Question 20
A man is facing towards the east. He turns towards north and walks for 5 km, and then turns 270 degrees anticlockwise and walks for 12 km more. What is the minimum distance between his initial and final position?
Solution
The man initially walked north for 5 km, then turned 270 degrees anticlockwise and walked towards east for 12 km.
Thus, minimum distance between his initial and final position
= $$\sqrt{(5)^2+(12)^2}$$
= $$\sqrt{25+144}=\sqrt{169}=13$$ km
=> Ans - (B)
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