Posted by **Lauren** on Thursday, September 10, 2009 at 5:08pm.

a particle starts at time t = 0 and moves along the x axis so that its position at any time t>= 0 is given by x(t) = ((t-1)^3)(2t-3)

a.find the velocity of the particle at any time t>= 0

b. for what values of t is the velocity of the particle negative?

c. find the value of t when the particle is moving and the acceleration is zero. explain your answer choice

- Calculus -
**Reiny**, Thursday, September 10, 2009 at 7:11pm
a) find x'(t) using the product rule, that will be your velocity

b) set x'(t) from above < 0 and solve

c) acceleration is the derivative of velocity, so differentiate x'(t) again, and set it equal to zero

- Calculus -
**drwls**, Thursday, September 10, 2009 at 7:12pm
a) Take the deriviative of x(t)

b) Set the derivative = 0 and solve the equation for t

c) Take the derivative of v(t) and set that equal to zero. If there is more than one answer, take the one for which v is not zero.

We will be glad to critique your work.

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