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

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

where the axes are parallel to the walls. Time $t$ is measured in minutes, and 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 7 minutes, the point moves with constant acceleration along the x-axis. The velocity of the point at $t \leq 7$ is $V_{x}=2(t-6)$ and becomes zero at $t_{1}=6$. The distance traveled will then be $L=x(0)+x(7)=1+36=37$.

The velocity vector $\bar{V}$ at any moment $t \geq 7$ is determined by its projections on the coordinate axes: $V_{x}=2(t-6), V_{y}=2(t-7)$. The change in velocity $\Delta \bar{V}$ over the eighth minute is determined by the change in the projections of the velocity $\Delta \bar{V}=\bar{V}(8)-\bar{V}(7)=(4 ; 2)-(2 ; 0)=(2 ; 2)$. The magnitude of the vector $\Delta \bar{V}$ is $2 \sqrt{2}$.

Answer: A) 37 ; B) $2 \sqrt{2}$.