The acceleration of the fox is due to the change in the direction of its velocity, as its speed is constant. This is a centripetal acceleration. The solution involves finding the state of the system at the moment of closest approach.

Step 1: Condition for Closest Approach The distance between the fox and the rabbit is at a minimum when the rate of change of the distance is zero, which means the component of their relative velocity along the line connecting them is zero. Let $\vec{v}_r$ be the velocity of the rabbit and $\vec{v}_f$ be the velocity of the fox. Let $\hat{u}$ be the unit vector pointing from the rabbit to the fox. The rate of change of the distance $d$ is given by $\frac{dd}{dt} = (\vec{v}_f - \vec{v}_r) \cdot \hat{u}$. The fox's velocity is always directed towards the rabbit, which is opposite to the direction of $\hat{u}$. So, $\vec{v}_f = -v_f \hat{u}$. Substituting this into the equation, we get:

$$ \frac{dd}{dt} = (-v_f \hat{u} - \vec{v}_r) \cdot \hat{u} = -v_f (\hat{u} \cdot \hat{u}) - \vec{v}_r \cdot \hat{u} = -v_f - \vec{v}_r \cdot \hat{u} $$

At the closest approach, $\frac{dd}{dt} = 0$. Therefore:

$$ -v_f - \vec{v}_r \cdot \hat{u} = 0 \implies \vec{v}_r \cdot \hat{u} = -v_f $$

Let $\theta$ be the angle between the rabbit's velocity $\vec{v}_r$ and the line of sight from the rabbit to the fox ($\hat{u}$). Then $\vec{v}_r \cdot \hat{u} = v_r \cos\theta$.

$$ v_r \cos\theta = -v_f $$ $$ \cos\theta = -\frac{v_f}{v_r} = -\frac{4}{5} $$

Step 2: Relative Velocity at Closest Approach The relative velocity of the fox with respect to the rabbit is $\vec{v}_{rel} = \vec{v}_f - \vec{v}_r$. The magnitude of the relative velocity can be found using the law of cosines. The angle between $\vec{v}_f$ and $\vec{v}_r$ is $(\pi - \theta)$, because $\vec{v}_f$ points from the fox to the rabbit, while $\theta$ is defined based on the vector from the rabbit to the fox.

$$ |\vec{v}_{rel}|^2 = v_f^2 + v_r^2 - 2 v_f v_r \cos(\pi - \theta) $$

Since $\cos(\pi - \theta) = -\cos\theta$:

$$ |\vec{v}_{rel}|^2 = v_f^2 + v_r^2 + 2 v_f v_r \cos\theta $$

Substituting $\cos\theta = -v_f/v_r$:

$$ |\vec{v}_{rel}|^2 = v_f^2 + v_r^2 + 2 v_f v_r \left(-\frac{v_f}{v_r}\right) = v_f^2 + v_r^2 - 2v_f^2 = v_r^2 - v_f^2 $$ $$ |\vec{v}_{rel}| = \sqrt{v_r^2 - v_f^2} = \sqrt{(5 \text{ m/s})^2 - (4 \text{ m/s})^2} = \sqrt{25 - 16} \text{ m/s} = \sqrt{9} \text{ m/s} = 3 \text{ m/s} $$

At the moment of closest approach, the relative velocity vector is perpendicular to the line connecting the fox and the rabbit.

Step 3: Angular Velocity of the Line of Sight The fox's velocity vector $\vec{v}_f$ always points towards the rabbit, so it rotates with the line of sight. The angular velocity of the line of sight, $\omega$, is given by the component of the relative velocity perpendicular to the line of sight, divided by the distance. At closest approach, the entire relative velocity is perpendicular to the line of sight.

$$ \omega = \frac{|\vec{v}_{rel}|}{d_{min}} = \frac{3 \text{ m/s}}{30 \text{ m}} = 0.1 \text{ rad/s} $$

Step 4: Calculate the Fox's Acceleration Since the fox's speed $v_f$ is constant, its acceleration is purely centripetal and is related to the rate of change of its velocity direction. This rate of change is determined by the angular velocity $\omega$. The magnitude of the fox's acceleration $a_f$ is given by:

$$ a_f = v_f \omega $$ $$ a_f = (4 \text{ m/s})(0.1 \text{ rad/s}) = 0.4 \text{ m/s}^2 $$

The acceleration vector is perpendicular to the fox's velocity vector, pointing towards the center of curvature of its path.

The magnitude of the fox's acceleration when it is closest to the rabbit is $0.4 \text{ m/s}^2$.