1. **State the problem:** A particle moves along the x-axis with position given by $x(t) = t e^{-at}$, where $a$ is a positive constant and $t > 0$. We need to find the time $t$ when the particle's position is farthest to the right. 2. **Identify the function:** The position function is $$x(t) = t e^{-at}$$ where $a > 0$. 3. **Find the maximum position:** To find when the particle is farthest to the right, we find the maximum of $x(t)$ by setting its derivative to zero. 4. **Compute the derivative:** Using the product rule, $$\frac{dx}{dt} = \frac{d}{dt}(t) \cdot e^{-at} + t \cdot \frac{d}{dt}(e^{-at}) = 1 \cdot e^{-at} + t \cdot (-a e^{-at}) = e^{-at} - a t e^{-at}$$ 5. **Set the derivative equal to zero:** $$e^{-at} - a t e^{-at} = 0$$ 6. **Factor out $e^{-at}$:** $$e^{-at}(1 - a t) = 0$$ Since $e^{-at} \neq 0$ for all $t$, we have $$1 - a t = 0$$ 7. **Solve for $t$:** $$t = \frac{1}{a}$$ 8. **Check the second derivative to confirm maximum:** $$\frac{d^2x}{dt^2} = \frac{d}{dt}(e^{-at} - a t e^{-at}) = -a e^{-at} - a \left(e^{-at} - a t e^{-at}\right) = -a e^{-at} - a e^{-at} + a^2 t e^{-at} = e^{-at}(-a - a + a^2 t) = e^{-at} (a^2 t - 2a)$$ Evaluate at $t = \frac{1}{a}$: $$\frac{d^2x}{dt^2} \bigg|_{t=\frac{1}{a}} = e^{-a \cdot \frac{1}{a}} \left(a^2 \cdot \frac{1}{a} - 2a\right) = e^{-1} (a - 2a) = e^{-1} (-a) < 0$$ Since $a > 0$, the second derivative is negative, confirming a maximum. **Final answer:** The particle's position is farthest to the right at time $$t = \frac{1}{a}$$.