440. A point moves along the x-axis according to the law $x=2 t^{3}-4 t^{2}+2 t+3$, where $t$ is time in seconds measured from $t=0$, and $x$ is the distance of the moving point from the origin in meters. It is required: a) to determine the law of variation of velocity and acceleration with time $t$; b) to find the initial velocity and the velocity at the moment $t=3$ s; c) to determine whether there are moments of time when the velocity is zero, and if so, what positions of the moving point correspond to these moments.
Solution. a) To determine the velocity of the motion, we find the derivative of the path with respect to time:

$$
v=\frac{d x}{d t}=6 t^{2}-8 t+2=2\left(3 t^{2}-4 t+1\right)
$$

and for determining the acceleration of the motion - the derivative of the velocity with respect to time:

$$
a=\frac{d v}{d t}=2(5 t-4)=4(3 t-2)
$$

b) If $t=0$, then $v=2$ (m/s) (initial velocity); if $t=3$, then $v=32(\mathrm{~m} / \mathrm{s})$.

c) The condition $v=0$ means that $3 t^{2}-4 t+1=0$. Solving this equation, we get $t=\frac{2}{3} \pm \sqrt{\frac{4}{9}-\frac{1}{3}}=\frac{2}{3} \pm \frac{1}{3}$. Therefore, the value $v=0$ is achieved twice: first at the moment $t=1 / 3$ s, and then at the moment $t=1$.

Let's find the abscissas of the moving point at these moments in time:

$$
\left.x\right|_{t=1 / 3}=\left(2 t^{3}-4 t^{2}+2 t+3\right)_{t=1 / 3}=3 \frac{8}{27} \mathrm{~m} ;\left.x\right|_{t=1}=3(\mathrm{~m})
$$