Answer

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**Hint:**We know that the radial acceleration is equal to the rate of change in velocity . The change which generates in speed during the non uniform circular motion known as tangential acceleration.

**Complete step-by-step solution:**

let's draw a diagram of this question to solve this question-

here in this problem , tangential acceleration = \[9m/{{s}^{2}}\]

Radius =\[4m\]

from the fig we get triangle ABC ,

\[\begin{align}

& \tan {{45}^{\circ }}\Rightarrow \dfrac{at}{ac} \\

& or,1\Rightarrow \dfrac{9}{ac} \\

& or,ac\Rightarrow 9m/{{s}^{2}} \\

\end{align}\]

Now , we know that tangential acceleration is

\[at\Rightarrow \dfrac{dv}{dt}\]

\[or,dv\Rightarrow at.dt\]

\[or,\int_{0}^{v}{dv\Rightarrow \int\limits_{0}^{t}{9dt}}\]

\[or,v\Rightarrow 9t\] \[m/s\]

Again , we know that centripetal acceleration \[\left( ac \right)=\dfrac{{{v}^{2}}}{r}\]

Now , using \[(ac)=(at)\] we get

\[\dfrac{{{v}^{2}}}{r}\Rightarrow \dfrac{dv}{dt}\]

\[or,\dfrac{{{(9t)}^{2}}}{4}\Rightarrow 9\]

\[or,\dfrac{81t}{4}\Rightarrow 9\]

\[or,{{1}^{2}}\Rightarrow \dfrac{4}{9}\]

\[or,t\Rightarrow \dfrac{2}{3}\sec \]

**Therefore, the correct option is B) \[\dfrac{2}{3}s\].**

**Note:**During circular motion the speed vector changes its direction at each point on the circle . This suggests that the tangential component of acceleration is always non-zero.

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