ( 0 ) 3 6 + C 2 C 2 = 0 Substitute 0 for C 2 in the equation (VII)n to find y ( t ) : y ( t ) = 1 m k 3 t 3 6 + 0 y ( t ) = 1 m k 3 t 3 6 Thus the position vector in y direction is 1 m k 3 t 3 6

University Physics with Modern Physics (14th Edition)

14th Edition

ISBN: 9780321973610

Author: Hugh D. Young, Roger A. Freedman

Publisher: PEARSON

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Chapter 4, Problem 4.56CP

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To determine: The position and velocity vector as the function of time.

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Chapter 4 Solutions

University Physics with Modern Physics (14th Edition)

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The position and velocity vector as the function of time.

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Chapter 4 Solutions