1. **State the problem:** We have a position function for a drone's height given by $$s(t) = -t^3 + 6t^2 + 15t + 10$$ for $$t \geq 0$$. We want to find the velocity $$v(t)$$ and acceleration $$a(t)$$ functions. 2. **Formulas and rules:** Velocity is the first derivative of position with respect to time: $$v(t) = \frac{ds}{dt}$$. Acceleration is the derivative of velocity with respect to time: $$a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}$$. 3. **Find velocity $$v(t)$$:** Differentiate $$s(t)$$ term-by-term: $$v(t) = \frac{d}{dt}(-t^3 + 6t^2 + 15t + 10) = -3t^2 + 12t + 15$$. 4. **Find acceleration $$a(t)$$:** Differentiate $$v(t)$$ term-by-term: $$a(t) = \frac{d}{dt}(-3t^2 + 12t + 15) = -6t + 12$$. 5. **State the second problem:** Given $$f(x) = \frac{5}{3} \sqrt{x + 6} + 2$$, find $$f(3)$$. 6. **Evaluate $$f(3)$$:** Substitute $$x=3$$: $$f(3) = \frac{5}{3} \sqrt{3 + 6} + 2 = \frac{5}{3} \sqrt{9} + 2 = \frac{5}{3} \times 3 + 2$$. 7. **Simplify:** $$f(3) = \cancel{\frac{5}{3}} \times \cancel{3} + 2 = 5 + 2 = 7$$. **Final answers:** - Velocity function: $$v(t) = -3t^2 + 12t + 15$$ - Acceleration function: $$a(t) = -6t + 12$$ - Function value: $$f(3) = 7$$