How do you find velocity and acceleration as functions of t?

Hint: We start the integration process for velocity and acceleration keeping in mind that the acceleration is a constant term. We assume the instantaneous changes as the velocity and acceleration. We integrate them to find the formulas.

Complete step-by-step solution:
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 $\dfrac{ds}{dt}=v$.
In combination we get $\dfrac{dv}{dt}=\dfrac{{{d}^{2}}s}{d{{t}^{2}}}=a$. The acceleration is a constant term.
We now integrate $\dfrac{dv}{dt}=a$ to get $\int{dv}=a\int{dt}$.
The integration gives $v=at+c$. We try to find the value of c. $u$ is the initial velocity.
We know at the start at $t=0$, the velocity $v=u$. So, $c=u$ which gives $v=u+at$.
Now we integrate $v=u+at$ and use $\dfrac{ds}{dt}=v$.
So, $\int{ds}=u\int{dt}+a\int{tdt}$ which gives $s=ut+\dfrac{a{{t}^{2}}}{2}+c$.
Again, we know at the start at $t=0$, the velocity $s=0$. So, $c=0$ which gives $s=ut+\dfrac{a{{t}^{2}}}{2}$.
Therefore, the expression of velocity and acceleration as functions of t are $v=u+at$ and $s=ut+\dfrac{a{{t}^{2}}}{2}$ respectively.

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.