What is the direction of wind?

Question

A man is walking in the north-east direction and wind appears to blow from north. If the man doubles his speed, wind appears at angle arccot 2 east of north. Find the actual direction of the wind.

Answer 1

ur question is very good

but i don't have this much of brain................

Answer 2

starting velocity = 0

Let east direction( in vector i)and north direction is also x (vector j)

So velocity = xi+ xj (as it is north east)



Let the wind speed be ai+bj

The relative velocity of wind = (a-x)i + (b-x) j

as it appears from norh so east component = 0

so a- x = 0 or x = a

wind speed = xi+bj ..2

it the speed of person is doubled then it is 2( xi+ xj)

now relative speed of wind = i(x- 2x) + (b- 2)x)j

= - ix + (b-2x) j

Now the angle is t means cot t = - x/(b-2x) = 2

This means the speed of the wind is proportional to the speed of man.

Say x = 1

-1/(b-2) = 2 or 2(b-2) = - 1 or b = 3/2

So speed of wind = 1i + 3/2 j

That is towards north east direction 1 in east and 3/2 in north direction

theta = tan^-(3/2)