A particle of mass $m$ and charge $q$ is accelerated through a potential difference $V$ over a distance $d$ - AQA - A-Level Physics - Question 16 - 2017 - Paper 2

Using the kinetic energy expression, we know that this energy manifests as kinetic energy when the particle reaches velocity $v$. Thus, we can equate:

$KE = \frac{1}{2}mv^2$

Setting the two equations for kinetic energy equal gives:

$qV = \frac{1}{2}mv^2$

Rearranging the equation to solve for velocity $v$ yields:

$v = \sqrt{\frac{2qV}{m}}$

The average acceleration $a$ of the particle while it travels a distance $d$ can be computed using the formula:

$a = \frac{v^2}{2d}$

Substituting $v^2$ from our equation:

$a = \frac{\frac{2qV}{m}}{2d} = \frac{qV}{md}$

Thus, the average acceleration of the particle is given by:

$a = \frac{qV}{md}$

This corresponds to option A in the multiple choice answers.