# Area under a velocity graph

If I took the definite integral of a velocity graph from 0 to 10 seconds, the answer would be the change in position over those 10 seconds correct? I am told by my teacher the area is change in displacement but that doesn't make sense.

You’re right. You’ll get the change in position $$\Delta x$$, which is the displacement. Maybe your teacher is confused about “displacement” ($$\Delta x$$) vs. “position” ($$x$$).

Both are correct as it is just a matter of wording although I would favour change in position and displacement most of the time.

Consider one dimensional motion along the x-axis with the unit vector $$\hat x$$ defining the positive x-direction.

A body starts at position $$+3 \,\hat x$$ with velocity $$+6\,\hat x$$ and after undergoing constant acceleration over a time interval of $$3$$ it has a velocity $$-3\,\hat x$$.

The area under the velocity time graph is $$+6 +(-3) = +3$$ and this is the change in position, so the new position is $$+6\,\hat x$$.

One could also say that the displacement of the particle during that time interval is $$+3 \,\hat x$$.

However going back to position $$+3 \, \hat x$$ that is a displacement of $$+3 \,\hat x$$ from the origin and position $$+6 \,\hat x$$ is a displacement of $$+6 \,\hat x$$ from the origin.

So the change in position is also the change in displacement if the displacements are measured from the origin.