Solution: The point is moving on the parabolic path which is given by \[ \gamma (t) = \left( x(t), y(t) \right), \] where $\vert \dot{\gamma }(t) \vert =1$. This implies that for $t\in \mathbb{R} $, \begin{equation}\label{eq:24Jun2023-1} \dot x(t)^2 + \dot y(t)^2 = 1. \end{equation} Let $\gamma \left( t_0 \right) = \left( \frac{1}{2}, \frac{1}{4} \right) $, so we need to find the magnitude of the acceleration, that is, \[ \left\lVert \ddot \gamma (t_0) \right\rVert = \sqrt{\ddot x\left( t_0 \right)^2 + \ddot y\left( t_0 \right) ^2}. \]

Note that \begin{align*} y(t) = x(t)^2 & \implies \dot y(t) = 2 x(t) \dot x(t) \\ & \implies \dot y\left( t_0 \right) = 2 x\left( t_0 \right) \dot x\left( t_0 \right) \\ & \implies \dot y \left( t_0 \right) = \dot x \left( t_0 \right) . \end{align*} Now using \eqref{eq:24Jun2023-1}, we have \[ \dot x\left( t_0 \right) ^2 + \dot y\left( t_0 \right) ^2 = 1 \implies \dot x\left( t_0 \right) = \dot y\left( t_0 \right) = \frac{1}{\sqrt{2} }. \]

Similarly, \begin{align*} \ddot y(t) = 2 \dot x(t)^2 + 2 x(t) \ddot x(t) & \implies \ddot y\left( t_0 \right) = 2 \dot x\left( t_0 \right) ^2 + 2 x\left( t_0 \right) \ddot x \left( t_0 \right) \\ & \implies \ddot y\left( t_0 \right) = 1 + \ddot x\left( t_0 \right). \end{align*} Now differentiating \eqref{eq:24Jun2023-1} at $t= t_0$, we have \begin{align*} 2 \dot x\left( t_0 \right) \ddot x\left( t_0 \right) + 2 \dot y\left( t_0 \right) \ddot y\left( t_0 \right) = 0 & \implies \ddot x\left( t_0 \right) = - \ddot y\left( t_0 \right) . \end{align*} Therefore, using $\ddot y\left( t_0 \right) = 1 + \ddot x\left( t_0 \right) $, we have \begin{align*} - \ddot x \left( t_0 \right) = 1 + \ddot x\left( t_0 \right) & \implies \ddot x\left( t_0 \right) = -\frac{1}{2} \\ & \implies \ddot y \left( t_0 \right) = \frac{1}{2}. \end{align*} Hence, the magnitude of the acceleration will be \[ \left\vert \ddot \gamma \left( t_0 \right) \right\vert = \sqrt{\frac{1}{4} + \frac{1}{4}} = \frac{1}{\sqrt{2}}. \]