Question

Ship A is sailing towards north-east with velocity $\overrightarrow{v}=30\hat{i}+50\hat{j}\frac{km}{h}$, where $\hat{i}$ points east and $\hat{j}$ north. Ship B is at a distance of 80 km east and 150 km north of ship A and is sailing towards west at 10 $\frac{km}{h}$. A will be at minimum distance from B in n hours. Report n. Answer upto 1 decimal place.

Answer 1

Considering the initial position of ship A as origin, so the velocity and position of ship will be
${\overrightarrow{v}}_A=(30\hat{i}+50\hat{j})$ and ${\overrightarrow{r}}_A=(0\hat{i}+0\hat{j})$
Now, as given in the question, 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 ships B and A are (80 - 10t, 150) and (30t, 50t).
So, distance between A and B after time t is
d = $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
d = $\sqrt{(80-10t-30t{)}^2+(150-50t{)}^2}$
$\Rightarrow d^2=(80-40t{)}^2+(150-50t{)}^2$
Distance is minimum when $\frac{d}{dt}(d{)}^2=0$
After differentiating, we get
$\Rightarrow \frac{d}{dt}\left[\right(80-40t{)}^2+(150-50t{)}^2]=0$
$\Rightarrow 2(80-40t)(-40)+2(150-50t)(-50)=0$
$\Rightarrow$ -3200 + 1600t - 7500 + 2500t = 0
$\Rightarrow 4100t=10700\Rightarrow t=\frac{10700}{4100}$ = 2.6 h