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.You need to login to perform this action.
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