Question

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

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

Answer 1

Answer: (D)

$2.6hrs.$

If we take the position of ship ${}^{\prime }{A}^{\prime }$ as origin then position and velocities of both ships can be given as:
${\overrightarrow{V}}_A=(30\hat{i}+50\hat{j})km/hr$
${\overrightarrow{V}}_B=-10\hat{i}km/hr$
${\overrightarrow{r}}_A=0\hat{i}+0\hat{j}$
${\overrightarrow{r}}_B=(80\hat{i}+150\hat{j})km$
Time after which distance between them will be minimum
$t=-\frac{{\overrightarrow{r}}_{BA}\cdot {\overrightarrow{V}}_{BA}}{|{\overrightarrow{V}}_{BA}{|}^2}$
where ${\overrightarrow{r}}_{BA}=(80\hat{i}+150\hat{j})km$
${\overrightarrow{V}}_{BA}=-10\hat{i}-(30\hat{i}+50\hat{j})$
$(-40\hat{i}-50\hat{j})km/hr$
$\therefore t=-\frac{(80\hat{i}+150\hat{j})\cdot (-40\hat{i}-50\hat{j})}{|-40\hat{i}-50\hat{j}{|}^2}$
$=\frac{3200+7500}{4100}hr=\frac{10700}{4100}hr=2.6hrs$.