An aircraft flies at `400 km//h` in still air. A wind of `200sqrt2 km//h` is blowing from the south towards north. The pilot wishes to travel from A to a point B north east of A. Find the direction he must steer and time of his journey if `AB=1000 km.`

Explanation

Given that, `v_(w)=200sqrt(2)km//h`
`v_(aw)=400km//hr` and `vecv_(a)`shou ld be along AB or in north-east direction. Thus, the direction of `vecv_(aw)` shou ld be such as the resu ltant of `vecv_(w)` and `vecv_(aw)` is along AB or in north -east direction.

Let `vecv_(aw)` makes an angles `alpha` with AB as shown in Fig. 5.35. Applying sine law in triangle ABC, we get
`(AC)/(sin 45^(@))=(BC)/(sin alpha)`
or `sin alpha=((BC)/(AC))sin 45^(@)`
`=((200sqrt(2))/(400))(1)/sqrt(2)=(1)/(2)`
`therefore alpha=30^(@)`
Therefore, the pilot shou ld steer in a direction at an angle of `(45^(@)alpha)` or `75^(@)` from north towards cast.
Further `(|vecv_(a)|)/(sin (180^(@)-45^(@)-30^(@)))=(400)/(sin 45^(@))`
or `|vecv_(a)|=(sin 105^(@))/(sin 45^(@))xx(400)km//h`
`=((cos 15^(@))/(sin 45^(@)))(400)(km)/(h)`
`=((0.9659)/(0.707))(400)(km)/(h)=546.47km//h`
`therefore` the time of journey from A to B is
`t=(AB)/(|vecv_(a)|)=(1000)/(546.4)h`
t=1.83h