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

Answer

c

Explanation

Net displacement of the car along x-direction
`d_(x)=(6-4)cos 45^(@)hati=2xx1/(sqrt(2))=sqrt(2)km`

Net displacement of the car along y-direction
`S_(y)=(6+4) sin 45^(@)hatj=10xx1/(sqrt(2))=5sqrt(2)km`
Net displacement from starting point
`|vecS|=sqrt(S_(x)^(2)+S_(y)^(2))=sqrt((sqrt(2))^(2)+(5sqrt(2))^(2))`
`=sqrt(52)km`
Angle which resultant makes with the east direction
`tan theta=(y-"component")/(x-"component")=(5sqrt(2))/(sqrt(2))`
`:. theta=tan^(-1)(5)`