The graph of an object's motion (along the $$x$$-axis) is shown in the figure. The instantaneous velocity of the object at points $$A$$ and $$B$$ are $$v_A$$ and $$v_B$$ respectively. Then

The graph provided is a displacement-time ($$x$$ vs $$t$$) plot. For such a graph, the instantaneous velocity at any point equals the slope of the tangent at that point:

$$v = \frac{dx}{dt} = \text{slope of the } x\text{-}t \text{ graph}$$

Case 1: Point $$A$$ lies on a straight-line segment of the curve. Two easily readable neighbouring points on that segment are, for example, $$(2\;\text{s},\,1\;\text{m})$$ and $$(4\;\text{s},\,2\;\text{m})$$. The slope (and hence the velocity at $$A$$) is

$$v_A \;=\; \frac{2\;\text{m} - 1\;\text{m}}{4\;\text{s} - 2\;\text{s}} \;=\; \frac{1}{2}\;\text{m s}^{-1} \;=\; 0.5\;\text{m s}^{-1}$$

Case 2: Point $$B$$ also falls on a straight portion of the graph. Picking two nearby points on that segment—for instance $$(10\;\text{s},\,3\;\text{m})$$ and $$(14\;\text{s},\,5\;\text{m})$$—gives

$$v_B \;=\; \frac{5\;\text{m} - 3\;\text{m}}{14\;\text{s} - 10\;\text{s}} \;=\; \frac{2}{4}\;\text{m s}^{-1} \;=\; 0.5\;\text{m s}^{-1}$$

Because both points lie on straight segments having the same slope, the instantaneous velocities are identical:

$$v_A = v_B = 0.5\;\text{m s}^{-1}$$

Therefore, the correct choice is
Option A which is: $$v_A = v_B = 0.5\ \text{m/s}$$