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(5) km and tan^-1 (sqrt(5))`

Answer

C

Explanation

Net movement along x-direction
`S_x = (6 - 4) cos 45^@ hat i`
=`2 xx (1)/(sqrt(2)) = sqrt(2) km`
Net movement along y-direction
`S_y = (6 + 4) sin 45^@ hat j`
=`10 xx (1)/(sqrt(2)) = 5 sqrt(2) km`
Net movement from starting point.
`|vec S| = sqrt(S_x^2 + S_y^2) = sqrt((sqrt(2))^2 + (5 sqrt(2))^2)`
=`sqrt(52) km`
Angle which resultant makes with the east direction
`tan theta = (y-"component")/(x-"component")`
=`(5 sqrt(2))/(sqrt(2))`
`:. theta = tan^-1 (5)`.