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**Hint:**First of all let us calculate the velocity of the particle. The velocity components will be there in both the directions $x$ and $y$. so first of all find out these components. Taking the resultant of these will give us the speed of the particle. These will be helping you in solving this question.

**Complete step-by-step answer:**

it has been mentioned in the question that the curve is mentioned in the equation as,

$y=\dfrac{{{x}^{2}}}{2}$

And also the $x$coordinates varies with the time which is given as,

$x=\dfrac{{{t}^{2}}}{2}$

As we all know, the velocity in a particular direction is the change in position of the object in that direction. Therefore we can write that,

The velocity of the particle in the $x$-direction can be written as,

$\therefore {{v}_{x}}=\dfrac{dx}{dt}$

As already mentioned the value of the $x$ can be written as,

$x=\dfrac{{{t}^{2}}}{2}$

Substituting this in the equation will give,

${{v}_{x}}=\dfrac{dx}{dt}=\dfrac{d}{dt}\left( \dfrac{{{t}^{2}}}{2} \right)$

Performing the differentiation will give,

${{v}_{x}}=\dfrac{d}{dt}\left( \dfrac{{{t}^{2}}}{2} \right)=\dfrac{2\times t}{2}=t$

In the similar way the velocity in the $y$direction can be written as,

${{v}_{y}}=\dfrac{d}{dt}y$

We already mentioned the equation of the curve.

$y=\dfrac{{{x}^{2}}}{2}$

And also,

$x=\dfrac{{{t}^{2}}}{2}$

Substituting this in the equation of curve will give,

$y=\dfrac{{{\left( \dfrac{{{t}^{2}}}{2} \right)}^{2}}}{2}=\dfrac{{{t}^{4}}}{8}$

Substitute this in the equation of velocity.

$\begin{align}

& {{v}_{y}}=\dfrac{d}{dt}y \\

& \Rightarrow {{v}_{y}}=\dfrac{d}{dt}\left( \dfrac{{{t}^{4}}}{8} \right)=\dfrac{4{{t}^{3}}}{8}=\dfrac{{{t}^{3}}}{2} \\

\end{align}$

As the value of time is mentioned in the question as,

$t=2s$

Let us substitute in the velocity equations will give,

\[{{v}_{y}}=\dfrac{{{t}^{3}}}{2}=\dfrac{{{2}^{3}}}{2}=4m{{s}^{-1}}\]

And the velocity in \[x\] direction will be given as,

\[{{v}_{x}}=t=2s\]

The magnitude of the velocity will be the speed. That is the resultant of the speed can be written as,

\[v=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}\]

Substituting the values in it will give,

\[v=\sqrt{{{2}^{2}}+{{4}^{2}}}=2\sqrt{5}m{{s}^{-1}}\]

Therefore the speed of the object will be given as,

\[s=2\sqrt{5}m{{s}^{-1}}\]

**So, the correct answer is “Option A”.**

**Note:**The velocity of the object can be shown in vector form as the \[\hat{i}\] and \[\hat{j}\]components. That is we can write that,

\[\vec{v}=2\hat{i}+4\hat{j}\]

The angle between velocity vectors will be the tangent of the velocity components. That is,

\[\begin{align}

& \tan \theta =\dfrac{4}{2}=2 \\

& \Rightarrow \theta =63.435{}^\circ \\

\end{align}\]

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