question_answer

Two persons A and B are located in 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 \[{{\mathbf{\vec{v}}}_{A}}=2\mathbf{j}m{{s}^{-1}}\] and \[{{\mathbf{\vec{v}}}_{B}}=2\mathbf{\hat{i}}m{{s}^{-1}}\]respectively. The time after which\[A\]and\[B\]are at their closest distance is

A)  2.5s             

B)  4 s

C)  1 s                

D)  \[\frac{10}{\sqrt{2}}\]s  

Correct Answer: A

Solution :

Let after the time (r) the position of A is  \[(0,\,{{v}_{A}}t)\]  and    position    of    \[B=({{v}_{B}}t,10)\]. Distance between them \[y=\sqrt{{{(0-{{v}_{B}}t)}^{2}}+{{({{v}_{A}}t-10)}^{2}}}\] or \[{{y}^{2}}={{(2t)}^{2}}+{{(2t-10)}^{2}}\] or \[{{y}^{2}}=l=4{{t}^{2}}+4{{t}^{2}}+100-40t\] \[\Rightarrow \] \[l=8{{t}^{2}}+100-40t\] Now, \[\frac{dl}{dt}=(16t-40)=0\] \[t=\frac{40}{16}=2.5s\] As \[\frac{{{d}^{2}}l}{d{{t}^{2}}}=16=(+ve)\] Hence, I will be minimum.