Chapter 3: Derivatives - Additional and Advanced Exercises - Page 183: 12

$1sec$, $2.5sec$.

Step 1. Given the equations for the two particles $s_1=3t^3-12t^2+18t+5$ and $s_2=-t^3+9t^2-12t$, we can get $v_1(t)=s'_1=9t^2-24t+18$ and $v_2(t)=s'_2=-3t^2+18t-12$. Step 2. Let $v_1(t)=v_2(t)$, we get $12t^2-42t+30=0$ or $2t^2-7t+5=0$ which in turn gives $t=1sec$ and $t=5/2=2.5sec$ (the times when the two particles have the same velocity).