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 $\hat{j}$ towards north. Ship $B$ is at a distance of $80km$ east and $150km$ north of Ship $A$ and is sailing towards west at $10km/hr$. Time in which $A$ will be at minimum distance from $B$ is

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

Answer 1

The correct option is C $2.6hrs$


${\overrightarrow{v}}_A=(30\hat{i}+50\hat{j})km/hr$
${\overrightarrow{v}}_B=-10\hat{i}km/hr;{\overrightarrow{r}}_A=0\hat{i}+0\hat{j}$
${\overrightarrow{r}}_B=(80\hat{i}+150\hat{j})km$
Time after which distance between them will be minimum
$t=-\frac{{\overrightarrow{r}}_{BA}.{\overrightarrow{v}}_{BA}}{|{\overrightarrow{v}}_{BA}{|}^2}$;
Where ${\overrightarrow{r}}_{BA}=(80\hat{i}+150\hat{j})km$
${\overrightarrow{v}}_{BA}=-10\hat{i}-(30\hat{i}+50\hat{j})=(-40\hat{i}-50\hat{j})km/hr$
$\therefore t=-\frac{(80\hat{i}+150\hat{j}).(-40\hat{i}-50\hat{j})}{|-40\hat{i}-50\hat{j}{|}^2}$
$=\frac{3200+7500}{4100}hr=\frac{10700}{4100}hr=2.6hrs$