A girl walks 4 km towards west, and then she walks 3 km in a direction `30^0` east of north and stops. Determine the girls displacement from her initial point of departure.

Explanation

Let the initial point of the position of the girl is O and end point is B, as shown in the following figure.
Now, `vec(OA)=-4hat(i)`
and `vec(AB) = hat(i)|vec(AB)|cos 60^(@) hat(j) |vec(AB)|sin 60^(@)`
`=hat(i) 3 xx (1)/(2) + hat(j) 3 xx (sqrt(3))/(2) = (3)/(2) hat(i) + (3sqrt(3))/(2)hat(j)`

From the triangle's law of vector addition,
`vec(OB) = vec(OA)+vec(AB) = (-4hat(i)) + ((3)/(2) hat(i) + (3sqrt(3))/(2) hat(j))`
`=(-4 + (3)/(2)) hat(i) + (3sqrt(3))/(2) hat(j)`
`=((-8+3)/(2)) hat(i) + (3sqrt(3))/(2)hat(j) = (-5)/(2)hat(j) + (3sqrt(3))/(2)hat(j)`
Therefore, displacement of the girl from the initial point is `(-5)/(2)hat(i)+ (3sqrt(3))/(2)hat
(j)`.