Question

Two particle A and B are located in x-y plane at points (0, 0) and (0, 4 m). They simultaneously start moving with velocities. $\overrightarrow{V_A}=2\hat{j}m/s$ and $\overrightarrow{V_B}=2\hat{i}m/s.$ Select the correct alternative(s).

• A. The distance between them is constant
• B. The distance between them first decreases and then increases
• C. The shortest distance between them is $2\sqrt{2}m$
• D. Time after which they are at minimum distance is 1s

Answer 1

Answer: (B), (C), (D)

The correct options are
B The distance between them first decreases and then increases
C The shortest distance between them is $2\sqrt{2}m$
D Time after which they are at minimum distance is 1s

$\overrightarrow{V_{AB}}=\overrightarrow{V_A}-\overrightarrow{V_B}=2(\overrightarrow{i}-\overrightarrow{j})\Rightarrow |\overrightarrow{V_{AB}}|=2\sqrt{2}.$
Assuming B to be at rest, A will move with velocity $\overrightarrow{V_{AB}}$ in the direction shown in figure. The distance between them will first decrease from A to C and then increase beyond C.
Minimum distance between them is BC which is equal to $\frac{4}{\sqrt{2}}$ or $2\sqrt{2}$ and the time after which they are at closest distance is :
$t=\frac{AC}{\overrightarrow{V_{AB}}}=\frac{2\sqrt{2}}{2\sqrt{2}}=1second.$