For any time t≥0, if the position of a particle in the xy-plane is given by x=t^2+1 and y=ln(2t+3), find the acceleration vector. Answer: (2, -4/(2t+3)^2) Please show all your work. Feb 1 | Sun from Los Angeles, CA | 2 Answers | 0 Votes Mark favorite Subscribe Comment
Position: r = [t^{2}+1, ln(2t+3)] Velocity: v = dr/dt = [2t, 2/(2t+3)] Acceleration: a = dv/dt = [2, -4/(2t+3)²] Feb 1 | Andre W. Comment Comments Thanks. Feb 1 | Sun from Los Angeles, CA Comment
V [(dx/ dt = 2t , dy/dt= 2/ ( 2t+3)] / velocity a [ d^{2}x / dt^{2}= 2 , d^{2}y/dt^{2} = -4/ (2t + 3) ^2 ^{ } Note : U = ln t du/dt = 1/t d^{2}u/dt^{2} = -1/ U^{2} ^{ } Feb 1 | Parviz F. Comment
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