1. **Problem statement:** Find the derivative of the velocity function $$v(t) = (t - 5)(t - 2)^2$$ for time $$t \geq 0$$. 2. **Formula used:** To differentiate a product of two functions, use the product rule: $$\frac{d}{dt}[f(t)g(t)] = f'(t)g(t) + f(t)g'(t)$$ 3. **Identify functions:** Let $$f(t) = (t - 5)$$ and $$g(t) = (t - 2)^2$$. 4. **Find derivatives:** $$f'(t) = 1$$ $$g'(t) = 2(t - 2)$$ (using the power rule and chain rule). 5. **Apply product rule:** $$v'(t) = f'(t)g(t) + f(t)g'(t) = 1 \cdot (t - 2)^2 + (t - 5) \cdot 2(t - 2)$$ 6. **Simplify:** $$v'(t) = (t - 2)^2 + 2(t - 5)(t - 2)$$ 7. **Factor out common term $$(t - 2)$$:** $$v'(t) = (t - 2)\big((t - 2) + 2(t - 5)\big)$$ 8. **Simplify inside parentheses:** $$(t - 2) + 2(t - 5) = t - 2 + 2t - 10 = 3t - 12$$ 9. **Final derivative:** $$v'(t) = (t - 2)(3t - 12) = 3(t - 2)(t - 4)$$ **Answer:** $$\boxed{v'(t) = 3(t - 2)(t - 4)}$$