Answer:
Let
particles A and B reach at A' and B'
So,
\[AA'={{\upsilon }_{1}}\]
t and \[BB'={{\upsilon }_{2}}t\].
Thus \[OA'=\left(
{{d}_{1}}-{{\upsilon }_{1}}t \right)\]
and
\[OB'=\left( {{d}_{1}}-{{\upsilon }_{2}}t \right)\]
If
\[A'B'=L\], then
\[{{L}^{2}}={{\left( {{d}_{1}}-{{\upsilon }_{1}}t
\right)}^{2}}+{{\left( {{d}_{2}}-{{\upsilon }_{2}}t \right)}^{2}}\] ??.
(i)
Differentiating
it w.r.t. time we get
\[2L\frac{dL}{dt}=2\left(
{{d}_{1}}-{{\upsilon }_{1}}t \right)\left( -{{\upsilon }_{1}} \right)+2{{\left(
{{d}_{2}}-{{\upsilon }_{2}}t \right)}^{2}}\]
For
L to be minimum, \[\frac{dL}{dt}=0\]
\[\therefore \] \[0=-2\left( {{d}_{1}}{{\upsilon
}_{1}}-\upsilon _{1}^{2}t \right)-2\left( {{d}_{2}}{{\upsilon }_{2}}-\upsilon
_{2}^{2}t \right)\]
or
\[\left( \upsilon _{1}^{2}+\upsilon _{2}^{2} \right)t={{d}_{1}}{{\upsilon
}_{1}}+{{d}_{2}}{{\upsilon }_{2}}\] or \[t=\frac{{{d}_{1}}{{\upsilon
}_{1}}+{{d}_{2}}{{\upsilon }_{2}}}{\left( \upsilon _{1}^{2}+\upsilon _{2}^{2}
\right)}\]
Putting
this values of t in (i), and simplifying it, we get,
\[{{L}_{\min
}}=\frac{{{d}_{1}}{{\upsilon }_{2}}-{{d}_{2}}{{\upsilon }_{1}}}{\sqrt{\upsilon
_{1}^{2}+\upsilon _{2}^{2}}}\]
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