question_answer

Two persons A and B are locarcd in the x-y plane at the points (0. 0) and (0, 10) respectively. (The distances are measured in MKS unit). At a time t=0, they start moving simultaneously with velocities \[{{\overrightarrow{v}}_{A}}=2\widehat{j}\,m{{s}^{-1}}\] and \[{{\overrightarrow{v}}_{B}}=2\widehat{i}\,\,m{{s}^{-1}}\].The time after which A and E are at their closest distance

A)  2.5 s                                     

B)  4 s

C)  1 s                                         

D)  \[\frac{10}{\sqrt{2}}s\]

Correct Answer: D

Solution :

 As the velocities are same,    so    the persons will move according to figure. When person A will reach the position of B, they will be at closest distance but then person B will be at point C. Here the covered distances by A and B are AB (covered by A) = 10 unit along y-axis BC (covered by B) = 10 unit along y-axis Velocity in covering these distances, \[{{t}_{A}}=\frac{10}{2}=5\,unit\] \[{{t}_{B}}=\frac{10}{2}=5\,unit\]                 Resultant time, \[t=\sqrt{t_{A}^{2}+t_{B}^{2}}\]                                                 \[=\sqrt{{{(5)}^{2}}+{{(5)}^{2}}}\]                                                 \[=5\sqrt{2}\]                                                 \[=\frac{10}{\sqrt{2}}\text{unit}\]