Coulomb's Law: Solving for Speed of Particle

[SupremeV]

1. The problem statement, all variables given/known data

A particle of charge 6 µC is held fixed while another particle of charge 8 µC is released
from rest at a distance of 1.4 m from the first particle. If the mass of the second particle is
3x10^-6 kg, what is its speed when it is very far away from the first particle?

Homework Equations

F=ma
F= kQ₁Q₂ / r²
V²=V_o +2ad

The Attempt at a Solution

I pretty much worked through most the problem. In the end, I used coulomb's law in combination with Newtons 2nd law to get a= 73469 m/s². However what's bugging me is the distance that the speed is "very far away". I'm under the assumption that means at a point where the force from the first particle no longer affects the second. However, when I back-tracked using the answer(454m/s) the distance ends up being 1.4( the original distance) with the velocity equations.

 

[SupremeV]

Apologies, I searched google and found a similar problem: rocket is launched straight up from the Earth's surface at a speed of 1.60×10^4 m/s.
What is its speed when it is very far away from the earth?

One of the helpers suggested conservation of energy, and I applied this to this problem it worked! Sorry!

 

[gneill]

You do realize that the acceleration is not constant? It decreases as the distance between the particles grows. So your V² = V_o + 2ad formula is not valid over the trajectory of the second particle.

Why not try a conservation of energy approach?