1. **State the problem:** We have the position function of an object given by $$s(t) = 6 + 8t - t^2$$ for $$0 \leq t \leq 5$$. We need to determine when the object is moving forward during this time interval. 2. **Recall the concept:** An object is moving forward when its velocity is positive. Velocity is the derivative of the position function with respect to time, $$v(t) = s'(t)$$. 3. **Find the velocity function:** Differentiate $$s(t)$$: $$ s'(t) = \frac{d}{dt}(6 + 8t - t^2) = 0 + 8 - 2t = 8 - 2t $$ 4. **Determine when velocity is positive:** Solve the inequality: $$ 8 - 2t > 0 $$ 5. **Solve the inequality:** $$ 8 > 2t $$ $$ \frac{\cancel{8}}{\cancel{2}} > \frac{2t}{2} \implies 4 > t $$ 6. **Interpret the result:** The object moves forward when $$t < 4$$. Since the time interval is $$0 \leq t \leq 5$$, the object moves forward for $$0 \leq t < 4$$. 7. **Final answer:** The object is moving forward during the time interval $$0 \leq t < 4$$.