= Solution
By symmetry, the resultant <Coulomb force> on each charge points along the line from the <centroid> to its vertex. Their <accelerations> and subsequent <speeds> are equal, so the <centroid> remains fixed and the <equilateral triangle> expands without rotating. Each charge follows a straight radial trajectory; the separation grows and the repulsive <force> decreases.
For example, if the instantaneous side is $s$ and the centroid-to-charge distance is $r=s/\sqrt3$, the two pair forces have a $60^\circ$ angle between them. Their resultant is $\sqrt3k_eq^2/s^2=k_eq^2/(\sqrt3r^2)$, along the outward radius. This confirms that the symmetric shape persists.
Assuming release from rest, nonrelativistic motion and negligible radiation losses, <energy conservation> gives
$$
\frac32mv_\infty^2=\frac{3k_eq^2}{a},\qquad\boxed{v_\infty=\sqrt{\frac{2k_eq^2}{ma}}=\sqrt{\frac{q^2}{2\pi\epsilon_0ma}}}.
$$
The three velocity vectors sum to zero, consistent with <momentum conservation>. Their magnitudes approach the displayed value as the remaining <electrostatic potential energy> tends to zero.
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