It is currently to my understanding that the area under a $v(t)$ graph is the displacement of an object because $$\int v(t)dt = s(t).$$

However, some of the problems I have attempted recently give you a $v(t)$ formula and ask you to find the total distance travelled. The solution to these problems would be the same solution to finding the position(displacement) of the object.

Are the methods of finding each value the same? If not, what is the difference between each process?

  • $\begingroup$ $s(t) = s(t_0) + \int_{t_0}^t v(t^\prime) dt^\prime$ $\endgroup$
    – Semoi
    Commented Jan 2, 2021 at 20:44
  • $\begingroup$ Your integral doesn't have a differential. Also the integral is undetermined (has no boundaries) $\endgroup$
    – Gert
    Commented Jan 2, 2021 at 20:49
  • $\begingroup$ @Semoi I think I understand now. s(t) is the displacement + the inital starting point and the total distance travelled is the area of the curve both over and under the curve(?) $\endgroup$
    – RudyGoburt
    Commented Jan 2, 2021 at 21:09
  • 1
    $\begingroup$ Yes. In other words, on a velocity vs. time graph, displacement is the net (or signed) area between some $t_i$ and $t_f$, whereas the distance is the total area between $t_i$ and $t_f$. $\endgroup$
    – user256872
    Commented Jan 3, 2021 at 0:29

2 Answers 2


If you're working with graphs, I will assume 1D problems. Let $v$ be the velocity.

On some time interval $t_i$ to $t_f$:

The displacement, $\Delta d$ is:$$\Delta d= \int_{t_i}^{t_f}v\ dt$$

The distance travelled is $$s= \int_{t_i}^{t_f}|v|\ dt$$

If $v=|v|\ \forall \ t\in[t_i,t_f]$ then the approaches will indeed be the same. The reason we use the absolute value (or in 2/3D, magnitude) of velocity (i.e. speed) is that when calculating distance, we do not care in which direction the object has moved, only that it moved with some speed.


$\vec v$ is a vector equal to $d \vec r/dt$. Your relationship is for the scalar speed $v(t)$. In rectangular coordinates the speed is $(v_x^2 + v_y^2 + v_z^2)^{1/2}$, where $v_x = dx/dt, v_y = dy/dt, v_z = dz/dt$. The scalar distance $s(t)$ can be evaluated using $ds/dt = v$; which gives $s(t) = s(t_0) + \int_{t_0}^{t} v(t')dt'$ as @Semoi says in the earlier comment.

For constant acceleration $\vec a$, say in the $x$ direction, using $dv/dt = a$ we have $v(t) = v(t_0) + at$, and using $dx/dt = v$ we have $x(t) = x(t_0) + v_0 t + 1/2 a t^2$. This is only true for constant acceleration


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