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.