A) 0 h
B) 5 h
C) \[5\sqrt{2}h\]
D) \[10\sqrt{2}h\]
Correct Answer: B
Solution :
It is clear from the diagram that the shortest distance between the ship A and B is PQ. Here \[\sin {{45}^{{}^\circ }}=\frac{PQ}{OQ}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,PQ=100\times \frac{1}{\sqrt{2}}=50\sqrt{2}m\] Also, \[{{v}_{AB}}=\sqrt{v_{A}^{2}+v_{B}^{2}}=\sqrt{{{10}^{2}}+{{10}^{2}}}=10\sqrt{2}km/h\] So, time taken for them to reach shortest path is \[t=\frac{PQ}{{{v}_{AB}}}=\frac{50\sqrt{2}}{10\sqrt{2}}=5h\]You need to login to perform this action.
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