1. **State the problem:** A particle moves along the x-axis with position given by a function involving a positive constant. We need to find the time when the particle's position is farthest to the right. 2. **Identify the function:** Let the position function be $x(t)$, where $t$ is time and the function depends on a positive constant $k$. 3. **Find the maximum position:** To find when the particle is farthest to the right, we need to find the maximum value of $x(t)$. 4. **Use the derivative:** The maximum occurs where the velocity (derivative of position) is zero, i.e., solve $$\frac{dx}{dt} = 0$$. 5. **Check the second derivative:** Confirm that the critical point is a maximum by checking $$\frac{d^2x}{dt^2} < 0$$ at that point. 6. **Solve for $t$:** Find the value(s) of $t$ satisfying the above conditions. Since the exact function $x(t)$ is not provided, the general method is: - Compute $\frac{dx}{dt}$. - Set $\frac{dx}{dt} = 0$ and solve for $t$. - Verify the nature of critical points with $\frac{d^2x}{dt^2}$. This will give the time(s) when the particle is farthest to the right. **Final answer:** The time $t$ at which $\frac{dx}{dt} = 0$ and $\frac{d^2x}{dt^2} < 0$ is when the particle's position is farthest to the right.