1. **State the problem:** We are given the acceleration function $$A(t) = t^3 - \frac{15}{2} t^2 + 12t + 10$$ and need to find the maximum acceleration on the interval $$0 \leq t \leq 6$$. 2. **Find the critical points:** To find maxima or minima, we differentiate $$A(t)$$ to get the jerk function $$J(t) = A'(t)$$. $$J(t) = \frac{d}{dt} \left(t^3 - \frac{15}{2} t^2 + 12t + 10\right) = 3t^2 - 15t + 12$$ 3. **Set the derivative equal to zero to find critical points:** $$3t^2 - 15t + 12 = 0$$ Divide both sides by 3: $$\cancel{3}t^2 - \cancel{3}5t + \cancel{3}4 = 0 \Rightarrow t^2 - 5t + 4 = 0$$ 4. **Factor the quadratic:** $$t^2 - 5t + 4 = (t - 4)(t - 1) = 0$$ So, $$t = 1$$ or $$t = 4$$. 5. **Evaluate acceleration at critical points and endpoints:** - At $$t=0$$: $$A(0) = 0^3 - \frac{15}{2} \cdot 0^2 + 12 \cdot 0 + 10 = 10$$ - At $$t=1$$: $$A(1) = 1 - \frac{15}{2} + 12 + 10 = 1 - 7.5 + 12 + 10 = 15.5$$ - At $$t=4$$: $$A(4) = 64 - \frac{15}{2} \cdot 16 + 48 + 10 = 64 - 120 + 48 + 10 = 2$$ - At $$t=6$$: $$A(6) = 216 - \frac{15}{2} \cdot 36 + 72 + 10 = 216 - 270 + 72 + 10 = 28$$ 6. **Determine the maximum acceleration:** Among $$10, 15.5, 2, 28$$, the maximum is $$28$$ at $$t=6$$. **Final answer:** The maximum acceleration is 28 meters per second squared at $$t=6$$ seconds, which corresponds to option (D).