Question

Ship $A$ is sailing towards north-east with velocity $\vec{v}=30\hat{i}+50\hat{j}$ 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. 4.2 hrs
• B. 2.2 hrs
• C. 3.2 hrs
• D. 2.6 hrs

Answer 1

Answer: (D)

If we take the position of ship 'A' as origin then positions and velocities of both ships can be given as : ${\vec{V}}_A=\left(30\hat{i}+50\hat{j}\right)km/hr$
${\vec{v}}_B=-10\hat{i}km/hr$
${\vec{r}}_A=0\hat{i}+0\hat{j}$
${\vec{r}}_B=\left(80\hat{i}+150\hat{j}\right)km$
Time after which distance between them will be minimum
$t=-\frac{{\vec{r}}_{BA}.{\vec{v}}_{BA}}{{|{\vec{v}}_{BA}|}^2}$
where ${\vec{r}}_{BA}=\left(80\hat{i}+150\hat{j}\right)km$
${\vec{v}}_{BA}=-10\hat{i}-\left(30\hat{i}+50\hat{j}\right)$
$\left(-40\hat{i}-50\hat{j}\right)km/hr$
$\therefore t=-\frac{\left(80\hat{i}+150\hat{j}\right).\left(-40\hat{i}-50\hat{j}\right)}{{|-40\hat{i}-50\hat{j}|}^2}$
$=\frac{3200+7500}{4100}hr=\frac{10700}{4100}hr=2.6hrs$