Search
Ask a question
0 0

Find the acceleration vector?

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.
Tutors, please sign in to answer this question.

2 Answers

Position: r = [t2+1, ln(2t+3)]
 
Velocity: v = dr/dt = [2t, 2/(2t+3)]
 
Acceleration: a = dv/dt = [2, -4/(2t+3)²]

Comments

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