What does it mean to be going 40 mph (or 64 kph, etc.) at a given moment?

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.)

Solution 1:

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.

Solution 2:

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.

Solution 3:

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.

Solution 4:

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.