$$ y=\frac{x^2}{2} $$ $$ \begin{aligned} & \text { at } \left.t=0, \begin{array}{l} x=0, y=0 \\ u=1 \end{array}\right\} \text { given } \\ & y=\frac{x^2}{2} \\ & \frac{d y}{d t}=\frac{1}{2} \cdot 2 x \frac{d x}{d t} \\ & \Rightarrow v_y=x v_x \end{aligned} $$ difference wrt time $$ \begin{aligned} & a_y=\frac{d x}{d t} \cdot V_x+x a_x \\ & a_y=v_x^2+x a_x \end{aligned} $$ Option (A) If $\mathrm{a}_{\mathrm{x}}=1$ and particle is at origin $$ (x=0, y=0) $$ $$ \begin{aligned} & a_y=v_x^2 \\ & a_y=1^2=1 \end{aligned} $$ At origin, at $\mathrm{t}=0 \mathrm{sec}$ speed $=1$ given (B) Option $$ a_y=v_x^2+x a_z $$ given in option $\mathrm{B}, \mathrm{a}_{\mathrm{x}}=0$ $$ \Rightarrow a_y=v_z{ }^2 $$ If $a_x=0, v_x=$ constant $=1$, (all the time) $\Rightarrow \mathrm{a}_y=\mathrm{I}^2=\mathrm{I}$ (all the time) (C) at $\mathrm{t}=0, \mathrm{x}=0 \quad \mathrm{v}_{\mathrm{y}}=\mathrm{x} \mathrm{v}_{\mathrm{x}}$ speed $=1$ $$ \begin{aligned} & v_y=0 \\ & v_z=1 \end{aligned} $$ (D) $$ \begin{aligned} & a_y=v_x^2+x a_2 \\ & v_y=x v_x \\ & a_x=0 \text { (given in D option) } \\ & \Rightarrow a_y=v_x^2 \end{aligned} $$ If $\mathrm{a}_{\mathrm{x}}=0 \Rightarrow \mathrm{~V}_{\mathrm{x}}=$ constant initially $\left(\mathrm{v}_{\mathrm{x}}=1\right)$ $$ \Rightarrow a_y=1^2=1 $$ at $\mathrm{t}=1 \mathrm{sec}$ $$ v_y=0+a_y \times t=1 \times 1=1 $$ $$ \tan \theta=\frac{v_y}{v_s}=x $$ ( $\theta \rightarrow$ angle with x axis) $$ \tan \theta=\frac{v_7}{v_s}=\frac{1}{1}=1 $$ $$ \theta=45^{\circ} $$