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 $1\lt t\lt 2$.
Your part $(ii)$ is wrong. For instance, assuming $t \geq 0$, $x'(t) < 0$ on the interval $(\frac{1}{2},+\infty)$, not $(0, \frac{1}{2})$. Have a look at this part again, then you can fix part $(iii)$.
Best Answer
A handout from my vector calculus class.
Consider the social network of seven individuals
with the unimaginative names $A,B,C,D,E,F$ and $G$. An edge connects each pair of friends. This network or graph consists of two smaller, distinct graphs or components.
Question: How to write an algorithm to suggest that person $B$ befriend $D$?
The computer program should analyze the two components $\{A,B,C,D\}$ and $\{E,F,G\}$, identify that person $B$ is in the first component and then step through that list to find people to whom person $B$ is not currently linked. To do this, the computer will be fed the graph Laplacian, a matrix defined via the formula: \begin{equation*} L = (a_{ij}) = \begin{cases} \text{degree of vertex $i$ along the diagonal} \\ \text{$-1$ when an edge connects vertices $i$ and $j$}. \end{cases} \end{equation*} For the network of seven friends, the Laplacian matrix looks like: \begin{equation} L = \begin{bmatrix} 3 & -1 & -1 & -1 & 0 & 0 & 0 \\ -1 & 2 & -1 & 0 & 0 & 0 & 0 \\ -1 & -1 & 3 & -1 & 0 & 0 & 0 \\ -1 & 0 & -1 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & -1 & 2 & -1 \\ 0 & 0 & 0 & 0 & 0 & -1 & 1 \end{bmatrix} \end{equation} where rows are in alphabetical order.
Question: How to determine the components of the graph using this matrix?
Note that the vector $\begin{bmatrix} 1 & 1 & 1 & 1 & 0 & 0 & 0 \end{bmatrix}^T$ is in the nullspace of $L$ and this vector corresponds to the first component. Can you find a second vector in the nullspace? In general, these vectors associated with the components form a basis for the nullspace (and this isn't difficult to prove). So if you find the basis for $N(L)$, you've found the components of the original graph.
In real life, graphs aren't as simple as the one pictured above. In fact, the graph may consist of one giant component with tightly clustered "approximate components" embedded within. (See any of the images in this search.) And if the graph does have a lot of components, there are more computationally efficient methods of finding them. So why introduce the graph Laplacian? It turns out that the graph Laplacian is a basic object in the field of spectral clustering, which has numerous "real life" applications. In fact, I actually used the technique at a previous job while analyzing a large dataset.
I should point out that in spectral graph theory, you analyze all eigenvalues of the graph Laplacian not just $\lambda=0$, as we have done.
Question: What does this have to do with derivatives?
If you take multi-variable calculus, you may learn about the Laplace operator $\Delta = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$, which I have written in two dimensions. You may not believe it, but there's actually a connection between the Laplace operator and the graph Laplacian which can be explained via the discrete Laplacian!