QJEEM18Ph8961: JEE Main 2018 Physics - Class 11 JEEM18Ph8961 Solution: A man in a car at location Q on a straight highway is moving…

Let the distance QM = l and distance RM = x.

Time to reach from Q to R is ${t_1} = {{l - x} \over v}$

Time to reach from R to P is ${t_2} = {{\sqrt {{x^2} + {d^2}} } \over {v/2}}$

Therefore, $t = {t_1} + {t_2} = {{l - x} \over v} + {{\sqrt {{x^2} + {d^2}} } \over {v/2}}$

On differentiating, we get

${{dt} \over {dx}} = {{0 - 1} \over v} + {1 \over {v/2}}{1 \over {2\sqrt {{x^2} + {d^2}} }} \times 2x$

$\Rightarrow {{dt} \over {dx}} = {{ - 1} \over v} + {{2x} \over {v\sqrt {{x^2} + {d^2}} }}$

Put ${{dt} \over {dx}} = 0$ , we get

${{ - 1} \over v} + {{2x} \over {v\sqrt {{x^2} + {d^2}} }} = 0$

$\Rightarrow {{2x} \over {v\sqrt {{x^2} + {d^2}} }} = {1 \over v}$

$\Rightarrow 2x = \sqrt {{x^2} + {d^2}} \Rightarrow 4{x^2} = {x^2} + {d^2}$

$\Rightarrow 3{x^2} = {d^2} \Rightarrow x = {d \over {\sqrt 3 }}$

Therefore, the distance $RM = x = {d \over {\sqrt 3 }}$ , the time taken to reach P is minimum.