Question
A ship $$A$$ is moving Westwards with a speed of $$10\,km\,{h^{ - 1}}$$ and a ship $$B$$ $$100\,km$$ South of $$A,$$ is moving Northwards with a speed of $$10\,km\,{h^{ - 1}}.$$ The time after which the distance between them becomes shortest, is
A.
$$5\,h$$
B.
$$5\sqrt 2 \,h$$
C.
$$10\sqrt 2 \,h$$
D.
$$0\,h$$
Answer :
$$5\,h$$
Solution :
$$\eqalign{ & {\overrightarrow V _A} = 10\left( { - \hat i} \right) \cr & {\overrightarrow V _B} = 10\left( {\hat j} \right) \cr & {\overrightarrow V _{BA}} = 10\hat j + 10\hat i \cr & = 10\sqrt 2 \,km/h \cr & {\text{Distance}}\,OB = 100\cos {45^ \circ } \cr & = 50\sqrt 2 \,km \cr} $$
Time taken to each the shortest distance between $$A$$ and $$B$$ $$ = \frac{{OB}}{{\overrightarrow {{V_{BA}}} }} = \frac{{50\sqrt 2 }}{{10\sqrt 2 }} = 5\,h$$
$$\eqalign{ & {\overrightarrow V _A} = 10\left( { - \hat i} \right) \cr & {\overrightarrow V _B} = 10\left( {\hat j} \right) \cr & {\overrightarrow V _{BA}} = 10\hat j + 10\hat i \cr & = 10\sqrt 2 \,km/h \cr & {\text{Distance}}\,OB = 100\cos {45^ \circ } \cr & = 50\sqrt 2 \,km \cr} $$
Time taken to each the shortest distance between $$A$$ and $$B$$ $$ = \frac{{OB}}{{\overrightarrow {{V_{BA}}} }} = \frac{{50\sqrt 2 }}{{10\sqrt 2 }} = 5\,h$$
