#### Answer

The detailed explanations are below.

#### Work Step by Step

$s=4-7t+6t^2-t^3$, $0\le t\le4$
1) Find $v$ and $a$:
$$v=\frac{ds}{dt}=\frac{d}{dt}(4-7t+6t^2-t^3)=-7+12t-3t^2$$
$$a=\frac{dv}{dt}=\frac{d}{dt}(-7+12t-3t^2)=12-6t$$
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 $4$ at $t=0$, goes down before reaching its minimum value at $t=0.709$. Then it goes up before being back to $s=4$ at $t=1.586$. This means that from $t=0$ to $t=0.709$, the object moved backwards, but then it stopped at $t=0.709$, changed direction and moved forwards continuously and was back to its original position at $t=1.586$.
- From $t=1.586$ to $t=3.291$, the curve continued to increase before reaching its maximum value at $t=3.291$. Then it went down again until $t=4$. This shows that the object continued to move forwards when it was back at origin at $t=1.586$ up until $t=3.291$. Then it stopped and changed direction one more time, moved backward until $t=4$.
Overall, the object is furthest from the origin at $t=3.291$.
The line $v$:
- The line $v$ started at $-7$ at $t=0$. From $t=0$ to $t=0.709$, $v$ is negative but it went up continuously and reached $0$ at $t=0.709$. Then it continued to go up (now $v$ is positive) until it reached its peak at $t=2$.
The object was moving backward at the highest speed at the starting time. Then as it continued to move backward $(v\lt0)$, it went more and more slowly before stopping at $t=0.709$, which is also when it changed direction the first time.
Then from $t=0.709$ to $t=2$, the object moved forwards $(v\gt0)$ at an increasing velocity, reaching its maximum at $t=2$.
- From $t=2$ to $t=3.291$, $v$ is positive but it went down continuously and reached $0$ at $t=3.291$. Then it continued to go down (now $v$ is negative) until it reached its minimum again at $t=4$.
After reaching its highest speed of moving forwards, the object still moved forward $(v\gt0)$ but its speed got slower and slower and became $0$ and the object stopped at $t=3.291$, which is where the object changed direction the second time.
Then from $t=3.291$ to $t=4$, the object moved backward $(v\gt0)$ at an increasing velocity, reaching its maximum speed again at $t=4$.
The line $a$:
The line $a$ goes down throughout the whole interval.
From $t=0$ to $t=2$, $a$ is positive but goes down continuously. This shows that the velocity of the object is going up but more and more slowly as $t$ increases.
From $t=2$ to $t=4$, $a$ is negative and goes down continuously. This shows that the velocity of the object is going down more and more quickly as $t$ increases.