Posted by LARA on .
please any on help me for this question ý need urgently.
The position of a particle moving along the xaxis is given by x = 3.49t2 – 2.21t3, where x is in meters and t is in seconds. What is the position of the particle when it achieves its maximum speed in the positive xdirection?

PHYSICS 
drwls,
x = 3.49t^2 – 2.21t^3
V = dx/dt = 6.98 t  6.63t^2
At maximum V, dV/dt = 0
6.98 = 13.26 t
t = 0.5264 s @ maximum V
X @t=0.5264 = 0.6447 m