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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 $x_\text{final}-x_\text{initial}= 6-(-2)= +8 \,\text{m}$.

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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.

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  • $\begingroup$ 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. $\endgroup$ Commented Feb 20, 2021 at 0:12

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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.

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