A man in a car at location Q on a straight hi | vedclass

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Question

(A) Let the car turn off the highway at a distance $x$ from point $M$. So,$RM = x$.Let the speed of the car on the highway be $v_h = v$ and the speed

Vedclass · Accepted Answer

$\frac{d}{\sqrt{3}}$

Vedclass · Answer

(A) Let the car turn off the highway at a distance $x$ from point $M$. So,$RM = x$.Let the speed of the car on the highway be $v_h = v$ and the speed in the field be $v_f = v/2$.The distance $QM$ is constant. Let $QM = L$. The distance covered on the highway is $QM - x = L - x$.The time taken to travel on the highway is $t_1 = \frac{L - x}{v}$.The distance $RP$ in the field is $\sqrt{d^2 + x^2}$.The time taken to travel in the field is $t_2 = \frac{\sqrt{d^2 + x^2}}{v/2} = \frac{2\sqrt{d^2 + x^2}}{v}$.Total time $t = t_1 + t_2 = \frac{L - x}{v} + \frac{2\sqrt{d^2 + x^2}}{v}$.For minimum time,$\frac{dt}{dx} = 0$.$\frac{d}{dx} \left( \frac{L - x}{v} + \frac{2\sqrt{d^2 + x^2}}{v} \right) = 0$.$\frac{1}{v} \left( -1 + 2 \cdot \frac{1}{2\sqrt{d^2 + x^2}} \cdot 2x \right) = 0$.$-1 + \frac{2x}{\sqrt{d^2 + x^2}} = 0 \implies \frac{2x}{\sqrt{d^2 + x^2}} = 1$.$4x^2 = d^2 + x^2 \implies 3x^2 = d^2 \implies x = \frac{d}{\sqrt{3}}$.

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