#### Two men A and B start with velocities v at the same time from the junction of two roads inclined at 45° to each other. If they travel by different roads, find the rate at which they are being separated.

Given: two men A and B start with velocities v at the same time from the junction of the two roads inclined at 45° to each other

To find: The rate of separation of the two men

Explanation:

The distance x travelled by A and B on any given time t will be same as they have velocity.

Hence

\begin{aligned} &\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v} \ldots \ldots (1)\\ &\text { In the } \triangle A O B \text { , after applying the cosine rule, we get }\\ &y^{2}=x^{2}+x^{2}-2 x \cdot x \cdot \cos 45^{\circ} \end{aligned}

\\ \begin{aligned} &\Rightarrow \mathrm{y}^{2}=2 \mathrm{x}^{2}-2 \mathrm{x}^{2} \cdot \frac{1}{\sqrt{2}}\\ &\Rightarrow \mathrm{y}^{2}=2 \mathrm{x}^{2}\left(1-\frac{1}{\sqrt{2}}\right)\\ &\Rightarrow \mathrm{y}^{2}=2 \mathrm{x}^{2}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)\\ &\text { Lets multiple and divide by } \sqrt{2} \text { , }\\ &\Rightarrow \mathrm{y}^{2}=2 \mathrm{x}^{2}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) \times \frac{\sqrt{2}}{\sqrt{2}}\\ &\Rightarrow \mathrm{y}^{2}=\sqrt{2} \mathrm{x}^{2}(\sqrt{2}-1)\\ &\Rightarrow \mathrm{y}=\sqrt{\sqrt{2} \mathrm{x}^{2}(\sqrt{2}-1)}\\ &\Rightarrow \mathrm{y}= \mathrm{x} \cdot \sqrt{(2-\sqrt{2})} \end{aligned}

Apply the derivatives with respect to t,

$\Rightarrow \frac{d y}{d t}=\frac{d( x \cdot \sqrt{(2-\sqrt{2}})}{d t}$

Now take out the constant terms, we get

$\Rightarrow \frac{d y}{d t}= \sqrt{(2-\sqrt{2})} \frac{d(x)}{d t}$

After value substitution for equation (i) we get

$\Rightarrow \frac{d y}{d t}= \sqrt{(2-\sqrt{2})}(v) \mathrm{m} / \mathrm{s}$
Thus, the above given amount is the rate of separation of the two roads.