Question: The position of a particle between t = 0 and t = 2.00 s is given by x(t) = (3.00m/S3)t3 - (10.0m/s2 )t2 + (9.00m/s)t.

The position of a particle between t = 0 and t = 2.00 s is given by x(t) = (3.00m/S3)t3 - (10.0m/s2 )t2 + (9.00m/s)t.
(a) Draw the x-t, v.-t, and az-f graphs of this particle.
(b) At what time(s) between f = 0 and t = 2.00 s is the particle instantaneously at rest? Does your numerical result agree with the Vz-f graph in part (a)?
(c) At each time calculated in part (b) is the acceleration of the particle positive or negative? Show that in each case the same answer is deduced from a. (f) and from the Vz-f graph.
(d) At what time(s) between t = 0 and t = 2.00 s is the velocity of the particle instantaneously not changing? Locate this point on the vz-t and az-t graphs of part (a).
(e) What is the particle's greatest distance from the origin (x = 0) between f = 0 and t = 2.00 s? (f) At what time(s) between t = 0 and t = 2.00 s is the particle speeding up at the greatest rate? At what time(s) between t = 0 and t = 2.00 s is the particle slowing down at the greatest rate? Locate these points on the v.-t and a.-t graphs of part (a).

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