Question

Ship $\text{A}$ is sailing towards north-east with velocity vector $\overrightarrow{v}=30\hat{i}+50\hat{j}\text{km/hr}$ where $\hat{i}$ points east and $\hat{j}$ north. Ship $B$ is at a distance of $80\text{km}$ east and $150\text{km}$ north of Ship $A$ and is sailing towards west at $10\text{km/hr}$. $A$ will be at minimum distance from $B$ in,

• A. $2.2hrs$
• B. $4.2hrs$
• C. $2.6hrs$
• D. $3.2hrs$

Answer 1

The correct option is C $2.6hrs$

Given data is represented in the figure,


From figure,
${\overrightarrow{V}}_{AB}={\overrightarrow{V}}_A-{\overrightarrow{V}}_B=(30\hat{i}+50\hat{j})-(-10\hat{i})$

$\Rightarrow {\overrightarrow{V}}_{AB}=40\hat{i}+50\hat{j}$ and

${\overrightarrow{r}}_{AB}=-80\hat{i}-150\hat{j}$

Now, $A$ will be at minimum distance from $B$ in time,

$t=\frac{{\overrightarrow{r}}_{AB}.{\overrightarrow{V}}_{AB}}{|{\overrightarrow{V}}_{AB}{|}^2}$

$\Rightarrow t=-\frac{(-80\hat{i}-150\widehat{)}.(40\hat{i}+50\hat{j})}{(\sqrt{{40}^2+{50}^2}{)}^2}$

$\Rightarrow t=-\frac{-3200-7500}{4100}=2.6\text{hr}$

Hence, $\left(C\right)$ is the correct answer.