Question

A man in a car at location $\text{Q}$ on a straight highway is moving with speed $v$. He decides to reach a point $\text{P}$ in a field at a distance $d$ from the highway (point $\text{M}$) as shown in the figure. Speed of the car in the field is half to that on the highway. What should be the distance $\text{RM}$, so that the time taken to reach $\text{P}$ is minimum ?

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

Answer 1

The correct option is A $\frac{d}{\sqrt{3}}$

Let the car take the turn of the highway at a distance ${}^{\prime }{x}^{\prime }$ from the point $\text{M}$.

So, $\text{RM}=x$

Given the speed of the car at $Q$ is $v$, so the time taken by the car to cover the distance $QR(=QM-x)$ on the highway is,

$t_1=\frac{QM-x}{v}.......\left(1\right)$

Time taken to travel the distance ${}^{\prime }R{P}^{\prime }$ in the field is,

$t_2=\frac{\sqrt{d^2+x^2}}{v/2}.......\left(2\right)$

Total time elapsed to move the car from $Q$ to $P$ is,

$t=t_1+t_2=\frac{QM-x}{v}+\frac{\sqrt{d^2+x^2}}{v/2}$

For $t$ to be minimum, $\frac{dt}{dx}=0$

$\frac{1}{v}[-1+\frac{2\times \left(2x\right)}{2\sqrt{d^2+x^2}}]=0$

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

$\Rightarrow d^2+x^2=4x^2$

$\therefore x=\frac{d}{\sqrt{3}}$

Hence, option $\left(\text{A}\right)$ is correct.