Understanding the Concept:
Velocity is the first derivative of position with respect to time ($v = \frac{d}{dt}$), and acceleration is the second derivative ($\vec{a} = \frac{d\vec{v}}{dt}$). For multi-dimensional paths tracked via functions like $y = f(x)$, we can evaluate the components by applying the chain rule for derivatives.
Step 1: Differentiate the position equation with respect to time to find velocity.
The equation of the path is:
\[
y = 9x^2
\]
Differentiating both sides with respect to time $t$:
\[
\frac{dy}{dt} = 9 \cdot \frac{d}{dt}(x^2) = 9 \cdot \left(2x \frac{dx}{dt}\right) \implies v_y = 18x \cdot v_x
\]
Step 2: Differentiate a second time to find acceleration components.
Now, differentiate the velocity expression with respect to time $t$. Since the question states that the x-component of velocity ($v_x$) is constant, its time derivative is zero ($\frac{dv_x}{dt} = a_x = 0$).
\[
\frac{dv_y}{dt} = 18 \cdot \frac{d}{dt}(x \cdot v_x) = 18 \cdot \left( \frac{dx}{dt} \cdot v_x + x \cdot \frac{dv_x}{dt} \right)
\]
\[
a_y = 18 \cdot (v_x \cdot v_x + x \cdot 0) \implies a_y = 18 v_x^2
\]
Step 3: Equate components and isolate $v_x$.
We are given that the acceleration vector is $\vec{a} = 2\hat{j}\text{ m s}^{-2}$, which means $a_y = 2\text{ m s}^{-2}$.
\[
2 = 18 v_x^2 \implies v_x^2 = \frac{2}{18} = \frac{1}{9}
\]
Taking the square root:
\[
v_x = \sqrt{\frac{1}{9}} = \frac{1}{3}\text{ m s}^{-1}
\]