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uate the position, velocity and acceleration when t = 3 seconds.

uate the position, velocity and acceleration when t = 3 seconds.

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
1. The velocity of a particle which moves along the a linear reference axis is given by v = 2—4t-6t³, † is in seconds while v is in meters per second. Eval- uate the position, velocity and acceleration when t = 3 seconds. Assume an your own initial position and initial point in time. Further, set a variable for position as you see fit.
Transcribed Image Text:1. The velocity of a particle which moves along the a linear reference axis is given by v = 2—4t-6t³, † is in seconds while v is in meters per second. Eval- uate the position, velocity and acceleration when t = 3 seconds. Assume an your own initial position and initial point in time. Further, set a variable for position as you see fit.
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can you please show the integration in the position? And can you please make it look like this? Thanks

 

GIVEN: V-2-4t - St t 3 SECONDS SOLUTION : v=2-4t-5t V= ds dt s=vSdt s. S(2-4t-st³) dt S= S-2t -2+²-5 (²3) + ² + c S = 2t-2+²-2t² + c S = 2(t-t²- + ³ + c) dv dt a = a : (2-4t-5t 2/2 ) dt 4-5 ( ²2/2) + 2/ - a = ·4-15 + 2/ REQUIRED: S = ? V = ? a = ? @t3 SECONDS 52 [3-(3) ² (3) ¾ ] +3 S= - 40.18 m v=2-[4(3)]-5 (3) ² v=-35.98 m/s a = -4-15/+2 a = - 4 - [ ¹1/21 (3) ¹² ] a = -16.99 m/s ² 2
Transcribed Image Text:GIVEN: V-2-4t - St t 3 SECONDS SOLUTION : v=2-4t-5t V= ds dt s=vSdt s. S(2-4t-st³) dt S= S-2t -2+²-5 (²3) + ² + c S = 2t-2+²-2t² + c S = 2(t-t²- + ³ + c) dv dt a = a : (2-4t-5t 2/2 ) dt 4-5 ( ²2/2) + 2/ - a = ·4-15 + 2/ REQUIRED: S = ? V = ? a = ? @t3 SECONDS 52 [3-(3) ² (3) ¾ ] +3 S= - 40.18 m v=2-[4(3)]-5 (3) ² v=-35.98 m/s a = -4-15/+2 a = - 4 - [ ¹1/21 (3) ¹² ] a = -16.99 m/s ² 2
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