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+### Statement
+
+$14.2.2.$ [Insert the problem statement]
+
+### Solution
+
+### Problem
+
+$14.2.2.$ Lifetime of a stationary particle \( \tau \). At what speed must this particle move in order to travel a distance \( l \) before decaying?
+
+
+### Solution
+
+In the laboratory frame, the lifetime of a moving particle is increased due to the relativistic time dilation effect:
+
+\[
+\tau' = \frac{\tau}{\sqrt{1 - \frac{v^2}{c^2}}}
+\]
+
+The particle must cover the distance \( l \) within the time \( \tau' \):
+
+\[
+l < v \tau'
+\]
+
+Substituting the expression for \( \tau' \):
+
+\[
+l < \frac{v \tau}{\sqrt{1 - \frac{v^2}{c^2}}}
+\]
+
+Let's solve this inequality for \( v \). First, divide both sides by \( \tau \) and square both sides:
+
+\[
+\frac{v^2}{1 - \frac{v^2}{c^2}} > \left( \frac{l}{\tau} \right)^2
+\]
+
+Multiply both sides by \( 1 - \frac{v^2}{c^2} \):
+
+\[
+v^2 > \left( \frac{l}{\tau} \right)^2 \left( 1 - \frac{v^2}{c^2} \right)
+\]
+
+Distribute and bring all terms to one side:
+
+\[
+v^2 + \left( \frac{l}{\tau c} \right)^2 v^2 > \left( \frac{l}{\tau} \right)^2
+\]
+
+Factor out \( v^2 \):
+
+\[
+v^2 \left( 1 + \left( \frac{l}{\tau c} \right)^2 \right) > \left( \frac{l}{\tau} \right)^2
+\]
+
+Now solve for \( v \):
+
+\[
+v^2 > \frac{\left( \frac{l}{\tau} \right)^2}{1 + \left( \frac{l}{\tau c} \right)^2}
+\]
+
+\[
+v > \frac{c}{\sqrt{1 + \left( \frac{\tau c}{l} \right)^2}}
+\]
+
+
+### Answer
+
+\[
+v > \frac{c}{\sqrt{1 + \left( \frac{\tau c}{l} \right)^2}}
+\]
+
+#### Answer
+
+[Insert a concise answer or boxed result]