\(O A\) and \(O B\) are two roads enclosing an angle of \(120^{\circ} . X\) and \(Y\) start from ' \(O\) ' at the same time. \(X\) travels along \(O A\) with a speed of \(4 \mathrm{~km} / \mathrm{h}\) and \(Y\) travels along \(O B\) with a speed of \(3 \mathrm{~km} / \mathrm{h}\). The rate at which the shortest distance between \(X\) and \(Y\) is increasing after \(1 \mathrm{~h}\) is

Given, speed of \(x\) is \(4 \mathrm{~km} / \mathrm{h}\) and \(y\) is \(3 \mathrm{~km} / \mathrm{h}\). After time \(t\) the distance covered by \(x\) is \(4 t\) and \(y\) is \(3 t\).



Let shortest distance between \(x\) and \(y=A\).
Then by cosine law
\(\quad A^{2}=(4 t)^{2}+(3 t)^{2}-(4 t)(3 t) 2 \cos 120^{\circ}\) \(\Rightarrow \quad A^{2}=16 t^{2}+9 t^{2}-24 t^{2}\left(-\frac{1}{2}\right)\) \(\Rightarrow \quad A^{2}=25 t^{2}+12 t^{2}\) \(\Rightarrow \quad A^{2}=37 t^{2}\) \(\Rightarrow \quad A=\sqrt{37} t\) If \(\quad t=1 \mathrm{~h}\), then \(\quad A=\sqrt{37} \mathrm{~km}\) Now, differentiating Eq. (i) w.r.t. \(t\), we get \(\quad 2 A A^{\prime}=37(2 t)\) After \(t=1 \mathrm{~h}\), we get \(2 \sqrt{37} A^{\prime}=2(37)\)
After \(t=1 \mathrm{~h}\), we get
\(\Rightarrow \quad A^{\prime}=\sqrt{37}\)
Thus, rate at which shortest distance \(A\) changes with time is \(\sqrt{37} \mathrm{~km} / \mathrm{h}\).