Question

Ship A is located $4km$ north and $3km$ east of ship B. Ship A has a velocity of $20km{h}^{-1}$ towards the south and ship B is moving at $40km{h}^{-1}$ in a direction ${37}^o$ north of east. X and Y axes are along east and north directions, respectively.

• A. Velocity of A relative to B is $-32\hat{i}-44\hat{j}$
• B. Position of A relative to B as a function of time is given by ${\overrightarrow{r}}_{AB}=(3-32t)\hat{i}+(4-44t)\hat{j}$
• C. Velocity of A relative to B is $32\hat{i}-44\hat{j}$
• D. Position of A relative to B as a function of time is given by $(32t\hat{i}-44t\hat{j})$

Answer 1

Answer: (A), (B)

The correct options are
A Velocity of A relative to B is $-32\hat{i}-44\hat{j}$
B Position of A relative to B as a function of time is given by ${\overrightarrow{r}}_{AB}=(3-32t)\hat{i}+(4-44t)\hat{j}$

Assuming B is at origin ${\overrightarrow{r}}_B=0$ at $t=0$ ,then,
position vector A is given as : ${\overrightarrow{r}}_A=3\hat{i}+4\hat{j}$
at $t=0$
From given data:
${\overrightarrow{V}}_A=(-20\hat{j})$
${\overrightarrow{V}}_B=\left(40cos{37}^o\right)\hat{i}+\left(40sin{37}^o\right)\hat{j}$
$=(32\hat{i}+24\hat{j})$
Relative velocity of A w.r.t. B is given as:
${\overrightarrow{V}}_{AB}=(-32\hat{i}-44\hat{j})km/h$
Option A is correct.
Position vector of A w.r.t. B is given as:
At time $t=0$
${\overrightarrow{r}}_{AB}={\overrightarrow{r}}_A-{\overrightarrow{r}}_B=(3\hat{i}+4\hat{j})$
Option B is correct.
$\therefore$ At time $t=t$,
${\overrightarrow{r}}_{AB}=\left({\overrightarrow{r}}_{ABatt=0}\right)+{\overrightarrow{V}}_{AB}t$
$=(3-32t)\hat{i}+(4-44t)\hat{j}$