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 $1\lt t\lt 2$.
$v=v_0+at$ since when $t=2$s we have $v=10$m/s then $v_0=4$m/s
So it took $2$s to increase from $4$m/s to $10$m/s
According to the formula
$s=\dfrac{1}{2}at^2+v_0t$
we plug the data and get
$s=14$m
Hope this helps
Best Answer
Consider the situation in the inertial frame of reference where you're initially at the origin with zero velocity. (Note that the acceleration constraint is the same in this moving frame.) The destination point is initially at $(d,0)$, and after time $t$ it will be at $(d,-V_0 t)$. On the other hand, the farthest you can get from the origin in time $t$ is $\frac{1}{2}a_0 t^2$, achievable through maximum acceleration in any fixed direction. So the minimum time is just the minimum positive solution to $\frac{1}{2}a_0t^2=\sqrt{d^2 + V_0^2 t^2}$. Squaring both sides, this becomes a quadratic in $t^2$: $\frac{1}{4}a_0^2 t^4 - V_0^2 t^2 - d^2 = 0$, with solutions $$ t^2 = \frac{V_0^2 \pm \sqrt{V_0^4 + a_0^2 d^2}}{a_0^2/2}. $$ If $d\neq 0$, then only the $+$ sign leads to a solution for $t$: $$ t^{*}=\sqrt{2\left(\frac{V_0}{a_0}\right)^2 + 2\sqrt{\left(\frac{V_0}{a_0}\right)^4 + \left(\frac{d}{a_0}\right)^2}} = \sqrt{2\alpha^2+2\sqrt{\alpha^4 + \beta^4}}=2\alpha\sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{1 + (\beta/\alpha)^4}} $$ where $\alpha=V_0/a_0$ and $\beta=\sqrt{d/a_0}$ are the two parameters in the problem with dimensions of time. And, of course, constant acceleration leads you along a parabolic trajectory in the original frame.