Question

A swimmer S is in the sea at a distance d km from the closest point A on a straight shore The house of the swimmer is on the shore at a distance $Lkm$ from A He can swim at a speed of $ukm/hr$ and walk at a speed of $vkm/hr(v>u)$ At what point on the shore should he land so that he reaches his house in the shortest possible time?

• A. $\frac{ud}{\sqrt{v^2-u^2}}$
• B. $\frac{vd}{\sqrt{v^2-u^2}}$
• C. $\frac{vd}{\sqrt{v^2+u^2}}$
• D. $\frac{ud}{\sqrt{v^2+u^2}}$

Answer 1

The correct option is A $\frac{ud}{\sqrt{v^2-u^2}}$

Let the house of the swimmer be at $B.$
$\therefore AB=Lkm$
Let the swimmer land at $C,$ on the shore and let
$AC=xkm$
$\therefore SC\sqrt{x^2+d^2}$ and $CB=(L-x)$
Therefore time=distance/speed
Time from $S$ to $B=$ time from $S$ to $C+$ time from $C$ to $B.$
$\therefore T=\frac{\sqrt{x^2+d^2}}{u}+\frac{L-x}{v}$
Let $f\left(x\right)=T=\frac{1}{u}\sqrt{x^2+d^2}+\frac{L}{v}-\frac{x}{v}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{u}.\frac{1,2x}{2\sqrt{x^2+d^2}}+0-\frac{1}{v}$
For maximum or minimum, put $\Rightarrow f^{\prime }\left(x\right)=0$
$\Rightarrow v^2{x}^2=u^2(x^2+d^2)$
$\Rightarrow x^2=\frac{u^2{d}^2}{v^2-u^2}$
$\therefore f^{\prime }\left(x\right)=0x=\pm \frac{ud}{\sqrt{v^2-u^2}},(v>u)$
But $x\ne \frac{-ud}{\sqrt{v^2-u^2}}$
$\therefore$ we consider $x=\frac{ud}{\sqrt{v^2-u^2}}$
Now, $f^{\prime \prime }\left(x\right)=\frac{1}{u}\frac{d^2}{\sqrt{x^2+d^2(x^2+d^2)}}>0$ for all $x$.
$\therefore$ f has minimum at $x=\frac{ud}{\sqrt{v^2-u^2}}$