# Need help interpreting displacement-time graphs

In this image is a displacement-time graph of the movement of an object. In the video (Khan Academy), the explanation is that between $$0\,\text{s}$$ and $$5\,\text{s}$$, the displacement of the object is $$x_\text{final}-x_\text{initial}= 6-(-2)= +8 \,\text{m}$$.

However, isn't displacement relative to the origin or starting point? What is the reason for having graphs start at a number other than zero? In the above example graph, I interpret it as the object is $$2\,\text{m}$$ in the negative direction of it's starting position. Thus wouldn't its displacement be $$6\,\text{m}$$ at $$5\,\text{s}$$? I really hope somebody can help me clarify this point.

• To get the displacement you need to have two time coordinates, it doesn't make any sense to say the displacement at time t=5, you would say the displacement between time t=0 and t=5. Commented Feb 20, 2021 at 0:12

Displacement is relative to the starting point and the starting point does not have to be at $$0$$; here the starting point is at $$-2$$ at time $$0$$. At time $$5$$ the iguana is at $$6$$; relative to the starting point the displacement is $$6 - (-2) = 8$$. The displacement is $$8$$ m in the first $$5$$ sec.
Here is a real world example of a negative starting point. A person stands erect in a hole dug out to $$1$$ m below the surface, and the top of the head of the person extends to $$1$$ m above the surface. The height of the person is $$1 - (-1) = 2$$ m.