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 $$y=x22$$. The x and y components of its acceleration are denoted by ax and ay, respectively. Then

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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 $$y=x22$$. The x and y components of its acceleration are denoted by ax and ay, respectively. Then

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$$y=x22$$ at $$t=0x=0$$,$$y=0u=1$$ given $$y=x22dydt=12$$⋅$$2xdxdt$$⇒$$vy=xvxdifferentiate$$ wrt $$timeay=dxdt$$⋅$$Vx+xaxay=vx2+xaxOption(A)$$ If ax $$=$$ $$1$$ and particle is at $$origin(x$$ $$=$$ $$0$$, y $$=0)ay=vx2ay$$ $$=$$ $$12$$ $$=$$ $$1At$$ origin, at t $$=$$ $$0$$ secspeed $$=$$ $$1$$ (given)(B)   $$Optionay=vx2+xax$$ given in option B,$$ax=0$$⇒$$ay=vx2$$ If $$ax=0$$,$$vx=$$ constant $$=1$$,$$($$ all the time $$)$$⇒$$ay=12=1$$ (all$$ the $$time)  (C)  at $$t=0$$,$$x=0$$ $$vy=xvx$$ speed $$=1vy=0vx=1D)$$ $$ay=vx2+xaxvy=xvxax=0$$ (given$$ in D $$option) ⇒$$ay=vx2$$ If $$ax=0$$⇒$$Vx=$$ constant initially $$vx=1$$⇒$$ay=12=1$$ at $$t=1secvy=0+ay$$×$$t=1$$×$$1=1tan$$⁡θ$$=vyvx=x($$θ→ angle with x axis $$)$$$$tan$$⁡θ$$=vyvx=11=1$$θ$$=45$$∘

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=x22. The x and y components of its acceleration are denoted by ax and ay, respectively. Then

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