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From point $A$ located on a highway as shown in figure, one has to get by car as soon as possible to point $B$ located in the field at a distance $ℓ$ from the highway. It is known that car moves in the field $η$ time slower on the highway. At what distance from point $D$ one must turn off the highway?

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$\frac{ℓ}{\sqrt{η^{2} - 2}}$

$\frac{ℓ}{\sqrt{η^{2} - 7}}$

$\frac{ℓ}{\sqrt{η^{2} - 4}}$

$\frac{ℓ}{\sqrt{η^{2} - 1}}$

answer is B.

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Detailed Solution

Suppose x distance from D the car turns off the highway. Let v be the speed of car on highway, then its speed on field will be $v / η$ . The time of motion,

$t = (\frac{A D - x}{v}) + \frac{\sqrt{x^{2} + ℓ^{2}}}{v / η}$

t to be minimum, $\frac{d t}{d x} = 0$

or $\frac{d t}{d x} = \frac{d}{d x} [\frac{A D - x}{v} + \frac{η}{v} {(x^{2} + ℓ^{2})}^{1 / 2}]$

[A D is constant ]

$0 = - \frac{1}{v} + \frac{η}{v} \times \frac{1}{2} {(x^{2} + ℓ^{2})}^{- 1 / 2} \times 2 x$

$η^{2} x^{2} = x^{2} + ℓ^{2} ∴ x = \frac{ℓ}{\sqrt{η^{2} - 1}} .$

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