**r**= [t

^{2}+1, ln(2t+3)]

**v**= d

**r**/dt = [2t, 2/(2t+3)]

**a**= d

**v**/dt = [2, -4/(2t+3)²]

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.

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Position: **r** = [t^{2}+1, ln(2t+3)]

Velocity: **v** = d**r**/dt = [2t, 2/(2t+3)]

Acceleration: **a** = d**v**/dt = [2, -4/(2t+3)²]

Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...

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}

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