That being said:

a) To describe the motion, the simplest thing to do is to graph the x(t) function. I'll do a Calculus based analysis of the motion, which pretty much answers all the subsequent questions.

x(t) = 2t^3 -13t^2 + 22t - 5

x'(t) = 6t^2 - 26t + 22

x"(t) = 12t - 26

The first thing to note is that at t = 0 s the particle is at x = -5 (m, I presume), so the particle is not starting from our origin. Also x'(0) = 22 so the particle is starting with a positive velocity, and since x"(0) = -26, this velocity is getting smaller.

b) x"(t) = 12t - 26

Set 12t - 26 = 0 ==> t = 13/6 s.

This is the time when the velocity switches over from being positive and goes negative. So the velocity is positive for [0, 13/6) and negative for (13/6, infinity).

c) To get a changing direction we need to see where the velocity is 0:

x'(t) = 6t^2 - 26t + 22

0 = 6t^2 - 26t + 22

0 = 3t^2 - 13t + 11

So t = (13 (+/-) sqrt(37))/6 s is when the velocity of the particle is 0. Now, we are taking t >=0 s, so we discard the negative solution and get t = (13 + sqrt(37))/6 s. So initially the velocity is positive and at this time the velocity switches and goes negative. Thus the particle changes direction at t = (13 + sqrt(37))/6 s.

(Note carefully I have not exactly proven a sign change exists in either b) or c). I leave that detail to you.)

d) The particle never "rests" as the velocity is always changing. However I suspect that the question is asking where v = 0. We already know this from the answer to c). The particle momentarily stops at t = (13 + sqrt(37))/6 s.

e) I'll leave the details of this to you. You have the velocity function x'(t) = 6t^2 - 26t + 22, which you can graph, find the zeros of, etc. The speed function is the magnitude of the velocity function: s(t) = |x'(t)| = |6t^2 - 26t + 22|

f) This problem can be solved numerically, but I wonder if we aren't looking for when x = -5? This has a much simpler solution.

Anyway, we are asking when x = 5. So solve

5 = 2t^3 -13t^2 + 22t - 5

0 = 2t^3 - 13t^2 + 22t - 10

I get t = 0.744721 s and and t = 1.62589 s and t = 4.12939 s as solutions.

-Dan