Question

Two particles A and B are located in X-Y plane at point $(0,0)$ and $0,8$m. If they start moving simultaneously with velocities ${\overrightarrow{V}}_A=10\hat{\text{j}}{\text{ms}}^{-1}$ and ${\overrightarrow{V}}_B=10\hat{\text{i}}{\text{ms}}^{-1}$, then

• A. the distance between them remains constant
• B. the distance between them will first decrease and then increase
• C. the shortest distance between them will be $4\sqrt{2}\text{m}$
• D. the time after which they are at a minimum separation distance is 0.4 s

Answer 1

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

The correct options are
B the distance between them will first decrease and then increase
C the shortest distance between them will be $4\sqrt{2}\text{m}$
D the time after which they are at a minimum separation distance is 0.4 s

Consider the relative motion of $A$ w.r.t observer on $B$
$\Rightarrow$ relative velocity,
$\Rightarrow {\overrightarrow{V}}_{AB}={\overrightarrow{V}}_{AG}-{\overrightarrow{V}}_{BG}=10\hat{\text{j}}-10\hat{\text{i}}\left(\text{in second quadrant}\right)$
$\left|{\overrightarrow{V}}_{AB}\right|=10\sqrt{2}{\text{ms}}^{-1}$
Here, minimum distance between them will be $BP$ which is equal to $8\sin {45}^{\circ }=8\left(\frac{1}{\sqrt{2}}\right)=4\sqrt{2}\text{m}$
Time after which their separation distance becomes a minimum will be
$\text{t}=\frac{AP}{\left|{\overrightarrow{V}}_{AB}|\right|}=\frac{8\cos {45}^{\circ }}{10\sqrt{2}}=\frac{8}{10}\left[\frac{1}{2}\right]$
$\text{t}=0.4$