Solution:

Given \(v_x = \frac{dx}{dt} = c\) (constant), hence \(a_x = 0\). Differentiating \(y = ax^2\) gives \(v_y = 2ax\frac{dx}{dt} = 2axc\). Differentiating again, \(a_y = 2ac\frac{dx}{dt} = 2ac^2\). Therefore, total acceleration vector is \(a = 2ac^2\hat{j}\).