Homework Help: PHYSİCS

Posted by LARA on Monday, February 27, 2012 at 8:05am.

please any on help me for this question ı need urgently.

The position of a particle moving along the x-axis 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 x-direction?

PHYSICS - drwls, Monday, February 27, 2012 at 8:52am
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

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