QJEEM19Ph9a27: JEE Main 2019 Physics - Class 11 JEEM19Ph9a27 Solution: Ship A is sailing towards north-east with velocity km/hr where…

Considering the initial position of ship A as origin, so the velocity and position of ship will be

${\overrightarrow v _A} = (30\widehat i + 50\widehat j)$ and ${\overrightarrow r _A} = (0\widehat i + 0\widehat j)$

Now, as given in the question, velocity and position of ship B will be, ${\overrightarrow v _B} = - 10\widehat i$ and ${\overrightarrow r _B} = (80\widehat i + 150\widehat j)$

Time after which the distance is minimum between A and B can be calculated as

$t = {{|{{\overrightarrow r }_{BA}}.\,{{\overrightarrow v }_{BA}}|} \over {|{{\overrightarrow v }_{BA}}{|^2}}}$

where, ${\overrightarrow r _{BA}} = {\overrightarrow r _B} - {\overrightarrow r _A} = 80\widehat i + 150\widehat j$

and ${\overrightarrow v _{BA}} = - 10\widehat i - (30\widehat i + 50\widehat j)$

$= - 40\widehat i - 50\widehat j$

$\Rightarrow t = {{|(80\widehat i + 150\widehat j)\,.\,( - 40\widehat i - 50\widehat j)|} \over {| - 40\widehat i - 50\widehat j{|^2}}}$

$= {{3200 + 7500} \over {4100}} = {{10700} \over {4100}} = 2.6$ h