# Does the area under a velocity-time graph represent distance or displacement?

In my textbook it says that the area under a velocity-time graph represents the distance, which I found strange because velocity if often related to displacement not distance. So which does it represent?

Is there even a difference when the velocity continues in one direction only? And is there is the velocity dips into the negatives?

• Consider circular motion - integrate over a long time and you get a big distance (total length traveled) with minimal displacement (its still going around in circles). Feb 1, 2022 at 15:49

If the velocity time graph depicts the magnitude of velocity (which can also be a vectorial magnitude), the area under the curve is distance (arc length): $$\mbox{distance}=\int |\vec v(t)| \mbox{d}t$$ If velocity is signed velocity (which can be generalized to a velocity vector), then the area (or areas) is displacement. $$\mbox{displacement}=\int \vec v(t) \mbox{d}t$$ If velocity is single-valued (1-dimensional) and only has positive sign, then there is no difference between the two because then $$v=|v|$$ everywhere.

• I wouldn't put a dot between a vector and a scalar. Feb 1, 2022 at 16:04
• @Newbie: you are probably right. I just hate it if the $dt$ is so obtrusive to the integrand. :-) The dot is not precisely wrong, it is just not an inner product, which should be clear from dt being a scalar differential. Feb 1, 2022 at 16:05
• How about $\int_{t}dt \vec v(t)$ or $\int_{t} \vec v(t) dt$? Feb 1, 2022 at 16:06
• Okay, I do as you wish. Feb 1, 2022 at 16:07
• Don't do it if you don't agree :) It's already a good answer :) Feb 1, 2022 at 16:08

If we consider a velocity-time graph, that means we consider movement in only one direction (there may be more, but in the graph we only consider one), and thus if the particle moves in the same direction all the time, the displacement matches the distance. So, for the case where the velocity is positive (i.e. moving in the same direction), both are correct.

To answer the general case, consider a graph like a sine curve representing the velocity of a particle and integrate over a full period. The integral becomes zero, which matches the displacement (the particle has gone back to its original position).

If you want to consider this more rigurously, let $$x(t)$$ be the position of the particle at time $$t$$ and let $$v = \frac{dx}{dt}$$ be its velocity. Then, the area under the curve of the graph of $$v$$ over a time interval $$[a,b]$$ is (note if $$v$$ is negative we add up negative areas):

$$\int_a^b v(t) dt = x(b)-x(a) = \Delta x$$

Which is indeed the displacement (this matches the result before).

This is not the full answer, though. If you were to consider the negative areas as positive, then you would slice the interval $$[a,b]$$ into intervals where $$v$$ has a constant sign and add up the absolute value of each integral (the areas). But we are back to the first paragraph! This would be adding up the motion of the particle over periods where it keeps a constant direction of motion, and this means you are calculating the distance.

So it all depends on what you define as "area under the curve". Is it the integral itself (displacement) or the sliced up version (distnace). That is probably the reason for the discrepancy of your sources.