To a car driver moving at `40 km*h^(-1)` towards south, wind appears to blow towards east. When the speed to the car is reduced to `20 km*h^(-1)` wind appears to blow from north -west . Find the magnitude and direction of the actual velocity of the wind.

Explanation

Let us choose : x-axis along east and y-axis along north.
Initial velocity of the car , `vecu_1 =-40 hatj km*h^(-1)`
Final velocity of the car , `vecu_2=-20hatjkm*h^(-1)`
Let , `vecv` = actual velocity of the wind.
Then , relative velocity of the wind respect to the car,
`vecw=vecv-vecu or, vecv=vecu+vecw`
Initially, `vecw_1=w_1hati, so vecv=vecu_1+vecw_1=w_1hati-40hatj .... (1)`
Finally , `vecw_2=w_2 cos 45^@ hati-w_2 sin 45^@ hatj` (as it is towards south -east)
`=(w_2)/(sqrt(2))hati-(w_2)/(sqrt(2))hatj`
`therefore vecv=vecu_2+vecw_2=(vecw_2)/(sqrt(2))hati-((w_2)/(sqrt(2))+20)hatj ..... (2)`
Comparing the coefficients of `hatj` in (1) and (2) ,
`(w^2)/sqrt(2)+20=40 or, (w_2)/sqrt(2) =20`
Then from (2) ,
`vecv=20 hati -(20+20)hatj=(20hati-40hatj) km *h^(-1)` (between east and south )
`therefore` Manitude of `vecv=v=sqrt((20)^2+(40)^2)=20sqrt(1+4)= 20 sqrt(5) km*h^(-1)`
If `vecv` is inclined at an angle `theta` with east, then
`tan theta =(-40)/(20)=-2=tan(-63.4^@) or, theta =-63.4^@`
So, the wind velocity is at an angle of `63.4^@` south of east.