Lots of questions! We have $s(t)=t^3-6t^2+9t$. So if the velocity is denoted by $v(t)$, we have
$$v(t)=s'(t)=3t^2-12t+9=3(t-1)(t-3).$$
The particle is moving to the right when the velocity is positive, and to the left when the velocity is negative.
Looking at $3(t-1)(t-3)$, we note that it is positive when $t\gt 3$, also when $t\lt 1$. So in the time interval $(-\infty,1)$ and in the time interval $(3,\infty)$, to the degree this makes physical sense, we have motion to the right. In the time interval $(1,3)$ we have motion to the left.
The acceleration $a(t)$ is the derivative of velocity. So $a(t)=6t-12$.
There is some possible ambiguity (or trick) in the question about speeding up. The velocity is increasing when the acceleration is positive, that is, when $t\gt 2$. The velocity is decreasing when $t\lt 2$.
You should be able to do the rest of the parts. But "total distance travelled in the first $5$ seconds" is tricky, so we do some detail.
The net change in displacement is easy, it is $s(5)-s(0)$. But for total distance travelled, we need to take account of the fact that we are travelling to the right when $t$ is between $0$ and $1$, also when $t$ is between $3$ and $5$, while between $1$ and $3$ we are travelling to the left. So while $s(1)-s(0)$ and $s(5)-s(3)$ are positive, the number $s(3)-s(1)$ is negative.
Thus the total distance travelled in the first $5$ seconds is
$$|f(1)-f(0)|+|f(3)-f(1)|+|f(5)-f(3)|.$$
Maybe Don't Read: Velocity is not the same thing as speed. The speed at time $t$ is the absolute value of velocity, so it is $3|(t-1)(t-3)|$. We may want to know when speed is increasing. That's a different question than asking when velocity is increasing.
To find out you where speed is increasing, you can find out where the derivative of $(3(t-1)(t-3))^2$ is positive. This derivative is $9(t-1)(t-3)(2t-4)$. It is not hard to find out where this is positive: for $t\gt 3$ and for $1\lt t\lt 2$.
You've got the right idea, but not the right execution. Let's call the direction of positive $x$ "up" and the opposed direction down.
It is moving upward--has a positive velocity--on $(0,1)$ and $(3,4)$. It is moving downward--has a negative velocity--on $(1,3).$ Put another way, the graph's height is increasing on $(0,1)$ as we move to the right--likewise in $(3,4)$--and decreasing on $(1,3)$ as we move to the right.
It is accelerating upward--has a positive acceleration--on $(2,4)$. It is accelerating downward--has a negative acceleration--on $(0,2)$. Put another way, the graph's slope is increasing on $(2,4)$ as we move to the right, and the graph's slope is decreasing on $(0,2)$ as we move to the right.
Now check where the signs match.
Best Answer
Velocity is a vector quantity, and indicates both speed (by its slope) and direction (by its sign). Speed is a scalar quantity, and represents, colloquially, how "fast" the particle is moving (distance over time). And because it doesn't matter in which direction the particle is moving, speed is represented by $|v|$. As Spencer commented, when velocity and acceleration are both positive or both negative, the speed is increasing. When they are different signs, then the speed is decreasing.
To see why, look at this portion of the graph of $x^3$ as x approaches 0. The particle's graph is going up for sure (positive velocity). However, the rate by which its increasing is decreasing (negative acceleration) -- hence why its increasing ever more gradually. In other terms, it's slowing down, because negative acceleration indicates a decreasing velocity.
The same would apply to the converse as well -- a positive acceleration and a negative velocity would mean a graph which is decreasing ever more slowly, because the velocity is still negative but is increasing. So the "graduating effects" are caused by opposite signs.
Conversely, when velocity and acceleration have the same sign, the graph either "shoots up" or "shoots down", which is what you are seeing here:
Either the velocity is increasing increasingly, or it is decreasing increasingly, depending on whether the signs are both positive or negative (this is a graph of positive signs).
The key point is that speed doesn't have regard for direction -- so whether or not the particle is going left or right is not of your concern. Hence the $|v|