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

$\begin{array}{l}\mathrm{y}=\frac{{\mathrm{x}}^2}{2}\\ \text{ at }\left.\mathrm{t}=0\begin{array}{c}\mathrm{x}=0,\mathrm{y}=0\\ \mathrm{u}=1\end{array}\right\}\text{ given }\\ \mathrm{y}=\frac{{\mathrm{x}}^2}{2}\\ \frac{\mathrm{dy}}{\mathrm{dt}}=\frac{1}{2}\cdot 2\mathrm{x}\frac{\mathrm{dx}}{\mathrm{dt}}\\ \Rightarrow {\mathrm{v}}_{\mathrm{y}}={\mathrm{xv}}_{\mathrm{x}}\end{array}$

differentiate wrt time

$\begin{array}{l}{a}_y=\frac{dx}{dt}\cdot V_x+x{a}_x\\ a_y=v_x^2+x{a}_x\end{array}$

Option

(A) If a_x = 1 and particle is at origin

(x = 0, y =0)

$a_y=v_x^2$
$a_y$ = 1² = 1
At origin, at t = 0 sec
speed = 1 (given)
(B)   Option

$\begin{array}{l}{a}_y=v_x^2+x{a}_x\\ \text{ given in option }\mathrm{B},{\mathrm{a}}_{\mathrm{x}}=0\\ \Rightarrow a_{\mathrm{y}}={\mathrm{v}}_{\mathrm{x}}^2\end{array}$

$\begin{array}{l}\text{ If }{a}_x=0,v_x=\text{ constant }=1,\left(\text{ all the time }\right)\\ \Rightarrow a_y=1^2=1\text{ (all the time) }\end{array}$

$\begin{array}{l}\text{ (C) at }\mathrm{t}=0,\mathrm{x}=0{\mathrm{v}}_{\mathrm{y}}={\mathrm{xv}}_{\mathrm{x}}\\ \text{ speed }=1\\ {\mathrm{v}}_{\mathrm{y}}=0\\ {\mathrm{v}}_{\mathrm{x}}=1\end{array}$

$\begin{array}{l}D)\hspace{0.17em}{a}_y=v_x^2+x{a}_x\\ v_{\mathrm{y}}=x{v}_{\mathrm{x}}\\ {\mathrm{a}}_{\mathrm{x}}=0\text{ (given in D option) }\\ \Rightarrow a_y=v_x^2\end{array}$

$\begin{array}{l}\text{ If }{\mathrm{a}}_{\mathrm{x}}=0\Rightarrow {\mathrm{V}}_{\mathrm{x}}=\text{ constant initially }\left({\mathrm{v}}_{\mathrm{x}}=1\right)\\ \Rightarrow {\mathrm{a}}_{\mathrm{y}}=1^2=1\\ \text{ at }\mathrm{t}=1\sec \\ {\mathrm{v}}_{\mathrm{y}}=0+{\mathrm{a}}_{\mathrm{y}}\times \mathrm{t}=1\times 1=1\\ \tan \theta =\frac{{\mathrm{v}}_{\mathrm{y}}}{{\mathrm{v}}_{\mathrm{x}}}=\mathrm{x}\\ (\theta \to \text{ angle with }\mathrm{x}\text{ axis })\\ \tan \theta =\frac{{\mathrm{v}}_{\mathrm{y}}}{{\mathrm{v}}_{\mathrm{x}}}=\frac{1}{1}=1\\ \theta ={45}^{\circ }\end{array}$