Question

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 $M$) as shown in the figure. 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 ?

• A. $d$
• B. $\frac{d}{\sqrt{2}}$
• C. $\frac{d}{2}$
• D. $\frac{d}{\sqrt{3}}$

Answer 1

Answer: (D)

Let the distance $QM=l$ and distance $RM=x$.
Time to reach from $Q$ to $R$ is $t_1=\frac{l-x}{v}$
Time to reach from $R$ to $P$ is $t_2=\frac{\sqrt{x^2+d^2}}{v/2}$
Therefore,$t=t_1+t_2=\frac{l-x}{v}+\frac{\sqrt{x^2+d^2}}{v/2}$
On differentiating, we get
$\frac{dt}{dx}=\frac{0-1}{v}+\frac{1}{v/2}\frac{1}{2\sqrt{x^2+d^2}}\times 2x$
$\Rightarrow \frac{dt}{dx}=\frac{-1}{v}+\frac{2x}{v\sqrt{x^2+d^2}}$

Put $\frac{dt}{dx}=0$, we get
$\frac{-1}{v}+\frac{2x}{v\sqrt{x^2+d^2}}=0$
$\Rightarrow \frac{2x}{v\sqrt{x^2+d^2}}=\frac{1}{v}$
$\Rightarrow 2x=\sqrt{x^2+d^2}\Rightarrow 4x^2=x^2+d^2$
$\Rightarrow 3x^2=d^2$
$\Rightarrow x=\frac{d}{\sqrt{3}}$
Therefore, the distance $RM=x=\frac{d}{\sqrt{3}}$,
the time taken to reach $P$ is minimum.