Explanation Given information: The function of time is x = 2 t 3 − 9 t 2 + 12 t + 10 . Calculation: Write the relation for the motion of a particle: x = 2 t 3 − 9 t 2 + 12 t + 10 (1). Here, position of the particle is x and time is t . Differentiate Equation (1) with respect to time. v = d x d t = 6 t 2 − 18 t + 12 = 6 ( t 2 − 3 t + 2 ) = 6 ( t − 2 ) ( t − 1 ) (2). Here, velocity of the boarder is v . Differentiate Equation (2) with respect to time. a = d v d t = 12 t − 18 (3). Here, acceleration of the board is a . Substitute 0 for v in Equation (2). 0 = 6 ( t − 2 ) ( t − 1 ) The values for time ( t ) are 2 sec and 1 sec. Calculate the position ( x 1 ) of the particle when t is 1 sec: Substitute 1 sec for t in Equation (1). x 1 = 2 ( 1 ) 3 − 9 ( 1 ) 2 + 12 × 1 + 10 = 2 − 9 + 12 + 10 = 15 ft Calculate the position ( x 2 ) of the particle when t is 2 sec: Substitute 2 sec for t in Equation (1)

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9781259639272

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 11.1, Problem 11.2P

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The time (t), position (x) and acceleration (a) of the particle when t is 5 seconds.

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Chapter 11.1, Problem 11.1P

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Can you solve it analytically using laplace transforms and with Matlab code as well please. Thank You

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Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics

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The time ( t ), position ( x ) and acceleration ( a ) of the particle when t is 5 seconds.

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The time (t), position (x) and acceleration (a) of the particle when t is 5 seconds.

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Chapter 11 Solutions