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 $1\lt t\lt 2$.
Best Answer
The quantity $x(t) = 10t - 2t^2$ represents the position of the particle at time $t$. To find the displacement from $t = 0$ to $t = 3$, we subtract $x(0)$ from $x(3)$. However, the displacement is not equal to the distance if the particle changes direction (it does). To determine the direction of the particle, look at the sign of the velocity, $x'(t)$. Determine the time interval in which the particle is moving to the left and calculate its displacement during this interval. Do the same for the time interval in which it is moving to the right. To find the total distance traveled by the particle, add the absolute values of the displacements to the left and right.