[Q1] Find the magnitude of the dog's acceleration at this instant.

The dog moves at a constant speed $v_2$, but its velocity vector changes direction. Therefore, its acceleration is purely centripetal (or normal), directed perpendicular to its velocity. The magnitude of this acceleration is given by:

$$a = v_2 \omega$$

where $\omega$ is the angular velocity of the dog's velocity vector. Since the dog always aims at the fox, its velocity vector is always aligned with the line segment DF. Thus, $\omega$ is the angular velocity of the line segment DF.

The angular velocity of the line segment connecting two points is given by the component of their relative velocity perpendicular to the line segment, divided by the distance between them.

$$\omega = \frac{v_{\text{rel}, \perp}}{r}$$

At the given instant, the distance is $r = L$. Let's set up a coordinate system where the fox is at the origin (0,0), its velocity $\vec{v}_F$ is along the x-axis ($\vec{v}_F = v_1 \hat{i}$), and the dog is at (0, -L). The dog's velocity $\vec{v}_D$ is aimed at the fox, so it is along the y-axis ($\vec{v}_D = v_2 \hat{j}$). The line segment DF is along the y-axis.

The relative velocity is $\vec{v}_{\text{rel}} = \vec{v}_F - \vec{v}_D = v_1 \hat{i} - v_2 \hat{j}$. The component of the relative velocity perpendicular to the line segment DF (i.e., perpendicular to the y-axis) is the x-component, so $v_{\text{rel}, \perp} = v_1$.

The angular velocity of the line segment DF at this instant is:

$$\omega = \frac{v_1}{L}$$

Therefore, the magnitude of the dog's acceleration is:

$$a = v_2 \omega = \frac{v_1 v_2}{L}$$

[Q2] Find the time it takes for the dog to catch up with the fox.

Let $r(t)$ be the distance between the dog and the fox, and let $\theta(t)$ be the angle between the line segment DF and the fox's path AB. The rate at which the distance between them decreases is the relative speed along the line of sight.

$$-\frac{dr}{dt} = v_2 - v_1 \cos\theta$$

Integrating this equation from the initial moment ($t=0$, $r=L$) to the catch time ($t_c$, $r=0$):

$$\int_L^0 -dr = \int_0^{t_c} (v_2 - v_1 \cos\theta) dt$$ $$L = v_2 t_c - v_1 \int_0^{t_c} \cos\theta dt \quad (*)$$

To solve this, we need to evaluate the integral. We can do this by considering the motion projected onto the fox's path (the line AB). For the dog to catch the fox, their displacements along AB must be equal at the time of capture.

The fox's displacement along AB is:

$$\Delta x_F = v_1 t_c$$

The dog's velocity component along AB is $v_{Dx} = v_2 \cos\theta$. The dog's displacement along AB is the integral of this component over time:

$$\Delta x_D = \int_0^{t_c} v_2 \cos\theta dt$$

Equating the displacements, $\Delta x_D = \Delta x_F$:

$$\int_0^{t_c} v_2 \cos\theta dt = v_1 t_c$$ $$\int_0^{t_c} \cos\theta dt = \frac{v_1}{v_2} t_c \quad (**)$$

Now, we substitute the result for the integral from $(**)$ into equation $(*)$:

$$L = v_2 t_c - v_1 \left( \frac{v_1}{v_2} t_c \right)$$ $$L = t_c \left( v_2 - \frac{v_1^2}{v_2} \right) = t_c \left( \frac{v_2^2 - v_1^2}{v_2} \right)$$

Solving for the catch time $t_c$:

$$t_c = \frac{L v_2}{v_2^2 - v_1^2}$$

This result is valid since the problem states $v_2 > v_1$, ensuring a positive, finite time.