$O A$ 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 hour is

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Aicharya · Accepted Answer

After time $t$ hour distance covered by $X$ and $Y$ are $4 t$ and $3 t$ respectively. Now, shortest distance between $X$ and $Y$ after time t hour is given by $\begin{aligned} A B^{2} &=O A^{2}+O B^{2}-2 O A \cdot O B \cos 	heta \ &=(4 t)^{2}+(3 t)^{2}-2 	imes 4 t 	imes 3 t \cos 120^{\circ} \ &=16 t^{2}+9 t^{2}-24 t^{2} 	imes\left(\frac{-1}{2}ight) \ &=16 t^{2}+9 t^{2}+12 t^{2}=37 t^{2} \ \Rightarrow \quad 	ext { Now, } \frac{d(A B)}{d t} &=\sqrt{37} \end{aligned}$ $	herefore$ Rate at which the shortest distance between $x$ and $y$ is increasing is $\sqrt{37} \mathrm{~km} / \mathrm{h}$.

$O A$ and $O B$ are two roads enclosing an angle of $120^{\circ} . X$ and $Y$ start from $O$ at the same…

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