$$\begin{cases} \text{Particle 1}=A\\ \text{Particle 2}=B\\ \end{cases} $$ $$\begin{cases} x_A=6 .00t^2 + 3 .00t + 2.00\\ a_B=-8.00t\\ v_B(0)=15\ut{m/s}\\ \end{cases} $$ $$ v_A(k)=v_B(k) =?$$ $$ \begin{aligned} v_A(t)&=\dot{x}_A=\dxt{x_A}\\ &=\dt(6 .00t^2 + 3 .00t + 2.00)\\ &=12t+3 \end{aligned} $$ $$ \begin{aligned} v_B(t)&=\int_0^ta\dd t\\ &=\int(-8.00t)\dd t\\ &=-4t^2+C\\ &=-4t^2+15 \ (\becau..