Why is the derivative of position velocity?

Dervative is the rate of change. So the rate of change of your position is your speed (velocity), and the rate of change of your velocity is your acceleration.

There's a reason behind it. Don't know if my explanation helped -- you've probably heard it before.

On 06 Aug 2006 21:42:58 GMT, snipped-for-privacy@cucumber.demon.co.uk (Andrew Gabriel) Gave us:

You forgot direction of travel. Hard to derive position without that as well.

Those are the definitions! How you define velocity and acceleration quantitatively? The ultimate rationale is that if you use those definitions, then Newton's laws work.

Bill

-- Ferme le Bush

That relationship of the derivatives and the physical parameters has nothing to do with caclulus - it has to do with newtonian mechanics

in calculus, you use an infinitely small element in the arithmetic operations, and in algebra you use a finite element in the arithmetic operations -

same arithmetic relationship, same arithmetic operations.

e.g. -

velocity = s/t = ds/dt == distance over time.

acceleration = v/t = dv/dt == velocity over time

The reasons would be covered in a class on dynamics. On the other hand, IMHO, a calculus class taught without grounding the abstract math in such simple principles as velocity and acceleration is pretty pointless.

Saying "derivative of position" is incomplete. You meant to say: "derivative of position with respect to time"

Although it is obvious, but it is the key to understand why it is like that. Think of it this way: Suppose you were moving for 20 seconds. At the 10nth second, you decided to figure out how fast you are moving at THAT EXACT SECOND. You do this: Measure the distant measured between second 10 and second 10.01 for example. Lets say you travelled 0.02 meters. Then your speed is

0.02/0.01 (change of distance with respect to time). For more accurate results, you can meaure distance tranvelled between the seconds 10 and 10.00001, and find that distance by 0.00001. . . . the smaller the "time window", the more accurate your CURRENT speed is. As this "time window" gets smaller and smaller, you'd be taking the derivate at that point. WHY? Simply because they called that process a derivative. Later on, they could prove that faster way than doing that process exist to find the same value!!

So, change of distance with respect to time is speed (they called it speed instead of saying change of distance with respect to time. Instead of saying: I cover 50 miles each hour, I say my speed is 50 miles/hour)

Now, same logic goes for acceleration. If between the seconds 10.00 and

10.01 your speed was lets say 10 meters/second. Then between 10.01 and 10.02 your speed was 11 meters/second or maybe 9 meters/second. So we are looking at a change in speed over a time of 0.01 second. This change of speed is called acceleration. Now, by how much did your speed was changing AT the second 10.01s ? What you can do is take a time window of 0.0000001 and measure your velocity at 10.01 and then at 10.0100001, then take that value and divide it by the "time window" which is 0.000001. For more accurate results, take a smaller window. As the time window is very very small, you'd be getting more accurate results. They called the process of making the "time window" to be very very small and finding the corresponding speeds, then dividing the change of speed by the very small time window a derviative. Later, they found out they can do shortcuts to make this process simpler.

I hope that helps, although boring.