1. **Problem 1:** A particle moves along the x-axis with acceleration $a(t) = 12t + 6$, velocity at $t=3$ is 36 cm/sec, and initial position $x(0) = 4$ cm. Find the position when velocity is zero. 2. **Formula:** Velocity is the integral of acceleration: $$v(t) = \int a(t)\,dt + C$$ Position is the integral of velocity: $$x(t) = \int v(t)\,dt + D$$ 3. **Find velocity function:** $$v(t) = \int (12t + 6) dt + C = 6t^2 + 6t + C$$ Use $v(3) = 36$: $$36 = 6(3)^2 + 6(3) + C = 54 + 18 + C = 72 + C \implies C = 36 - 72 = -36$$ So, $$v(t) = 6t^2 + 6t - 36$$ 4. **Find time when velocity is zero:** $$0 = 6t^2 + 6t - 36$$ Divide both sides by 6: $$0 = \cancel{6}t^2 + \cancel{6}t - \cancel{6}6 \implies 0 = t^2 + t - 6$$ Solve quadratic: $$t = \frac{-1 \pm \sqrt{1^2 - 4(1)(-6)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm 5}{2}$$ Positive root: $$t = \frac{-1 + 5}{2} = 2$$ 5. **Find position function:** $$x(t) = \int v(t) dt + D = \int (6t^2 + 6t - 36) dt + D = 2t^3 + 3t^2 - 36t + D$$ Use initial position $x(0) = 4$: $$4 = 2(0)^3 + 3(0)^2 - 36(0) + D = D \implies D = 4$$ 6. **Find position at $t=2$:** $$x(2) = 2(2)^3 + 3(2)^2 - 36(2) + 4 = 2(8) + 3(4) - 72 + 4 = 16 + 12 - 72 + 4 = -40$$ **Answer:** The position when velocity is zero is $\boxed{-40}$ cm. --- **Problem 2:** A particle moves along the y-axis with velocity $v(t) = 12t^2 - 24t$, initial position $y(0) = 10$ cm. Find position at minimum velocity. 7. **Find time of minimum velocity:** Velocity minimum occurs where acceleration $a(t) = v'(t) = 0$. $$a(t) = \frac{d}{dt}(12t^2 - 24t) = 24t - 24$$ Set to zero: $$0 = 24t - 24 \implies t = 1$$ 8. **Find position function:** $$y(t) = \int v(t) dt + C = \int (12t^2 - 24t) dt + C = 4t^3 - 12t^2 + C$$ Use $y(0) = 10$: $$10 = 4(0)^3 - 12(0)^2 + C = C \implies C = 10$$ 9. **Find position at $t=1$:** $$y(1) = 4(1)^3 - 12(1)^2 + 10 = 4 - 12 + 10 = 2$$ **Answer:** The position at minimum velocity is $\boxed{2}$ cm.