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 $1\lt t\lt 2$.
Let denote the way the runner is passing while he's accelerating $s_1$, because we want to find the average acceleration we have:
$$s_1 = v_0t + \frac{at_1^2}{2}$$
The initial velocity is 0, so we left out the first term. Also we make a change $a = \frac{\Delta v}{t_1}$
$$s_1 = \frac{\Delta vt_1}{2}$$
We know that $\Delta v = v - v_0$, where $v$ is the maximum reached velocity. We know that $v_0 = 0$, so $\Delta v = v$. Now we get:
$$s_1 = \frac{vt_1}{2}$$
Know let's denote the second part of the way, when the velocity is constant as $s_2$. We have:
$$s_2 = v(t-t_1)$$
We know that the sum of the two distances is $100m$ so we have:
$$s_1 + s_2 = 100$$
$$\frac{vt_1}{2} + v(t-t_1) = 100$$
$$\frac{4.1 \cdot v}{2} + v(9-4.1) = 100$$
$$2.05 \cdot v + 4.9 \cdot v = 100$$
$$6.95 \cdot v = 100$$
$$v = \frac{100m}{6.95s}$$
$$v \approx 14.39m/s$$
Now we substitute back:
$$a = \frac{\Delta v}{t_1} = \frac{v}{t_1} = \frac{14.39}{4.1} \approx 3.51 \frac{m}{s^2}$$
Best Answer
Acceleration is the derivative of the velocity, $a(t) = \frac{dv}{dt}$, so you can find an expression for the velocity by integrating and then do the same averaging procedure that you did for the acceleration.