Two particles A and B located as shown start moving simultaneously with velocities V 4i and B 4j

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. $$\vec{V_{A}}=2\hat{j}\ m/s$$ and $$\vec{V_{B}}=2\hat{i}\ m/s.$$ 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

$$\overrightarrow { V_{AB} } =\quad \overrightarrow { V_A } -\overrightarrow { V_B } =2(\overrightarrow { i } -\overrightarrow { j } )\quad \Rightarrow |\overrightarrow { V_{AB} } |=2\sqrt { 2 } .$$

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 $$\dfrac { 4 }{ \sqrt { 2 } } $$ or $$2\sqrt { 2 } $$ and the time after which they are at closest distance is :

$$t=\dfrac { AC }{ \overrightarrow { V_{AB} } } =\dfrac { 2\sqrt { 2 } }{ 2\sqrt { 2 } } =\quad 1\quad second.$$

Text Solution

the shortest distance between them is `4sqrt(2)m`the shortest distance between them first decreases and then increases the distance between them increases from the beginning the magnitude of relative velocity of A w.r.t. B is 4m/s

Answer : a,c

Solution : The shortest distane h/w them is <br> `AB=4sqrt(2)m` <br> Because the distance b/w them increases from start. <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/GRB_AM_PHY_C05_E01_199_S01.png" width="80%">