Question

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

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

Answer 1

The correct option is B $2.6h$

Considering the initial position of ship $A$ as origin, the velocity of ship will be
${\overrightarrow{v}}_A=30\hat{i}+50\hat{j}$
Position of ship A will be,
${\overrightarrow{r}}_A=0\hat{i}+0\hat{j}$
Velocity and position of ship $B$ will be
${\overrightarrow{v}}_B=-10\hat{i}$ and ${\overrightarrow{r}}_B=(80\hat{i}+150\hat{j})$
Hence, the given situation can be represented graphically as


After time $t$, coordinates of ship $A$ and $B$ will be
$(30t,50t)$ and $(80-10t,150)$
So, distance between $A$ and $B$ after time $t$ is
$D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
$D=\sqrt{(80-30t-10t{)}^2+(150-50t{)}^2}$
$D^2=(80-40t{)}^2+(150-50t{)}^2$
Distance will be minimum when $\frac{d}{dt}\left(D^2\right)=0$
$⟹2(80-40t)(-40)+2(150-50t)(-50)=0$
$⟹-6400+3200t-15000+5000t=0$
$\therefore t=\frac{21400}{8200}=2.6h$