# Why is the first derivative velocity?

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1. Oct 9, 2015

### Niaboc67

Just something that has been bugging me. Can someone bestow why the first derivative is velocity and the second derivative is acceleration. I want to conceptually understand this.

Thank you

2. Oct 9, 2015

### Staff: Mentor

The first and second derivative of position with respect to time?
They are named "velocity" and "acceleration" by definition.

3. Oct 9, 2015

### Staff: Mentor

You could look at it via unit analysis so on a distance / time chart the first derivative the slope at any point on the curve and slope is delta distance / delta time or in units say meters / second aka velocity. Similary for 2nd derivative ony now you have a velocity / time chart hence (meters/second)/second or acceleration.

4. Oct 9, 2015

### Staff: Mentor

The first and second derivatives aren't necessarily velocity and acceleration, respectively. These are only two interpretations of these derivatives in a specific context; i.e., where you have a function that represents the position of something at a given time.
With regard to velocity, if you start at one place (call it 0) and end up at a point 50 miles away an hour later, what would your average velocity be for the entire trip? How would you calculate your average velocity over the period t = 45 min. to t = 46 min.?

5. Oct 9, 2015

### Svein

In a very limited sense this is true. If you have a formula describing a position with respect to time: $\vec{r} = \vec{f(t)}$, then the mean velocity between t1 and t2 is given by $\frac{\vec{f}(t_{2})-\vec{f}(t_{1})}{t_{2}-t_{1}}$ Taken to the limit, the instantaneous velocity at $t$ is given by $\vec{v}(t)=\frac{d\vec{r}}{dt}=\frac{d\vec{f}(t)}{dt}$.

In the same way: If the velocity changes, there is an acceleration involved. Thus the mean acceleration between t1 and t2 is given by $\frac{\vec{v}(t_{2})-\vec{v}(t_{1})}{t_{2}-t_{1}}$. Taken to the limit, the instantaneous acceleration at $t$ is given by $\vec{a}(t)=\frac{d\vec{v}(t)}{dt}=\frac{d^{2}\vec{f}(t)}{dt^{2}}$.

6. Oct 9, 2015

### HallsofIvy

If x(t) is the distance (in, say km) something has traveled in time t (hours), then x(t+ h) is the distance it has traveled in time t+ h hours, then x(t+ h)- x(t) is the distance traveled in that time h hours. So (x(t+h)- x(t))/h is the average speed in km/hour. Taking the limit as h goes to 0 (taking the average speed over shorter and shorter time intervals) gives the speed at a specific time.

However, as others have taking the derivative to be "speed" or "velocity" and the second derivative to be "acceleration" is an application of the derivative. Given a function f(x), df/dx is the rate of change of f. It is "velocity" only if f(x) is a position function that df/dx is the rate of change of position so "speed" or "velocity". If f is a function describing some other quantity, then df/dx is the rate of change of that quantity, not necessarily speed.

7. Oct 9, 2015

### rcgldr

Given a function for position versus time, p(t), or if just one dimension x(t), the function could be integrated, reulsting in units like meter second, in which case position could be considered as the first derivative, but I don't see how the integral of position versus time would be useful.

8. Oct 10, 2015

### CWatters

This assumes your starting point is distance or displacement. I would draw some graphs to convince yourself. Lets assume we are talking about a moving car. Plot a graph of distance vs time.

The derivative is the slope of the graph. The slope is equal to the change in distance/change in time (=dx/dt). That's the very definition of velocity. Typical units would be meters per second, miles per hour.

Then plot a new graph of velocity vs time. Once again the derivative is the slope of the graph. That's equal to the change in velocity/change in time (=dv/dt). That's the definition of acceleration. Typical units would be meters per second per second.