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 constantB
The distance between them first decreases and then increasesC
The shortest distance between them is $$2\sqrt{2}m$$D
Time after which they are at minimum distance is 1s
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 :
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="//d10lpgp6xz60nq.cloudfront.net/physics_images/GRB_AM_PHY_C05_E01_199_S01.png" width="80%">