Answer
$r = 2.92~m$
Work Step by Step
In uniform circular motion, the acceleration vector is always directed toward the center of the circle.
We can find the angle of the acceleration vector relative to the positive x axis at $t_1 = 2.00~s$:
$tan~\theta = \frac{4.00}{6.00}$
$\theta = tan^{-1}~\frac{4.00}{6.00}$
$\theta = 33.7^{\circ}$
We can find the angle of the acceleration vector relative to the positive x axis at $t_2 = 5.00~s$:
$tan~\theta = \frac{-6.00}{4.00}$
$\theta = tan^{-1}~\frac{-6.00}{4.00}$
$\theta = -56.3^{\circ}$
In uniform circular motion, the position vector is directed opposite the acceleration vector.
At $t_1 = 2.00~s$, the direction of the position vector is $~~180^{\circ}+33.7^{\circ} = 213.7^{\circ}$
At $t_2 = 5.00~s$, the direction of the position vector is $~~180^{\circ}-56.3^{\circ} = 123.7^{\circ}$
The particle moves counterclockwise. We can find the angle the particle moves in a time of $3.00~s$:
$\theta = (360^{\circ}-213.7^{\circ})+(123.7^{\circ}) = 270^{\circ}$
Then the particle completes a full circle in $4.00~s$
We can find an expression for the speed:
$v = \frac{2\pi~r}{4.00~s}$
We can find the magnitude of the acceleration:
$a = \sqrt{(4.00~m/s^2)^2+(6.00~m/s^2)^2} = 7.21~m/s^2$
We can find the radius:
$a = \frac{v^2}{r}$
$a = \frac{(\frac{2\pi~r}{4.00~s})^2}{r}$
$a = \frac{4\pi^2~r}{(4.00~s)^2}$
$r = \frac{a~(4.00~s)^2}{4\pi^2}$
$r = \frac{(7.21~m/s^2)~(4.00~s)^2}{4\pi^2}$
$r = 2.92~m$
