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 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 $R M$ , so that the time taken to reach $P$ is minimum?

Accepted Answer

A man in a car at location $Q$ on a straight highway is moving with speed $v$ .

Let distance $Q R = x$ and $Q M = L$ and $R M = L - x$

Time taken to travel from $Q$ to $R$ is $t_{Q R} = \frac{x}{v}$ .

Speed of the car in the field is half to that on the highway.

Time taken to travel from $R$ to $P$ is $t_{R P} = \frac{\sqrt{d^{2} + {(L - x)}^{2}}}{\frac{v}{2}} = \frac{2 \sqrt{d^{2} + {(L - x)}^{2}}}{v}$

Total time $t = t_{Q R} + t_{R P}$

$\Rightarrow t = \frac{x}{v} + \frac{2 \sqrt{d^{2} + {(L - x)}^{2}}}{v} . . . . . (i)$

For time $t$ to be a minimum,

$\frac{d t}{d x} = 0$

Now, differentiate the equation $(i)$ with respect to $x$ ,

$\frac{d t}{d x} = \frac{d}{d x} (\frac{x}{v}) + \frac{d}{d x} (\frac{2 \sqrt{d^{2} + {(L - x)}^{2}}}{v})$

$\Rightarrow \frac{d t}{d x} = \frac{1}{v} + \frac{2}{v \times 2 \sqrt{d^{2} + {(L - x)}^{2}}} \times \frac{d}{d x} (d^{2} + {(L - x)}^{2})$

$\Rightarrow \frac{d t}{d x} = \frac{1}{v} + \frac{2}{v \times 2 \sqrt{d^{2} + {(L - x)}^{2}}} \times 2 (L - x) \times (- 1)$

$\Rightarrow \frac{d t}{d x} = \frac{1}{v} - \frac{2 (L - x)}{v \times \sqrt{d^{2} + {(L - x)}^{2}}} . . . . . (ii)$

Now, $\frac{d t}{d x} = 0$

$\Rightarrow \frac{1}{v} - \frac{2 (L - x)}{v \times \sqrt{d^{2} + {(L - x)}^{2}}} = 0$

$\Rightarrow \frac{1}{v} = \frac{2 (L - x)}{v \times \sqrt{d^{2} + {(L - x)}^{2}}}$

$\Rightarrow 2 (L - x) = \sqrt{d^{2} + {(L - x)}^{2}}$

Squaring both sides,

$\Rightarrow 4 {(L - x)}^{2} = d^{2} + {(L - x)}^{2}$

$\Rightarrow 3 {(L - x)}^{2} = d^{2}$

$\Rightarrow {(L - x)}^{2} = \frac{d^{2}}{3}$

$\Rightarrow (L - x) = \sqrt{\frac{d^{2}}{3}} = \frac{d}{\sqrt{3}}$

$\Rightarrow R M = \frac{d}{\sqrt{3}}$

Differentiating equation $(ii)$ with respect to $x$ ,

$\frac{d^{2} t}{d x^{2}} = \frac{2 d^{2}}{v {[{(L - x)}^{2} + d^{2}]}^{\frac{3}{2}}} > 0$

So $L - x = \frac{d}{\sqrt{3}}$ gives the point of minima.

Hence, the time taken to reach $P$ is minimum when the distance $R M = \frac{d}{\sqrt{3}}$ .

Important Questions on Application of Derivatives

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