What if there is acceleration for the unit positive charge while measuring the electrostatic potential? Adding the general definition of electrostatic potential:
" Electrostatic potential (V )
at any point in a region with electrostatic field is
the work done in bringing a unit positive
charge (without acceleration) from infinity to
that point "
What if there is acceleration for unit +ve charge?
 A: If there is acceleration of the test charge then the charge would have kinetic energy along with the potential energy. 
Potential is defined as the work done by the external force per unit charge against the electric field of the reference charge (the source of electric field). So if the charge is accelerating then it means that the external force is not equal to the force due to the electric field. Thus, we won't get the correct potential this way.
While calculating the potential due a charge, we only consider the change in potential energy of the test charge when bringing the charge from infinity towards the reference charge.
We only want to consider the potential energy per unit charge to calculate the potential, so that if we wish the find the potential energy of an another charge, we just have to multiply the charge of that particle and the potential of the point where we want to find it's potential energy.
A: We consider a test charge $q$, the magnitude of which is infinitesimely small  so that it wouldn't affect the electric field due to source charge. Hence we can calculate potential due to large charge (suppose $x$) easily just by multiplying $x$ with $W/q$ (which is the actual potential).
Now If acceleartion comes into play then accelerated charge particle emits electromagnetic radiation or wave; a wave is always an energy carrier which carry vibrational energy, this energy in turn do work by applying force on source charge causing it to displace to another position thereby changing the configuration of original electric field due to source charge. And at this point if we analyse the accelerated charge then it can be well understood that it violates the assumption that the test charge doesn't affect the electric field due to source charge and we can't further proceed to derive the expression of electric potential in the way which is done above. So it is better to do work on test charge by applying external force but without accelerating it.
