Question

Ship $A$ is moving Westwards with a speed of $20km{h}^{-1}$ and another ship $B$ which is at $200km$ South of $A$ is moving Northwards with a speed of $10km{h}^{-1}$. The time after which the distance between them is shortest and the shortest distance between them respectively

• A. $4h,80\sqrt{5}km$
• B. $50\sqrt{2}h,\sqrt{10}km$
• C. $100\sqrt{2}h,2\sqrt{10}km$
• D. $80\sqrt{2}h,4km$

Answer 1

Answer: (A)

According to the question,

Let ship $A$ travel $x_A$ distance and ship $B$ travel $x_B$ distance, in time $t$.
So, $u_A=\frac{x_A}{t}\Rightarrow t=\frac{x_A}{20}$
and $u_B=\frac{x_B}{t}\Rightarrow t=\frac{x_B}{10}$
$\Rightarrow \frac{x_A}{20}=\frac{x_B}{10}\Rightarrow x_A=2x_B$
So, $AB=\sqrt{x_A^2+{\left(200-x_B\right)}^2}$
$=\sqrt{4x_B^2+40000+x_B^2-400x_B}$
$=\sqrt{5x_B^2-400x_B+40000}$
Differentiate distance $AB$ w.r.t. $x_B$ for finding value of $x_B$
$\frac{d(AB)}{d{x}_B}=\frac{1}{2\sqrt{5x_B^2-400x_B+40000}}\left(10x_R-400\right)=0$
or $x_B=40m$
Again differentiating,
So, ${\left(\frac{d^2(AB)}{d{x}^2}\right)}_{x_B=40m}>0$
$[\because$ Distance always greater than zero $]$
Hence $x_B$ at point, $x_B=40m$ distance, $AB$ will be shortest.
So, $AB=\sqrt{5\times 40^2-400\times 40+40000}$
$AB=\sqrt{32000}$
or $AB=80\sqrt{5}km$
The time after which the distance $AB$ is shortest,
$t=\frac{x_A}{20}=\frac{x_B}{10}=\frac{40}{10}=4hr$.