How do you solve for a in $y={{v}_{0}}t+\dfrac{1}{2}a{{t}^{2}}$?

Hint: This equation $y={{v}_{0}}t+\dfrac{1}{2}a{{t}^{2}}$is the kinematic equation of physics. To solve for we will simply add and subtract some terms from both sides of the equation which will manipulate the equation and find the required value of a.

Complete step by step answer:
According to the question,
On the right side, we have the sum of ${{v}_{0}}t+\dfrac{1}{2}a{{t}^{2}}$ . We want to solve for a for that first we subtract ${{v}_{0}}t$ from both sides of the equation, we get
\[\begin{align}
  & y-{{v}_{0}}t={{v}_{0}}t+\dfrac{1}{2}a{{t}^{2}}-{{v}_{0}}t \\
 & \Rightarrow y-{{v}_{0}}t=\dfrac{1}{2}a{{t}^{2}} \\
\end{align}\]
Now we divide the above equation by ${{t}^{2}}$ on both sides of the equation, we get
$\begin{align}
  & \dfrac{y-{{v}_{0}}t}{{{t}^{2}}}=\dfrac{\dfrac{1}{2}a{{t}^{2}}}{{{t}^{2}}} \\
 & \Rightarrow \dfrac{y-{{v}_{0}}t}{{{t}^{2}}}=\dfrac{1}{2}a \\
\end{align}$
Now in order to find the final value of a we divide the above equation by$\dfrac{1}{2}$on both sides of the equation, we get
$\dfrac{\dfrac{y-{{v}_{0}}t}{{{t}^{2}}}}{\dfrac{1}{2}}=\dfrac{\dfrac{1}{2}a}{\dfrac{1}{2}}$
Now as the above equation contains fractions in both denominator and numerator on both sides so the reciprocal of the fraction in the denominator is multiplied by the numerator on both sides of the equation.
Therefore, we get
$\Rightarrow \dfrac{2(y-{{v}_{0}}t)}{{{t}^{2}}}=\dfrac{1}{2}a\times 2$
And we know that the same term in the denominator and numerator of the fraction gets canceled therefore we cancel 2 on the right-hand side of the equation. After canceling 2 we are only left with an on the right-hand side of the equation, so the equation is written as
$\Rightarrow \dfrac{2(y-{{v}_{0}}t)}{{{t}^{2}}}=a$
Therefore the value of a is
$\Rightarrow a=\dfrac{2(y-{{v}_{0}}t)}{{{t}^{2}}}$.

Note:
 Here, we should note that we have to solve for a therefore we should start by eliminating terms from the side of a to avoid doing tedious calculations as they can lead to calculation mistakes.