I've read and (I think) understood that instantaneous velocity is the velocity of an object for an "infinitesimally" small time interval: $$ \lim_{\Delta t\to0} \frac{\Delta s}{\Delta t} = \frac{ds}{dt} $$ This makes sense in my mind. But what does instantaneous displacement mean? Let's say we have an expression for velocity, $v$ in terms of time $t$: $$v=t +1$$

If we find the antiderivative of this, we get an expression for displacement, $s$: $$s = \int{(t+1)dt}$$ $$ s = \frac{t^2}{2} + t + c $$

Let us say we somehow find out the value of $c$ to be $0$. So: $$s = \frac{t^2}{2} + t $$ If we plug in, let's say, $t = 2$ into this expression, we get s = 4. What does this 4 mean/represent?

Also, is it different to take the definite integral from $t=0$ to $t=2$:

$$ \int_{0}^{2}(t+1)\space dt$$ and using this to find displacement at $t=2$, compared to what we did using the indefinite integral?

Is there a difference between saying "displacement at $t=2$" and "displacement from $t=0$ to $t=2$"?

  • $\begingroup$ Since you are integrating the velocity, the function $s(t)$ is your current position at time $t$ (the area accumulated under the velocity vs time curve) $\endgroup$
    – RC_23
    Jul 30, 2022 at 16:03
  • $\begingroup$ In that case, is it different to take the definite integral from $t = 0$ till $t = 2$ and then finding out the displacement, as opposed to somehow finding the value of $c$ and then plugging $t=2$ into the expression obtained from the indefinite integral? $\endgroup$
    – AVS
    Jul 30, 2022 at 16:06
  • $\begingroup$ Try that in your example and see what happens $\endgroup$
    – RC_23
    Jul 30, 2022 at 16:53
  • $\begingroup$ I get the same answer since i've taken c as zero, but if c has any other value, my answer differs by $c$. So is displacement really the change in position from the origin? I've never thought of it in that way, rather the change in position from the original position of the object. $\endgroup$
    – AVS
    Jul 30, 2022 at 16:55
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    $\begingroup$ This is a good observation. The thing you are integrating is velocity $v$, which does not tell you anything about the original position as you say. You are free to decide that. Instead of $c$ you can call it $s_0$ which is position if you plug in $0$ for $t$. In general, you can say $\int v dt = s(t)$ (absolute position), and $\int_0^2 v dt = \Delta s$ (displacement from $t=0$ to $2$) $\endgroup$
    – RC_23
    Jul 30, 2022 at 17:05

2 Answers 2


Answer in response to the original question and your following chat message:

"The answer on the link I've sent says that displacement is not the position, and it is only the difference between final and initial position (change in position). You have said "you can say $\int v dt = s(t)$ (absolute position)". In my mind, there are two possibilities here: either the s here refers to position, or the s refers to displacement and you are saying that the displacement is the absolute position. If it is the latter, that is what the contradiction. And yes, the specific problem is: "v = t+1, find s (displacement) at the end of 4 seconds if initially body is at origin." My teacher said that I cannot use definite integration as we do not know the boundaries for time, so I have to first use indefinite integration and find the value of c. Aren't the boundaries t=0 and t=4 (they used the word initially). When I tried both approaches, my answer differed by the value of c, as I said."


Your teacher is creating unnecessary distinctions, but possibly for a good reason to create the least confusion for students new to the topic. In my personal experience, I would call a function s(t) that describes a particles trajectory "position," and I would call a subtraction of two positions $Δs = (s_2–s_1) = s(t_2)–s(t_1)$ "displacement."  I would say the majority of users in Western Industry and Academia would agree with those terms. I would not call s(t) "displacement" in normal usage, but you should recognize that the concept of "position" simply means "displacement from an origin which is agreed to be zero." 

As for your specific problem, you can use either indefinite or definite integral to achieve your answer, as long as you use them correctly. This is because mathematically they are related. If there is a function $F(t)$ whose derivative is $f(t)$, which is to say

$$\frac{dF}{dt} = f(t)$$ then the following are true:

$$ \int f(t) dt = F(t) + c$$

$$ \int_a^t f(t) dt = F(t)-F(a)$$

So we get that $F(a)=-c$. If you want to obtain $F(t)$ in isolation, choose $a$ such that $F(a)=0$.

When learning these topics, it is best to learn to understand the physical concepts involved rather than be married to strict definitions of terms. You can find "how far the object with velocity $v$ has traveled," and "where it's current location with respect to an origin is," and call those whatever you want.

One last thing, you do know the boundaries for time because the problem says "at the end of 4 seconds." Therefore you are starting your clock at $t=0$ and stopping it at $t=4$. You also know the initial position because it says "if initially [$t=0$] the body is at origin [$s=0$]."


The instantaneous displacement is just the displacement of the system evaluated at some instant $t$, $x(t)$.

As we intuitively think the speed of a body in rest should be zero, it is not so intuitive to think the displacement of a body from a reference point should be zero too. Since our intuitive reference for speed is zero, we should also expect our intuitive reference for displacement to be zero too.

If you think in the mass+spring+damper system: $$ \ddot x+2\zeta \omega\dot x+\omega^2x=0, x(0)=1 $$ Its zero energy state is at $x=0$ and $\dot x=0$, when both the instantaneous velocity and the instantaneous displacement are zero. In this case, the intuition is somehow more explicit.

In your specific example, if $x(t)=\frac{t^2}2+t$, the instantaneous displacement at $t=2$ is $x(2)=4$, meaning the instantaneous displacement measured from the reference is 4.

In the case of the mass spring damper, for example, if we have an underdamped system $x(t)=e^{-\zeta\omega t}\cos(\omega\sqrt{1-\zeta^2}t)$, the same concept applies, the instantaneous displacement at $t=0$ is $x(0)=1$ but in here it is more evident that the displacement is a deviation from the reference.

In contrast, check the following: Suppose we are measuring a displacement with respect to the last measured position, i.e. suppose $dx'(t)=dt'$ represents the change of displacement from the last time with a resolution of $dt'$, that is, after each $dt'$ interval, the displacement increases in that same measure. In this case, we should, again, integrate to obtain the displacement: $$ x(t)=\int_0^{t} dx'=\int_0^t dt'=t $$ That is, since the position is constantly changing, we expect the position to be linearly increasing. Again, $x(t)$ represent a measurement wrt to a reference, and $\delta x(t)$ is a mathematical object to depict what you could understand by change of displacement. Note, how this last concept is by far less used, hence the use of that notation, less used too.

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    $\begingroup$ So is displacement actually change in position w.r.t. to the original position, or is it change in position from a reference point i.e. origin? $\endgroup$
    – AVS
    Jul 30, 2022 at 16:14
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    $\begingroup$ Just like any measured variable, displacement, velocity, acceleration, etc. are w.r.t. a reference. x(t), s(t), v(t) are changes wrt that reference. The displacement x(t) is measured wrt the origin (x(t)=0), as the velocity is measured wrt the rest (v(t)=0). Sounds more complicated that what really is. By contrast, we never measure variables wrt to the last known variable, since sensors, as the mass spring system, measure wrt to a reference. $\endgroup$
    – Brethlosze
    Jul 30, 2022 at 16:21
  • $\begingroup$ Check the edit for clarification $\endgroup$
    – Brethlosze
    Jul 30, 2022 at 16:31
  • $\begingroup$ physics.stackexchange.com/questions/24662/… Doesn't the (accepted) answer to this question contradict what you are saying? Sorry if I am confused. $\endgroup$
    – AVS
    Jul 30, 2022 at 17:02
  • $\begingroup$ That answer is misleading. Review the case of the spring mass system, which repeats in multiple physics situations and will give you a good insight of what you do need to measure: If you need to measure a position from a fixed point, or, at every time step, you need to measure a change. You will realize, the first case is what you need, and the second case is more seldom used and require a different computation. $\endgroup$
    – Brethlosze
    Jul 30, 2022 at 18:25

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