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?

Options

A. `sqrt(50)km` and `tan^(-1)(5)`
B. `10km` and `tan^(-1)(sqrt5)`
C. `sqrt(52)km` and `tan^(-1)(5)`
D. `sqrt(52)km` and `tan^(-1)(sqrt5)`

Answer

C

Explanation

Net movement along x-direction
`S_x=(6-4)cos45^@hati=2xx(1)/(sqrt(2))=sqrt(2)km`
Net movement along y-direction
`S_y=(6+4)sin45^@hatj=10xx(1)/(sqrt2)=5sqrt(2)km`
Net movement from starting point
`|vecS|=sqrt(S_x^(2)+S_y^(2))=sqrt((sqrt(2))^(2)+(5sqrt(2))^(2))=sqrt(52)km`
Angle which makes with the east direction
`tantheta=(Y-"component")/(X-"component")=(5sqrt(2))/(sqrt(2))becausetheta=tan^(-1)(5)`