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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.

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Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
5.0 5.0 (3 lesson ratings) (3)
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Position: r = [t2+1, ln(2t+3)]
 
Velocity: v = dr/dt = [2t, 2/(2t+3)]
 
Acceleration: a = dv/dt = [2, -4/(2t+3)²]

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Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...
4.8 4.8 (4 lesson ratings) (4)
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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