Hello, nikk!
A particle $\displaystyle P$ moves along straight line and passes through a fixed point $\displaystyle O$.
Its velocity is given by: .$\displaystyle v(t) \:=\: 6t^24t2$
where $\displaystyle t$ is the time, in seconds, after passing through $\displaystyle O$.
(a) The time interval during which the particle moves towards the left. "Moves towards the left" means that the velocity is negative.
The velocity function is an upopening parabola.
It is negative between its $\displaystyle x$intercepts.
$\displaystyle x$intercepts: .$\displaystyle 6t^2  4t  2 \:=\:0 \quad\Rightarrow\quad 3(3t+1)(t1) \:=\:0 \quad\Rightarrow\quad x \;=\;\text{}\tfrac{1}{3},\;1$
The particle moves to the left on the interval: .$\displaystyle t \in \left(\text{}\tfrac{1}{3},\;1\right)$
(b) the distance, in $\displaystyle m$, travelled by the particle during the first 3 seconds. Let $\displaystyle c$ be the location of point $\displaystyle O$.
This is the position of the particle when $\displaystyle t = 0.$
The position function is given by: .$\displaystyle s(t) \:=\:\int v(t)\,dt$
We have: .$\displaystyle s(t) \;=\;\int(6t^2  4t  2)\,dt \quad\Rightarrow\quad\boxed{s(t) \;=\;2t^3  2t^2  2t + c}$
The particle stops and changes directions when $\displaystyle v = 0.$
In (a), we found that this happens at: .$\displaystyle t = \text{}\frac{1}{3}$ and $\displaystyle t = 1.$
At $\displaystyle t = 0\!:\;\;s(0) \:=\:c$
At $\displaystyle t = 1\!:\;\;s(1) \:=\:2+c$
In the first second, it has moved: .$\displaystyle (2+c)  c \:=\:2$ . . . 2 units to the left.
At $\displaystyle t = 3\!:\;\;s(3) \:=\:30+c$
In the next two seconds, it has moved: .$\displaystyle (30+c)(2+c) \:=\: +32$ . . . 32 units to the right.
Therefore, in the first 3 seconds, it moves: .$\displaystyle 2 + 32 \:=\:34$ units.
(c) Sketch the velocitytime graph of the motion of the particle for 0 to 3 We already know that it is a parabola, but we can plot some points.
. . $\displaystyle \begin{array}{cc}
t & v(t) \\ \hline
0 & \text{}2 \\ \frac{1}{3} & \text{}\frac{8}{3} \\ 1 & 0 \\ 2 & 14 \\ 3 & 40 \end{array}$ . . I dropped the "c"
Code:

 o (3,40)




 o (2,14)


  +   o   +   +  
 (1,0)
(0,2)o
 o (1/3,8/3)
