1. **State the problem:** We are given the position function of a ball rolling down an inclined plane as $$s(t) = 2t^3 + 3t^2 + 4$$ where $s(t)$ is the distance in cm and $t$ is the time in seconds for $0 \leq t \leq 3$. We need to find the velocity of the ball at $t=2$ seconds. 2. **Recall the formula:** Velocity is the rate of change of position with respect to time, which is the first derivative of the position function: $$v(t) = \frac{ds}{dt}$$ 3. **Differentiate the position function:** $$v(t) = \frac{d}{dt}(2t^3 + 3t^2 + 4) = 6t^2 + 6t + 0 = 6t^2 + 6t$$ 4. **Evaluate the velocity at $t=2$:** $$v(2) = 6(2)^2 + 6(2) = 6 \times 4 + 12 = 24 + 12 = 36$$ 5. **Interpretation:** The velocity of the ball at 2 seconds is 36 cm/s. **Final answer:** $$\boxed{36 \text{ cm/s}}$$