Imagine an object is moving on a perfect circle as shown at below. If the radius of circle is 4.50km and the angle between AOB is 120 degree. What is distance the object traveled

Distance is the arc $\overset{\huge\frown}{AB}$,

Distance $\overset{\huge\frown}{AB}$ = $\frac{120}{360} \times 2 \pi r$ = 9.42 km

Displacement is the vector by connecting A and B with direction pointing toward B.

In triangle AOB,
AO = BO = 4.50 km, ∠AOB = 120°

From O we draw a line OD, so OD ⊥ AB. As AO = OB. OD bisects AB and ∠AOB.

∠AOB = 120° so, ∠BOD = 60°

Using Pythagoras theorem,
sin60° = $\frac{BD}{OB}$
or, BD = OB × $\frac{\sqrt 3}{2}$ = 3.90 km

So AB = 2BD = 2 × 3.90 = 7.80 km