Question

Ship $A$ is moving Westwards with a speed of $20\, km\, h ^{-1}$ and another ship $B$ which is at $200\, km$ South of $A$ is moving Northwards with a speed of $10\, km\, h ^{-1}$. 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. $80 \sqrt{2} h , 4 \, km$

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