1. **State the problem:** We have a particle moving along the y-axis with velocity function $v(t) = 12t^2 - 24t$ for $t \geq 0$. The initial position is $x(0) = 10$ cm. We need to find the position $x(t)$ at the time when the velocity is minimum. 2. **Recall the relationship between position and velocity:** Velocity is the derivative of position, so $$v(t) = x'(t) = 12t^2 - 24t.$$ The position function is given by integrating velocity: $$x(t) = \int v(t) dt = \int (12t^2 - 24t) dt.$$ 3. **Find the position function:** Integrate term-by-term: $$x(t) = 12 \cdot \frac{t^3}{3} - 24 \cdot \frac{t^2}{2} + C = 4t^3 - 12t^2 + C.$$ Use initial condition $x(0) = 10$: $$x(0) = 4 \cdot 0^3 - 12 \cdot 0^2 + C = C = 10.$$ So, $$x(t) = 4t^3 - 12t^2 + 10.$$ 4. **Find the time when velocity is minimum:** Velocity minimum occurs where its derivative (acceleration) is zero and changes from negative to positive. Calculate acceleration: $$a(t) = v'(t) = \frac{d}{dt}(12t^2 - 24t) = 24t - 24.$$ Set acceleration to zero: $$24t - 24 = 0 \implies 24t = 24 \implies t = 1.$$ 5. **Verify minimum velocity at $t=1$:** Check second derivative of velocity: $$v''(t) = a'(t) = 24 > 0,$$ so velocity has a minimum at $t=1$. 6. **Find position at $t=1$:** Substitute $t=1$ into $x(t)$: $$x(1) = 4(1)^3 - 12(1)^2 + 10 = 4 - 12 + 10 = 2.$$ **Final answer:** The position of the particle at minimum velocity is $\boxed{2}$ cm.