1. **State the problem:** We are given the position function of an object as $x(t) = t e^{-at}$ where $a$ is a positive constant and $t > 0$. 2. **Understand the function:** This function describes the position $x$ at time $t$. The term $e^{-at}$ represents exponential decay because $a > 0$. 3. **Find the velocity:** Velocity is the derivative of position with respect to time, so we use the product rule: $$x(t) = t \cdot e^{-at}$$ $$x'(t) = \frac{d}{dt}(t) \cdot e^{-at} + t \cdot \frac{d}{dt}(e^{-at})$$ 4. Calculate each derivative: $$\frac{d}{dt}(t) = 1$$ $$\frac{d}{dt}(e^{-at}) = -a e^{-at}$$ 5. Substitute back: $$x'(t) = 1 \cdot e^{-at} + t \cdot (-a e^{-at}) = e^{-at} - a t e^{-at}$$ 6. Factor out $e^{-at}$: $$x'(t) = e^{-at}(1 - a t)$$ 7. **Interpretation:** Velocity depends on time $t$ and constant $a$. For $t > 0$, velocity changes sign at $t = \frac{1}{a}$. This completes the analysis of the position function and its velocity.