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Question
A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops. Determine the girl’s displacement from her initial point of departure.
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Solution
Let the headings O and B be the girl's initial and final positions. Then, the position of the girl can be shown.

now we have:
`vec(OA) = -4hati`
`vec(AB) = hati |vec(AB)| cos60° + hatj |vec(AB)|sin60°`
= `hati xx 3 xx (1/2) + hatj xx 3 xx (sqrt3/2)`
= `(3/2)hati + ((3sqrt3)/2)hatj`
By the triangle rule of vector addition, we have:
`vec(OB) = vec(OA) + vec(AB)`
= `(-4hati) + (3/2hati + (3sqrt3)/2hatj)`
= `(-4 + 3/2)hati + (3sqrt3)/2hatj`
= `((-8 + 3)/2)hati +(3sqrt3)/2hatj`
= `(-5)/2hati + (3sqrt3)/2hatj`
Therefore, the girl's displacement from her initial point of departure is `(-5)/2hati + (3sqrt3)/2hatj`.
