question_answer

Ship A is sailing towards north-east with velocity \[\overset{\to }{\mathop{v}}\,=30\overset{\wedge }{\mathop{i}}\,+50\overset{\wedge }{\mathop{j}}\,km/hr\] where \[\overset{\wedge }{\mathop{i}}\,\] points east and \[\hat{j},\] north. Ship B is at a distance of 80 km east and 150 km north of Ship A and is sailing towards west at 10 km/hr. A will be at minimum distance from B in : [JEE Main 8-4-2019 Morning]

A) 4.2 hrs.                       

B) 2.2 hrs.

C) 3.2 hrs.                       

D) 2.6 hrs.

Correct Answer: D

Solution :

If we take the position of ship 'A' as origin then positions and velocities of both ships can be given as : \[{{\text{\vec{v}}}_{A}}=\left( 30\hat{i}+50\hat{j} \right)km/hr\] \[{{\text{\vec{v}}}_{B}}=-10\hat{i}\,km/hr\] \[{{\vec{r}}_{A}}=0\hat{i}+0\hat{j}\] \[{{\vec{r}}_{B}}=\left( 80\hat{i}+150\hat{j} \right)km\] Time after which distance between them will be minimum \[t=-\frac{{{{\vec{r}}}_{BA}}.{{{\text{\vec{v}}}}_{BA}}}{{{\left| {{{\text{\vec{v}}}}_{BA}} \right|}^{2}}};\] where\[{{\vec{r}}_{BA}}=\left( 80\hat{i}+150\hat{j} \right)km\] \[{{\text{\vec{v}}}_{BA}}=-10\hat{i}-\left( 30\hat{i}+50\hat{j} \right)\] \[\left( -40\hat{i}-50\hat{j} \right)km/hr\] \[\therefore \]\[t=-\frac{\left( 80\hat{i}+150\hat{j} \right).\left( -40\hat{i}-50\hat{j} \right)}{{{\left| -40\hat{i}-50\hat{j} \right|}^{2}}}\] \[=\frac{3200+7500}{4100}hr=\frac{10700}{4100}hr=2.6hrs\]