From a light-house an observer two ships `A` and `B`. Ship `A` proceeding towards north at a speed `20sqrt(2) km//h` and ship `B` proceeding towards north-east at a speed of `20 km//h`. Find in which direction and at what speed the ship `B` would appear to move to an observer standing on the deck of the ship `A`.

Explanation

In this problem, the velocity of `B` with respect to `A` is required. `v_(B//A)`=resultant of (velocity of `B` and velocity of `A` in opposite direction)


`v_(B//A)=sqrt((20)^(2)+(20sqrt(2))^(2)+2xx20xx20sqrt(2)cos 135^(@))`
`=sqrt(400+800-800)=20 km//h`
`tan alpha=(20 sin 135^(@))/(20sqrt(2)+20cos 135^(@))`
`=(20/(sqrt(2)))/(20sqrt(2)-20/(sqrt(2)))=1 implies alpha=45^(@)`
For an observer on the deck of `A`, ship `B` is moving with a speed `20km//h` along the sourth-east.
OR
`vec(v)_(A//g)=20sqrt(2) hatj,vec(v)_(B//g)=20cos45^(@) hati+20sin45^(@) hatj`
`=10sqrt(2) hati+10sqrt(2)hatj`
`vec(v)_(B//A)=vec(v)_(B//g)-vec(v)_(A//g)=10sqrt(2)hati-10sqrt(2)hatj m//s`
`|vec(v)_(B//A)|=v_(B//A)=20 m//s`