It's just a little fight, not breakup

The couple are walking at the same velocity of $\sqrt{2+\sqrt 2}$ on two roads at $45^\circ$ . The rate they are being separated is given by:

$\begin{aligned} r & = 2 \sqrt{2+\sqrt 2} \color{#3D99F6} \cos 67.5^\circ & \small \color{#3D99F6} \text{Note that }\cos \theta = \sin (90^\circ - \theta) \\ & = 2 \sqrt{2+\sqrt 2} \color{#3D99F6} \sin 22.5^\circ \\ & = 2 \sqrt{2+\sqrt 2} \cdot \color{#3D99F6} \frac 12 \sqrt{2-\sqrt 2} & \small \color{#3D99F6} \text{See note.} \\ & = \sqrt{4-2} = \sqrt 2 \approx \boxed{1.414} \end{aligned}$

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Note:

$\begin{aligned} 1 - 2 \sin^2 22.5^\circ & = \cos 45^\circ = \frac 1{\sqrt 2} \\ \implies \sin 22/5^\circ & = \frac 12 \sqrt{2-\sqrt 2} \end{aligned}$

Building upon Chew-Seong's isosceles triangle diagram, let $s(t) = \sqrt{2+\sqrt{2}} \cdot t$ be a function of side length vs. time for each person. Their separation, L(t), can be computed from the diagram using the Law of Cosines:

$L(t)^{2} = 2s(t)^{2} - 2s(t)^{2} \cdot cos(\frac{\pi}{4}) = (2 - 2 \cdot (\frac{1}{\sqrt{2}})) \cdot s(t)^{2} = (2 - \sqrt{2}) \cdot s(t)^{2}$ ;

or $L(t) = \sqrt{2 - \sqrt{2}} \cdot s(t)$ . Since we are interested in the rate of separation (I.e. $\frac{dL}{dt})$ , differentiating with respect to $t$ gives:

$\frac{dL}{dt} = \sqrt{2-\sqrt{2}} \cdot \frac{ds}{dt} = \sqrt{2-\sqrt{2}} \cdot \sqrt{2+\sqrt{2}} = \sqrt{2^2 - (\sqrt{2})^2} = \sqrt{4-2} = \boxed{\sqrt{2}}$ m/s.