question_answer

Two particles A and B move with constant velocities \[{{\upsilon }_{1}}\]and\[{{\upsilon }_{2}}\] along two mutually perpendicular straight lines towards the intersection point O. At moment t = 0, the particles were located at distance \[{{d}_{1}}\]and\[{{d}_{2}}\] from O respectively. Find the time, when they are nearest and also this shortest distance.

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, \[{{L}_{\min }}=\frac{{{d}_{1}}{{\upsilon }_{2}}-{{d}_{2}}{{\upsilon }_{1}}}{\sqrt{\upsilon _{1}^{2}+\upsilon _{2}^{2}}}\]