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{d t}{d x}=\frac{0-1}{v}+\frac{1}{v / 2} \frac{1}{2 \sqrt{x^{2}+d^{2}}} \times 2 x $
$\Rightarrow \frac{d t}{d x}=\frac{-1}{v}+\frac{2 x}{v \sqrt{x^{2}+d^{2}}}$

Put $\frac{d t}{d x}=0$, we get
$\frac{-1}{v}+\frac{2 x}{v \sqrt{x^{2}+d^{2}}}=0 $
$\Rightarrow \frac{2 x}{v \sqrt{x^{2}+d^{2}}}=\frac{1}{v}$
$ \Rightarrow 2 x=\sqrt{x^{2}+d^{2}} \Rightarrow 4 x^{2}=x^{2}+d^{2}$
$\Rightarrow 3 x^{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.