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Question Number 123212 by 676597498 last updated on 24/Nov/20

Commented by 676597498 last updated on 24/Nov/20

$$plsineedhelp$$

Answered by MJS_new last updated on 24/Nov/20

$$\mathrm{mouse}$$$$\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}0\\ 1\end{array}\right)+\left(\begin{array}{c}6p\\ -2p\end{array}\right)$$$$\mathrm{cat}$$$$\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}0\\ -1\end{array}\right)+\left(\begin{array}{c}8q\\ q\end{array}\right)$$$$\mathrm{the}\mathrm{paths}\mathrm{intersect}\mathrm{at}$$$$6p=8q\wedge 1-2p=-1+q\Rightarrow p=\frac{8}{11}\wedge q=\frac{6}{11}\Rightarrow$$$$\Rightarrow \left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}48/11\\ -5/11\end{array}\right)$$$$\mathrm{but}\mathrm{they}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{meet}\mathrm{because}p\ne q.p>q\mathrm{means}$$$$\mathrm{the}\mathrm{mouse}\mathrm{is}\mathrm{late}$$$$\mathrm{for}\mathrm{the}\mathrm{time}\mathrm{when}\mathrm{they}\mathrm{are}\mathrm{the}\mathrm{nearest}\mathrm{let}$$$$p=q=t$$$$\mathrm{the}\mathrm{position}\mathrm{of}\mathrm{the}\mathrm{mouse}\mathrm{is}\left(\begin{array}{c}6t\\ 1-2t\end{array}\right)$$$$\mathrm{the}\mathrm{position}\mathrm{of}\mathrm{the}\mathrm{cat}\mathrm{is}\left(\begin{array}{c}8t\\ -1+t\end{array}\right)$$$$\mathrm{their}\mathrm{distance}\mathrm{is}$$$$\sqrt{{(6t-8t)}^2+(1-2t-{(-1+t)}^2}=$$$$=\sqrt{13t^2-12t+4}$$$$\mathrm{the}\mathrm{minimum}\mathrm{distance}\mathrm{is}\mathrm{at}t=\frac{6}{13}\mathrm{and}\mathrm{is}$$$$\frac{4\sqrt{13}}{13}\approx 1.11\mathrm{m}$$

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