posted by Lauren on .
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
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
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.
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