Find the acceleration vector?

Question

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.

Answer 1

Position: r = [t²+1, ln(2t+3)]

 

Velocity: v = dr/dt = [2t, 2/(2t+3)]

 

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

Answer 2

V [(dx/ dt = 2t , dy/dt= 2/ ( 2t+3)] / velocity        

 

 

 

a [ d²x / dt²= 2  , d²y/dt² = -4/ (2t + 3) ^2

                         

 

 

 Note :  U = ln t    du/dt = 1/t      d²u/dt² =  -1/ U²