Two men `Pa n dQ` start with velocity `u` at the same time from the junction of two roads inclined at `45^0` to each other. If they travel by different roads, find the rate at which they are being separated.

Explanation

Let two men start from the point C with velocity v each at the same time.
Also, `angleBCA=45^(@)`
Since, A and B are moving with same velocity v. so they will cover same distance in same time.
Therefore, `DeltaABC` is an isosceles triangle with AC=BC.

Now, draw CD `bot` AB.
Let at any instant t, the distance between them is AB.
Let AC=BC=x and AB=y
In `DeltaACD` and `DeltaDCB`,
`angleCAD = angleCBD`
`angleCDA = angleCDB=90^(@)`
`angleACD=angleDCB`
or `angleACD=1/2 xx angleACB`
`rArr angleACD=1/2 xx 45^(@)`
`rArr angleACD = pi/8`
`therefore sinpi/8 = (AD)/(AC)`
`rArr sinpi/8 = (AD)/(AC)`
`rArr sinpi/8 = (y//2)/(x)`
`rArr y/2 = xsinpi/8`
`rArr y=2x.sinpi/8`
Now, differentiating both sides w.r.t., we get
`(dy)/(dt) = 2.sinpi/8.(dx)/(dt)`
`=2.sinpi/8.v` `[therefore v=(dx)/(dt)]`
`=2v.(sqrt(2)-sqrt(2))/(2)` `[therefore sinpi/8 = (sqrt(2)-sqrt(2))/(2)]`
`=sqrt(2-sqrt(2))` v unit/s
Which is the rate at which A and B are bieng separated.