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

To a car driver moving at 40 km*h^(-1) towards south, windappears to b

1.	To a car driver moving at 40 kmh^(-1) towards south, windappears to blow towards east. When the speed to the car is reduced to 20 kmh^(-1) wind appears to blow fromnorth -west . Findthe magnitude and direction of the actual velocity of the wind.
Answer» SOLUTION :LET us choose : x-axisalong east and y-axis along north. Initial velocity of the car , `vecu_1 =-40 HATJ kmh^(-1)` Final velocity of the car , `vecu_2=-20hatjkmh^(-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) kmh^(-1)` If `vecv` is inclined at anangle `THETA` with east,then `tan theta =(-40)/(20)=-2=tan(-63.4^@) or, theta =-63.4^@` So, the wind velocity is at anangle of `63.4^@` south of east.