According to given information, we have the following figure Now, from $\triangle A C D$ and $\triangle B C D$, we have $\tan 30^{\circ}=\frac{h}{x+y}$ and $\tan 60^{\circ}=\frac{h}{y}$ $\Rightarrow h=\frac{x+y}{\sqrt{3}}\quad...(i)$ and $h=\sqrt{3} y\quad...(ii)$ From Eqs. (i) and (ii), $\frac{x+y}{\sqrt{3}}=\sqrt{3} y \Rightarrow x+y=3 y$ $\Rightarrow x-2 y=0 \Rightarrow y=\frac{x}{2}$ $\because$ Speed is uniform. $\therefore$ Distance $y$ will be cover in $5 \mathrm{~min}$. $\because$ Distance $x$ covered in $10 \mathrm{~min}$. $\therefore$ Distance $\frac{x}{2}$ will be cover in $5 \mathrm{~min}$.