I was coming back from my Driver's Education class, and something mathsy really stuck out to me.

One of the essential properties of a car is its current speed. Or speed at a current time. For example, at a given point in time in my drive, I could be traveling 40 mph. But what does that mean?

From my basic algebra classes, I've learned that speed = distance/time. So if I travel ten miles in half an hour, my average speed would be $20$ mph ($\frac{10 mi}{25 h}$).

But instantaneous velocity...you aren't measuring average speed for a given amount of time. You're measuring instantaneous speed over an...instantaneous amount of time.

That would be something like (miles) / (time), where time = $0$? Isn't that infinite?

And perhaps, in a difference of time = $0$, then I'd be travelling $0$ miles. So would I be said to be going $0$ mph at an instantaneous moment in time? I'd like to be able to tell that to any cops pull me over for "speeding"!

But then if miles = $0$ and time = $0$, then you have $\frac00$?

This is all rather confusing. What does it mean to be going $40$ mph at a given moment in time, exactly?

I've heard this explained using this strange art called "calculus" before, and it's all gone over my head. Can anyone explain this using terms I (a High School Algebra and Geometry and Driving student) will understand?

(I figured that my problem had numbers in it, and therefore has to do with Maths.)

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    $\begingroup$ ^^^Seed pretext of the week^^^ :) $\endgroup$ Jul 26, 2010 at 22:51
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    $\begingroup$ Welcome to calculus! That's what it's for. $\endgroup$
    – badp
    Jul 27, 2010 at 10:51
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    $\begingroup$ A very informal definition of current speed: It's the speed that matters when you crash into a tree. See Kline's Mathematics and the physical world: books.google.com/… $\endgroup$ May 31, 2011 at 4:13
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    $\begingroup$ Does someone know what the speedometer in a car actually measures? $\endgroup$
    – GEdgar
    May 31, 2011 at 13:11
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    $\begingroup$ @GEdgar: I believe that a speedometer measures the rotation speed of the output shaft of the transmission. This is scaled to linear speed. $\endgroup$
    – Tpofofn
    Jun 26, 2011 at 4:20

6 Answers 6


I think there is a very clear meaning in the physical world: If, at some moment, you were going 40 mph, if you were to stop de/accelerating and just hold that velocity, you would cover 40 miles in 1 hour.

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    $\begingroup$ In a similar sense, you could equate instantaneous speed to kinetic energy. This allows us to quietly bypass the concept of limits and derivatives. $\endgroup$
    – Tpofofn
    Jun 26, 2011 at 4:22
  • $\begingroup$ It's also the integral of accelleration ($\int acc \delta{}time$) and the derivative of position ($\frac{\delta}{\delta{}time} pos$) $\endgroup$
    – Cole Tobin
    Mar 28, 2014 at 16:18

If you were to try to measure instantaneous speed as you described, you would in fact have traveled 0 miles in 0 time and 0/0 is undefined If, however, you look at your average speed over smaller and smaller periods of time around the instant you care about--that is, (distance traveled from $t=t_0$ to $t=t_0+\epsilon$ for various small values of ε--and these average speeds "converge" (they all get closer to a single value as ε gets closer to 0), then we say that the instantaneous speed is that single value upon which the average speeds around that point converge. This is, of course, a somewhat informal explanation; to be more precise requires getting into differential calculus.


This is similar to other answers. Imagine another car beside yours. That car is covering a distance of 40 miles over the next hour at a constant velocity. At the point at which you keep pace with that car (relative velocity = 0), you are traveling at 40 mph.

  • $\begingroup$ Actually I like this answer a lot, probably as much as the selected answer. $\endgroup$
    – Justin L.
    Jan 31, 2014 at 22:32

Hmm, apparently the other answerers' algebra classes were a lot more intense than mine...My answer's based on what I would have been comfortable with after basic algebra and geometry.

Basically, if you think graphically, instantaneous velocity is the slope of a line at a single point rather than over an interval. At least that's what you want. So, if you happen to be going at a constant velocity (any straight line on a position graph), you can just use the slope formula. Where calculus comes in is if you're dealing with an inconstant velocity (a curved line).

If that's not making sense right off, just try to realistically graph the movement of a car as it accelerates/decelerates, where the y-axis is position and the x-axis is time.

Using algebra, you can't take the slope of a curved line. What you can do is take the slope between two points on the curve. The closer these points get to each other, the closer the slope between them will approximate the actual slope of the curve. So, if the slope between t=5.001 min. and t=5.01 min. is 40, then that approximates the actual slope, and instantaneous velocity, at t=5.0055 min.

I don't think I can get much more specific/accurate without going into (pre-)calculus.

  • $\begingroup$ +1: Great way to answer question w/o resorting to Calc, limits, etc. $\endgroup$
    – Tpofofn
    Jun 26, 2011 at 3:48

in the real world you need a bit of time to brake or accelerate. This means that in a small amount of time (let's say 1/10 of a second) your speed is more or less constant. So, instead of dividing by 0 seconds, just take a small elapsed time and do the division.

In calculus you will learn how to be more precise, noticing that using a smaller and smaller elapsed time the ratio between distance and time changes less and less; at that point you postulate that in this scenario only you may say that the limit (which is 0/0 and therefore indefinite) is the instantaneous speed. But the odometers don't care for instantaneous speed, so for the moment you shouldn't worry.

  • $\begingroup$ "You may say that the limit (which is 0/0 and therefore indefinite)", but its not $\endgroup$
    – Casebash
    Jul 26, 2010 at 21:28
  • $\begingroup$ which is of the form 0/0, which is indefinite. The whole purpose of calculus is to find a sound way to give a meaning ot such "ghosts of departed quantities", as Berkeley said. $\endgroup$
    – mau
    Jul 27, 2010 at 7:27

A state-variable kind of approach: (represents the physical states of a system)

Newton's first law of motion says that without external forces, masses will move at constant velocity. Every change in velocity (for a car, whether a change in speed due to braking, or a change in direction due to friction of the tires and the steering mechanism) is due to an external force.

So the "state" of the car at any instant in time is its velocity; distance traveled can be derived from velocity:

While you are going 60mi/hr = 26.8m/sec (see conversion factors) in a particular direction at constant velocity, you can multiply a change in time times velocity and get distance: 26.8m/sec = 26.8 meters in one second. Or 26.8mm in one millisecond. Or 26.8 micrometers in one microsecond.

The process of going from velocity to (directed) distance is called integration, and for those two variables (velocity and directed distance) is just the summing up of distance increments (= velocity * time increments) over infinitesimally short periods of time. For smoothly varying velocities, coarse time increments are a good approximation to reality -- differential equation solvers use this or similar approaches (e.g. Simpson's Rule) to perform integration.

But instantaneous velocity...you aren't measuring average speed for a given amount of time

Well, instantaneous velocity is just the limit of average velocity as the time period becomes infinitely small.


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