Task 1. A vacuum robot is programmed to move on the floor according to the law:

$\left\{\begin{array}{l}x=t(t-6)^{2} \\ y=0,0 \leq t \leq 7 ; y=(t-7)^{2}, t \geq 7\end{array}\right.$

where the axes are parallel to the walls and the movement starts from the origin. Time $t$ is measured in minutes, and the coordinates in meters. Find the path traveled by the robot in the first 7 minutes and the magnitude of the change in velocity vector during the eighth minute.
Solution. For the first seven minutes, the point moves along the $x$-axis. The velocity of the point for $t \leq 7$ is $\dot{x}=3(t-2)(t-6)$ and equals zero at times $t_{1}=2$ and $t_{2}=6$. The distance traveled will be $L=x(7)+2(x(2)-x(6))=7+2 \cdot 32=71$.

The velocity vector $\bar{V}$ at each moment $t \geq 7$ is determined by its projections on the coordinate axes $\bar{V}(\dot{x} ; \dot{y})$, where $\dot{y}=2(t-7)$. The change in velocity $\Delta \bar{V}$ over the eighth minute is determined by the change in the projections of the velocity $\Delta \bar{V}=\bar{V}(8)-\bar{V}(7)=(36 ; 2)-(15 ; 0)=(21 ; 2)$. The magnitude of the vector $\Delta \bar{V}$ is $\sqrt{445}$.

Answer: A) 71; B) $\sqrt{445}$.

Answer to variant 212: A) 12 ; B) $\sqrt{229}$.