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A man walking eastward at 5 m/s observes that wind is blowing from the north. On doubling his speed eastward, he observes that wind is blowing from north-east. Find the velocity of the wind.Solution:Let velocity of the wind is,vw=(v1i+v2j)m/sAnd velocity of the man is,vm=5i So, vwm = vw- vm=(v1-5)i + v2jIn first case,v1- 5 = 0 => v1= 5 m/s.In the second case, tan 45o = v2/(v1- 10)=> v2= v1 - 10 = -5 m/s.=> vw= (5i - 5j) m/s.Can you tell how can we consider tan 45 as the angle and how the tan45 is v2/v1-10

A man walking eastward at 5 m/s observes that wind is blowing from the north. On doubling his speed eastward, he observes that wind is blowing from north-east. Find the velocity of the wind.Solution:Let velocity of the wind is,vw=(v1i+v2j)m/sAnd velocity of the man is,vm=5i So, vwm = vw- vm=(v1-5)i + v2jIn first case,v1- 5 = 0 => v1= 5 m/s.In the second case, tan 45o = v2/(v1- 10)=> v2= v1 - 10 = -5 m/s.=> vw= (5i - 5j) m/s.Can you tell how can we consider tan 45 as the angle and how the tan45 is v2/v1-10

Grade:11

1 Answers

Shaswata Biswas
132 Points
7 years ago
In the question since it is not mentioned about the angle of the wind with any axis, but it is given "NORTH-EAST", It is considered to be equally inclined to both the axes I.e, North and East. So, in this case the angle is taken to be 45০. 
In the second case, it is mentioned in the question that "the man doubles his speed eastward". So in this case his speed, v = 2×5\widehat{i} = 10\widehat{i}. That is why the horizontal component has been taken as (v1-10>
THANKS

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