A man in a car at location $Q$ on a straight highway is moving with speed $v$. He decides to reach a point $P$ in a field at a distance $d$ from the highway (point $M$ on the highway is the foot of the perpendicular from $P$). He leaves the highway at a point $R$ between $Q$ and $M$ and drives straight to $P$. Speed of the car in the field is half to that on the highway. What should be the distance $RM$, so that the time taken to reach $P$ is minimum?
Answer: (A) $\dfrac{d}{\sqrt3}$
Let $QM = D$ and $RM = x$. The total time is
$$t = \frac{D - x}{v} + \frac{\sqrt{x^2 + d^2}}{v/2}$$
For the minimum,
$$\frac{dt}{dx} = -\frac1v + \frac{2x}{v\sqrt{x^2 + d^2}} = 0 \quad\Rightarrow\quad \sqrt{x^2 + d^2} = 2x$$
$$3x^2 = d^2 \quad\Rightarrow\quad x = \frac{d}{\sqrt3}$$
Solution by Sreeraj P, M.Sc Physics