1. **State the problem:** We want to derive the velocity $v$ from the time dilation formula in special relativity. 2. **Recall the time dilation formula:** The time dilation formula relates the proper time interval $\Delta t_0$ (time measured in the moving frame) to the dilated time interval $\Delta t$ (time measured in the stationary frame) as $$\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$ where $c$ is the speed of light. 3. **Goal:** Solve for velocity $v$ in terms of $\Delta t$, $\Delta t_0$, and $c$. 4. **Isolate the square root term:** Multiply both sides by the denominator: $$\Delta t \sqrt{1 - \frac{v^2}{c^2}} = \Delta t_0$$ 5. **Divide both sides by $\Delta t$:** $$\sqrt{1 - \frac{v^2}{c^2}} = \frac{\Delta t_0}{\Delta t}$$ 6. **Square both sides to remove the square root:** $$1 - \frac{v^2}{c^2} = \left(\frac{\Delta t_0}{\Delta t}\right)^2$$ 7. **Rearrange to isolate $\frac{v^2}{c^2}$:** $$\frac{v^2}{c^2} = 1 - \left(\frac{\Delta t_0}{\Delta t}\right)^2$$ 8. **Multiply both sides by $c^2$ to solve for $v^2$:** $$v^2 = c^2 \left[1 - \left(\frac{\Delta t_0}{\Delta t}\right)^2\right]$$ 9. **Take the square root to find $v$:** $$v = c \sqrt{1 - \left(\frac{\Delta t_0}{\Delta t}\right)^2}$$ **Final answer:** $$\boxed{v = c \sqrt{1 - \left(\frac{\Delta t_0}{\Delta t}\right)^2}}$$ This formula shows how velocity $v$ relates to the ratio of proper time to dilated time in special relativity.