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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.
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Solution
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.
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