Question

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 seperated.

Answer 1

Let P be any point at which the two roads are inclined at an angle of 45°.

Two men A and B are moving along the roads PA and PB respectively with the same speed ‘V’

Let A and B be their final positions such that AB = y

∠APB = 45° and they move with the same speed.

∴ ΔAPB is an isosceles triangle.

Draw PQ ⊥ AB

AB = y

∴ AQ = `y/2` and PA = PB = x  ...(Let)

∠APQ = ∠BPQ

= `45/2`

= `22 1/2^circ`

[∵ In an isosceles Δ, the altitude drawn from the vertex, bisects the base]

Now in right ΔAPQ,

`sin 22 1/2^circ = "AQ"/"AP"`

⇒ `sin 22 1/2^circ = 2/x = y/(2x)`

⇒ y = `2x * sin 22 1/2^circ`

Differentiating both sides w.r.t, t, we get

`"dy"/"dt" = 2 * "dx"/"dt" * sin 22 1/2^circ`

= `2 * "V" * sqrt(2 - sqrt(2))/2`  ......`[because sin 22 1/2^circ = sqrt(2 - sqrt(2))/2]`

= `sqrt(2 - sqrt(2))` V m/s

Hence, the rate of their separation is `sqrt(2 - sqrt(2))` V unit/s.