I am new to this topic and was just wondering about the use of instantaneous speed. I mean, we use to calculate the speed of car let us say at 5 sec. So we take the distance travelled in 4.9 to 5.0 seconds and divide it by time. We get instantaneous speed. We could simply as well have had taken distance travelled from 0 to 5 seconds and then divide it by time. So what is the use of instantaneous speed then?

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    $\begingroup$ What if the car accelerates in the meantime? $\endgroup$
    – Bernhard
    Commented Jan 3, 2021 at 13:04
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    $\begingroup$ The calculations you describe result in the average speed over the two given time intervals, not instantaneous speed. (Indeed, they actually result in average velocities rather than average speeds.) $\endgroup$ Commented Jan 3, 2021 at 21:08
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    $\begingroup$ Try telling that to a judge when you're defending yourself in traffic court: "But your honor, my average speed was only 50 miles per hour" $\endgroup$ Commented Jan 4, 2021 at 3:40
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    $\begingroup$ @HovercraftFullOfEels: Fun fact : the average speed can be used in court, but not for your defence: ec.europa.eu/transport/road_safety/specialist/knowledge/… If your average speed on a road section was above the speed limit, your instantaneous speed must have been above the speed limit somewhere on this section. You don't need radar anywhere, just two webcams. $\endgroup$ Commented Jan 4, 2021 at 13:11
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    $\begingroup$ @user253751: No, it's not. Unless time is actually quantized, it's always possible to improve the technology to measure shorter intervals. (That's really the basis of calculus.) So in the bike example, you could put a sensor on each spoke of the wheel, which would give a closer approximation to the instantaneous speed, but still wouldn't give it exactly. $\endgroup$
    – jamesqf
    Commented Jan 4, 2021 at 17:19

8 Answers 8


Because instantaneous speed affects physics. Imagine a wall $10~\textrm m$ in front of you. You walk towards it smoothly over a timeframe of, say, $20~\textrm s$, and without getting slower, you walk into the wall. You'll feel a slight bonk, but nothing serious is going to happen. Now imagine the same 20 seconds going differently: You wait for 17 seconds, then you sprint towards the wall at full speed.

Both scenarios will give you the same average speed over the 20 seconds, but you better be wearing a helmet for the second one. The difference lies in the fact that the instantaneous speed at the end of the 20 second interval is different. It's a quantity that affects things. So it makes sense to talk about it.

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    $\begingroup$ It's worth noting that instantaneous speed is always better at measuring the size of the 'bonk' than the average speed. Say I took the average speed over an interval of $5$ seconds—that might give me a better understanding of the size of the bonk than an interval of $20$ seconds, but again, we only care about what happens when we hit the wall. Even an interval of $1$ second is not as good. It's only when we measure the average speed over $0.0001$ seconds that we feel confident that we know how hard we will hit the wall. From a physicists point of view, that's what instantaneous speed is. $\endgroup$
    – Joe
    Commented Jan 4, 2021 at 19:16
  • $\begingroup$ There's an actual hypothesis that turns up from time to time that there's no such thing as instantaneous speed, and time is discreet. $\endgroup$
    – Joshua
    Commented Jan 5, 2021 at 21:26

It is really simple:

Average speed is as good as instantaneous speed only if speed does not change with time. If you are studying a body with rapidly changing speed then using average speed to describe that body gives you an inaccurate depiction of the phenomena taking place.

Of course we can only truly measure average speeds, but measuring those over really short periods of time helps getting us closer to the true value of the instantaneous speed. In fact remember that the definition of instantaneous speed ($v_i$) is precisely average speed ($v_a(\Delta t)$) measured over an interval of time that tends to zero:

$$\lim _{\Delta t \to 0} v_a(\Delta t)=v_i$$

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    $\begingroup$ I think your instinct is right that the OP might be picturing a constant-speed situation. $\endgroup$ Commented Jan 3, 2021 at 21:09
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    $\begingroup$ I'm not sure I agree that we can only "truly" measure average speed. This may be true in a distance divided by time measurement, but using alternative means like Doppler Radar or other velocity derived measures can and do measure instantaneous speed. $\endgroup$
    – throx
    Commented Jan 4, 2021 at 1:07
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    $\begingroup$ @throx Not true. Doppler radar measures average speed. It's over a very short time, true, but it's still an average over time. Doppler works by measuring the frequency change of a radio wave. Meaning, it has to measure the distance between peaks. if the speed of the target changes during one wave, then the answer it gets will be the average over the time between the two peaks. $\endgroup$ Commented Jan 4, 2021 at 3:16
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    $\begingroup$ @HiddenWindshield Assuming you're measuring the energy of the photons received then it is a measure of instantaneous velocity, with an uncertainty reflected by Heisenberg limits. You can hand-wave it away as an average, but that implies that there is actually finer grained measurement possible, when that may or may not be the case. At that level of precision, you're not going to measure any reasonable instantaneous position to begin with. :) $\endgroup$
    – throx
    Commented Jan 4, 2021 at 3:33
  • $\begingroup$ @GregMartin I think I totally misread the question as well. $\endgroup$
    – Hank Igoe
    Commented Jan 4, 2021 at 5:04

Instantaneous speed is what the police officer fines you for. On a trip to the bakery it doesn't really matter that your average speed is 40 km/hr if you during the trip reached 120 km/hr for just a moment.

In some scenarios an average is useful. In others the instantaneous speed is useful. An average is lacking full detail but "represents" the full trip concisely,** whereas having the instantaneous speed at every moment is accurate but a huge amount of data. Which you will be using depends on what you need and thus varies form case to case.

** Also note that there are various types of averages with different ways of "representing" an amount of data in one single and more tangible number, such as arithmetic average or mean, geometric average, RMS value and the like.


One way of looking at it is that instantaneous speed gives you more details about your journey, especially when your journey consists of variable speeds. The smaller the time interval in which you measure, the more information you have about your journey. But if your speed were constant, instantaneous speed would make no difference as it would be the same as the average speed at all times.

You can think about it like pixels on a computer screen. The smaller the pixels and the more pixels you have, the clearer and more detailed the picture will be. But if your picture is just the color red, it makes no difference how many pixels there are, even if the entire screen were just one big pixel, which is analogous to a car having constant speed.

Another thing to note is that even when you say instantaneous speed, it is not truly instantaneous but rather the average speed within a much smaller time interval


It was precisely this problem that led Isaac Newton (and Gottfried Leibnitz) to develop the Calculus. Based on Galileo's and Copernicus's measurements, Newton and others were able to calculate the speed and acceleration of the planets in their orbits. And of course they know that $F= ma$ so could calculate $F/m$. But what sort of Force would produce the ever changing acceleration? It had to be due to the changing distances of planets from whatever the center of that force was. It did not take much to guess that the center of that force was the sun. But if that distance is constantly changing, how could it be equal to the $a= (v_1- v_0)/2$? It couldn't, of course. In order to make any sense of that at all, it was necessary to define "instantaneous acceleration" which then required "instantaneous velocity".

By the way, the original post says "So we take the distance travelled in 4.9 to 5.0 seconds and divide it by time. We get instantaneous speed."

No, we don't! The distance traveled in 4.9 to 5.0 seconds and then divided by 0.1 seconds is just as much an AVERAGE speed as calculating the distance traveled between 4 and 5 seconds and dividing by 1 second. To get "instantaneous speed" at 5 seconds you have to take the LIMIT of $(d(5+h)- d(5-h))/2h$ as $h$ goes to $0$, where $d(t)$ is the distance at time $t$ seconds.

  • $\begingroup$ Greetings. Please note you should use MathJax to format equations, and the GUI or markup tools to add bold and italics. $\endgroup$
    – kaylimekay
    Commented Jan 4, 2021 at 4:40

I'd argue that we don't need instantaneous speed. What we actually need is momentum, which is often introduced as defined by $$ p := m\cdot v $$ but actually it is the more fundamental physical quantity (namely, it's a conserved quantity and meanigful also in quantum mechanics, unlike velocity). One could go the other way around and write $$ v = \frac{p}{m} $$ but we can also skip velocity completely and instead observe only that $$ x(t_1) - x(t_0) = \int\limits_{t_0}^{t_1}\!\mathrm{d}t\: \frac{p(t)}m $$ holds for any point mass with time-variable position $x$ and momentum $p$. If you don't understand what the $\int$ symbol means, see below. The momentum can be time-variable because the object can interact with its surrounding via forces, but in particular for an isolated object, $p$ is constant and then we get $$ x(t_1) - x(t_0) = (t_1-t_0)\cdot \frac{p}m, $$ which is equivalent to saying velocity is equal to instantaneous velocity. But that's the “boring case”: all interesting physics is about when you do have interaction between particles, i.e. the momentum of individual objects is not constant (though the sum over all objects will be), and then the integral does not boil down to a mere product with the time-interval anymore.

A brief introduction to integrals

Already in the example above we saw that the integral of a constant function $f(t) = f_0$ over the interval $[a,b]$ is simply that constant multiplied with the size of the interval: $$ \int\limits_a^b\!\mathrm{d}t\: f_0 = (b-a)\cdot f_0 $$ Integration generalises this idea to non-constant functions. For example, say you're driving your car 10 minutes at $30\:\mathrm{km/h}$ to get out of town, and then another 50 minutes at $90\:\mathrm{km/h}$. What is the total distance? Well, you multiply the speed for each of the time-interval with how long it was: $$\begin{align} x(60\:\mathrm{min}) - x(0) =& 10\:\mathrm{min}\times 30\:\mathrm{km/h} + 50\:\mathrm{min}\times 90\:\mathrm{km/h} \\ =& \tfrac16\:\mathrm{h}\times 30\:\mathrm{km/h} + \tfrac56\:\mathrm{h}\times 90\:\mathrm{km/h} \\ =& 5\:\mathrm{km} + 75\:\mathrm{km} \\ =& 80\:\mathrm{km} \end{align} $$ A general function won't be constant on any interval at all, but if you subdivide the intervals ever and ever smaller then it is eventually almost constant on each of them. That process is called Riemann integration. And this velocity-on-a-very-small-time-interval is then again our instantaneous velocity.

  • $\begingroup$ Yeah, but try building a usable momentumometer for your car :-) $\endgroup$
    – jamesqf
    Commented Jan 4, 2021 at 17:22
  • $\begingroup$ @jamesqf And you need to know mass =/ $\endgroup$
    – DKNguyen
    Commented Jan 4, 2021 at 21:04
  • $\begingroup$ @jamesqf that's not really more impractical than building a true instantantaneous-speed-ometer. And “momentumometers” are actually used for some applications: inertial navigation is based on accelerometers. A (mechanical) accelerometers actually measures rather the $\partial_t p$ for a proof mass, than directly acceleration. $\endgroup$ Commented Jan 4, 2021 at 22:22
  • $\begingroup$ @leftaroundabout: But it's measuring that value for the tiny mass (or masses, I think, for a multiple axis one) that's in the chip. It's irrelevant whether the chip is installed in your cell phone or a 550K ton oil tanker. $\endgroup$
    – jamesqf
    Commented Jan 5, 2021 at 6:05
  • $\begingroup$ How would you define the speed of light as a momentum? $\endgroup$
    – Fax
    Commented Jan 5, 2021 at 15:00

If the speed of an object varies, then its average speed over an interval of time can still be calculated, but it will be an approximation that may or may not be meaningful.

from 4.9 to 5.0 seconds and divide it by time. We get instantaneous speed. We could simply as well taken distance travelled from 0 to 5 seconds and then divide it by time

This is correct. Distance over an interval of time gives an average speed, which is an approximation of instantaneous speed. Note both time intervals, 5s or 0.1s result in average speed measurements. But since 0.1s feels "short" it "feels" more like we measured instantaneous speed. It just feels like it, which doesn't mean it is true.

What actually matters is how much the instantaneous speed changes during the measurement interval. If you want to measure instantaneous speed, you should first decide how accurate your measurement must be, say a percentage of error, then estimate how much the speed of the object can change over time, and decide on a time interval that is small enough to give you the accuracy you want.

Say for example you ride a bicycle over hilly terrain. Downhill, you'll go fast, but uphill, not so much. If you spend 50 minutes uphill and 10 minutes downhill, then measuring over an hour will give a meaningless result. Maybe you did 12 km/h average, but that really was 10 km/h uphill and 50 km/h downhill.

If your downhill speed is constant, then measuring over 10 minutes would be more accurate. If you use brakes, which will change speed, then you need to measure over a shorter interval.

Note the same applies to the velocity vector. If you run around a circle and average the velocity vector over one turn, then the average will be a zero vector no matter what the speed is, because the change in vector direction as you turn makes the whole vector sum cancel out to zero. To measure the velocity vector (including direction) you have to pick an interval where not only the speed but also the direction stays "constant enough" to give a meaningful average.

wondering about the use of instantaneous speed

First, it exists: the object is moving at a certain speed, and if this speed varies you need a way to represent it mathematically if you want to write down equations.

Then it is useful for initial conditions: you throw a rock, how far will it go? This depends on its instantaneous speed when it leaves your hand, among other things.

It is also required for calculating kinetic energy.

  • $\begingroup$ You might want to add a note about the difference between average speed and average velocity. $\endgroup$ Commented Jan 3, 2021 at 21:09
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    $\begingroup$ Thanks, I just learned what "velocity" means lol. In French we call it "vecteur vitesse" ie "speed vector" which makes it quite obvious it's a vector... $\endgroup$
    – bobflux
    Commented Jan 3, 2021 at 21:12

Mathematically, speed is defined as the first derivation of place as a function of time:

$$\vec{v}(t) := \frac{\partial\vec{x}(t)}{\partial t}$$

This is what you call instantaneous speed. This speed is then used to derive other physically important variables like impulse and energy:

$$\vec{p}(t) := m\cdot\vec{v}(t)$$ $$E_{kin} := \frac{m}{2}|\vec{v}(t)|^2$$

These variables have a lot of physical meaning to them, and they can change continuously as a car accelerates, for instance. Most importantly, they are conserved quantities, so that you can easily answer the question: "At which speed will a stone hit the pavement if I drop it from 1m height?" You do not need to compute the speed at any other point in time, you only take the potential energy of the stone before it's dropped, realize that this energy is equal to the kinetic energy that the stone has when it hits the pavement, and solve the above equation to get the speed and momentum with which the stone will hit the pavement.

Since speed is defined as a mathematical derivation of a function, it is by definition an instantaneous quantity. The average speed (that you compute for instance for a trip to another city) has virtually no physical meaning. It does not give you handles on the energy or momentum that is used to do the trip, both of these can vary all over the place. Much in the same way that you cannot judge road distance by measuring distance on the map with a ruler.

  • $\begingroup$ "speed is defined as the first derivation of place as a function of time:" I thought the first derivative is the slope of the curve, which is acceleration, not speed. (It's been 36 years, though, since I took calculus, and haven't used it since, so I might be wrong.) $\endgroup$
    – RonJohn
    Commented Jan 5, 2021 at 16:47
  • $\begingroup$ @RonJohn You are both correct. Speed is defined as the (first) derivative of position to time, which indeed is the slope of the position graph. The first derivative provides us with the tangent at every point on the graph; the tangent tilt is the slope at that point. $\endgroup$
    – Steeven
    Commented Jan 5, 2021 at 22:31
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    $\begingroup$ @RonJohn Speed is the first derivative of place in time, which is the slope in a diagram showing place over time. Acceleration is the second derivation of place in time, which shows as the curvature in the place over time diagram. $\endgroup$ Commented Jan 5, 2021 at 22:39
  • $\begingroup$ That's right... slope is the current speed, and acceleration is the change in that slope. Like I said, it's been a long time. $\endgroup$
    – RonJohn
    Commented Jan 5, 2021 at 22:59

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