Question

At an instant particle A is at origin and moving with constant velocity $(3\hat{i}+4\hat{j})$ m/s and particle B is at (4, 4)m and moving with constant velocity $(4\hat{i}-3\hat{j})$ m/s. Then at this instant

• A. Relative velocity of B w.r.t. A is $(4\hat{i}-3\hat{j})$m/s
• B. Approach velocity of A & B is $3\sqrt{2}$m/s
• C. Relative velocity of B w.r.t. A remains constant.
• D. Approach velocity of A & B remains constant.

Answer 1

Answer: (A), (B), (C)

The correct options are
A

Relative velocity of B w.r.t. A is $(4\hat{i}-3\hat{j})$m/s

B

Approach velocity of A & B is $3\sqrt{2}$m/s

C

Relative velocity of B w.r.t. A remains constant.

$\overrightarrow{V_A}=3\hat{i}+4\hat{j}\overrightarrow{V_B}=4\hat{i}-3\hat{j}\overrightarrow{V_{BA}}=\overrightarrow{V_B}-\overrightarrow{V_A}=\hat{i}-7\hat{j}\overrightarrow{r_A}=0{\overrightarrow{r}}_B=4\hat{i}+4\hat{j}{\overrightarrow{r}}_{BA}=4\hat{i}+4\hat{j}$
$\left|V_{app}\right|=|{\overrightarrow{V}}_{BA}.{\hat{r}}_{BA}|$
$\left|\right(\hat{i}-7\hat{j}).(\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j})|$
$|\frac{1}{\sqrt{2}}-\frac{7}{\sqrt{2}}|=3\sqrt{2}m/s$