A car travles 6km towards north at an angle of `45^(@)` to the east and then travles distance of 4km towards north at an angle of `135^(@)` to east (figure). How far is the point from the starting point? What angle does the straight line joining its initial and final position makes with the east?

Explanation

Net movement along X- direction
`=(6-4)cos 45^(@)hat(i)`
`2xx(1)/(sqrt(2))=sqrt(2)` km

Net movement along Y-direction
`=(6+4) sin 45^(@) hat(j)`
`=10xx(1)/(sqrt(2))=5sqrt(2)` km
Net movement form starting point (Total distance travelled)
`6+4=10km`
Angle which makes with the east direction
`tan theta =(Y"-component")/(X "-component")`
`=(5sqrt(2))/(sqrt(2))`
`theta=tan^(-1)(5)`