A stone is falling freely and describes a distance $s$ in $t$ seconds given by equation $s=\dfrac{g{{t}^{2}}}{2}$. The acceleration of the stone is
A. uniform
B. 0
C. non-uniform
D. indeterminate

Hint: We start the differentiation process for velocity and acceleration keeping in mind that the given function is of distance. We assume the instantaneous changes as the velocity and acceleration represented by $\dfrac{ds}{dt}=v,\dfrac{dv}{dt}=a$ respectively. We differentiate the function to find the solution.

Complete step by step answer:
We have to find the velocity and acceleration as functions of t where t represents time.
We know that velocity is considered as the rate of change of displacement and acceleration is considered as the rate of change of velocity.
These rates of change will be considered for instantaneous.
If we consider displacement, velocity and acceleration as $s,v,a$ respectively, then the instantaneous changes will be considered with respect to time.
Therefore, instantaneous change of velocity is acceleration which gives $\dfrac{dv}{dt}=a$ and instantaneous change of displacement is velocity which gives $\dfrac{ds}{dt}=v$.
It is given $s=\dfrac{g{{t}^{2}}}{2}$. We now differentiate $s=\dfrac{g{{t}^{2}}}{2}$ to get $v,a$. Here $g$ is constant.
The differentiation gives $v=\dfrac{ds}{dt}=\dfrac{g}{2}\times 2t=gt$ and $a=\dfrac{dv}{dt}=g$.
Therefore, the acceleration of the stone is uniform.

So, the correct answer is “Option A”.

Note: We completed the formulas assuming that no external forces were applied. To find velocity, we take the derivative of the original position equation. To find acceleration, we take the derivative of the velocity function.