A person travelling east wards at the rate of ` 4 km h^(-1)` finds that the wind seems to blow directl from the borth . On dubling ins speed, the wind appears to come from ` 45^@` north of west. Find the actual velocity of the wind.

Explanation

Let ` vec u` be the velocity of person and ` vec v ` that og wind . Let ` hat I and hat j` be the unit vectors along east and north directions respectively. As per question
` vec u = u hat I and vec v= u_x = hat I = v_y hat j`
If ` vec v_r` is the velocity of wind relatve to person, then
` vec v_r = vec v- vec u = ( vec v_x hat i + v_y hat j ) u hat i`
`= ( v_x -u ) hat i + v_y hat j`
As wind seems to blow directly from north, so
` vec v_r =- v_r hat j`
:. ` - v_r hat j = ( v_x - u) hat i + v_y = v_y`
Comparing coefficients of ` hat i and hat j` , we have
` v-x -u =0 or v_x =u and -v_r =v_y`
When person doubles his speed, then ` vec u' = 2 u hat i`
Let the relative velocity of wind w.r.t., person be ` vec v'-r`
` vec v'_r =vec v - vecu' = (v_x hat i v-y + v_y 0 -2 u hat i`
` = ( u hat i + v_y hat j) - 2 u hat i =- hat i + u-y hat j`
As the wind appears to come from ` 45^@` north east so,
` tan 45^@= v_y/ (-u) or u_y =- u tan 45^@ =- xx 1 =- u`
Thus the actual velocity of wind
` vec v = vec v_x hat i + = u hat j = u hat i -u hat j = 4 hat i - 4 hat j`
` | vec v| = sqrt( 4^2 =(-4)^2 ) = 4 sqrt 2 km h^(-1)`
tan theta = u_y/v_x = (-4)/4 =- 1 or theta = 135^@`
It means the wind is blowing from borth-west direction.