1. **Problem statement:** A particle moves along the x-axis with acceleration $a(t) = 12t + 6$. Given velocity at $t=3$ is 36 cm/sec and initial position $x(0) = 4$ cm, find the position when velocity is zero. 2. **Formula and rules:** Velocity is the integral of acceleration: $$v(t) = \int a(t) dt + C$$ Position is the integral of velocity: $$x(t) = \int v(t) dt + D$$ Constants $C$ and $D$ are determined by initial conditions. 3. **Find velocity function:** $$v(t) = \int (12t + 6) dt + C = 6t^2 + 6t + C$$ Given $v(3) = 36$, substitute: $$36 = 6(3)^2 + 6(3) + C = 54 + 18 + C = 72 + C$$ Solve for $C$: $$C = 36 - 72 = -36$$ So, $$v(t) = 6t^2 + 6t - 36$$ 4. **Find time when velocity is zero:** Set $v(t) = 0$: $$6t^2 + 6t - 36 = 0$$ Divide both sides by 6: $$\cancel{6}t^2 + \cancel{6}t - \cancel{36} = 0 \Rightarrow t^2 + t - 6 = 0$$ Factor: $$(t + 3)(t - 2) = 0$$ Solutions: $$t = -3 \text{ (discard since } t \geq 0), \quad t = 2$$ 5. **Find position function:** Integrate velocity: $$x(t) = \int (6t^2 + 6t - 36) dt + D = 2t^3 + 3t^2 - 36t + D$$ Given $x(0) = 4$: $$4 = 2(0)^3 + 3(0)^2 - 36(0) + D = D$$ So, $$x(t) = 2t^3 + 3t^2 - 36t + 4$$ 6. **Find position at $t=2$:** $$x(2) = 2(2)^3 + 3(2)^2 - 36(2) + 4 = 2(8) + 3(4) - 72 + 4 = 16 + 12 - 72 + 4 = -40$$ **Final answer:** The position when velocity is zero is $\boxed{-40}$ cm.