#### Answer

The detailed explanations are below.

#### Work Step by Step

$s=t^2-3t+2$, $0\le t\le5$
1) Find $v$ and $a$:
$$v=\frac{ds}{dt}=\frac{d}{dt}(t^2-3t+2)=2t-3$$
$$a=\frac{dv}{dt}=\frac{d}{dt}(2t-3)=2$$
The graphs of $s, v$ and $a$ are shown below.
2) We examine each element of the graph:
The curve $s$:
- The curve $s$ starts from $2$ at $t=0$, decreases constantly before reaching its minimum value at $t=1.5$. So from $t=0$ to $t=1.5$, the object moves backwards.
- From $t=1.5$ to $t=5$, the curve increases constantly before reaching its maximum value, which is $12$ at $t=5$. So during these times, the object moves forward. That being said, at $t=1.5$, the object has changed its direction.
Overall, the object is furthest from the origin at $t=5$, when its position reaches the value $12$.
The line $v$:
- The line $v$ increases constantly on $[0,5]$. From $t=0$ to $t=1.5$, $v$ is negative and increases constantly before reaching $0$ at $t=1.5$.
The object was moving backwards at the highest speed at the starting time. Then as it continued to move backwards $(v\lt0)$, it went more and more slowly before stopping at $t=1.5$, which is also when it changed direction.
- From $t=1.5$ to $t=5$, $v$ is positive and increases constantly. After the object stopped moving backwards at $t=1.5$, it started to move forward $(v\gt0)$ and its velocity rose constantly and reached its maximum at $t=5$.
The line $a$:
The line $a$ is a horizontal line $a=2$ throughout the whole interval. Since $a$ is positive, we know that the movement trend of velocity is increase on the whole interval, and since $a$ does not change its value at all, this increase trend is constant.