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(A) The total acceleration vector makes an angle of $135^{\circ}$ with the positive $x$-axis (tangent direction). The centripetal acceleration $a_c$ i

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$4$

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(A) The total acceleration vector makes an angle of $135^{\circ}$ with the positive $x$-axis (tangent direction). The centripetal acceleration $a_c$ is directed towards the center (negative $x$-direction). The component of the total acceleration $a = 8\sqrt{2} \ m/s^2$ along the radial direction is $a_c = a \cos(180^{\circ} - 135^{\circ}) = 8\sqrt{2} \cos(45^{\circ}) = 8\sqrt{2} 	imes \frac{1}{\sqrt{2}} = 8 \ m/s^2$. We know that the centripetal acceleration is given by $a_c = \frac{v^2}{r}$. Substituting the values,$8 = \frac{v^2}{2}$. $v^2 = 16$. $v = 4 \ m/s$.

The acceleration of a particle performing non-uniform circular motion is shown in the diagram at point $P$. If the radius of the circular path is $2 \ m$,then what is the velocity at point $P$ in $m/s$?

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