Starting at time 𝑡=0 from the origin with speed 1 ms$$^{-1}$$, a particle follows a two-dimensional trajectory in the x-y plane so that its coordinates are related by the equation $$y = \frac{x^2}{2}$$. The x and y components of its acceleration are denoted by $$𝑎_𝑥$$ and $$𝑎_y$$, respectively. Then

The trajectory equation is $$y=\frac{x^{2}}{2}$$, which holds at every instant of time.

Step 1 : Relation between the velocity components
Differentiate the path equation w.r.t. time $$t$$:
$$\frac{dy}{dt}=x\,\frac{dx}{dt}$$
Hence the velocity components satisfy
$$v_{y}=x\,v_{x}\qquad -(1)$$

Step 2 : Relation between the acceleration components
Differentiate $$(1)$$ once more:
$$a_{y}= \frac{d}{dt}(x\,v_{x}) = v_{x}^{2}+x\,a_{x}\qquad -(2)$$

Step 3 : Initial velocity direction (Option C)
At $$t=0$$ the particle is at the origin, so $$x=0,\;y=0$$. Equation $$(1)$$ gives $$v_{y}=0\cdot v_{x}=0$$. The speed is given as $$|{\mathbf v}|=1\;{\rm m\,s^{-1}}$$, therefore $$v_{x}= \pm1$$ and the velocity points along the $$x$$-axis. Thus Option C is correct.

Step 4 : Verifying Option A
At the origin $$x=0$$, so from $$(2)$$
$$a_{y}=v_{x}^{2}+0\cdot a_{x}=v_{x}^{2}=1$$ (because $$v_{x}= \pm1$$). Therefore, if in addition it is given that $$a_{x}=1\;{\rm m\,s^{-2}}$$, we still get $$a_{y}=1\;{\rm m\,s^{-2}}$$. Option A is true.

Step 5 : Verifying Option B
Suppose $$a_{x}=0$$ for all time. Then $$v_{x}$$ is a constant. From the initial condition $$v_{x}= \pm1$$, so $$v_{x}^{2}=1$$. Using $$(2)$$ with $$a_{x}=0$$ gives
$$a_{y}=v_{x}^{2}=1\;{\rm m\,s^{-2}}$$ at every instant. Hence Option B is correct.

Step 6 : Verifying Option D
Continue with the case $$a_{x}=0$$. Since $$v_{x}=1$$ (choose the positive sign for definiteness),
$$\frac{dx}{dt}=1 \;\Rightarrow\; x=t$$ (taking $$x=0$$ at $$t=0$$).
From $$(1)$$: $$v_{y}=x\,v_{x}=t\cdot1=t$$.
At $$t=1\;{\rm s}$$ we have $$v_{x}=1,\;v_{y}=1$$. The angle $$\theta$$ made by the velocity with the $$x$$-axis is therefore
$$\tan\theta=\frac{v_{y}}{v_{x}}=\frac{1}{1}=1\;\Longrightarrow\;\theta=45^{\circ}$$. Thus Option D is also correct.

Final result: Option A, Option B, Option C, and Option D are all correct.