Write $|v(t)|$ as a piecewise function:
$$
|v(t)|= \cases{5-3t, & $0\le t\le 5/3$ \cr 3t-5, & $5/3\le t\le 3$ }
$$
Then, split the integral $\int_0^3 |v(t)|\,dt $ into two pieces.
$$
\int_0^3 |v(t)|\,dt =\int_0^{5/3} |v(t)|\,dt +\int_{5/3}^3 |v(t)|\,dt =\int_0^{5/3} 5-3t\,dt +\int_{5/3}^3 3t-5\,dt,
$$
and evaluate.
Essentially, find the set over which $v$ is positive and the set over which $v$ negative; then integrate $|v|$ over these sets separately. On the set where $v$ is negative, you'd integrate $-v$ and over the set where $v$ is positive, you'd integrate $v$.
The difference between displacement and total distance should be clear: here the point travels to the left the first $5/3$ seconds then travels to the right from $t=5/3$ to $t=3$.
The displacement is the difference between the final and initial position of the point. Since the point traveled left and then right, this will be less than the total distance traveled (which is the sum of the distance traveled left with the distance traveled right).
For example if you move along a line left $4$ units then right $5$ units, the displacement is $1$ and the total distance traveled is $9$.
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
You should integrate the absolute value of velocity from 0 to 3. Than you get the desired result.