Average Speed vs. Velocity: Explained

[Huzaifa]

Though average speed over a finite interval of time is greater or equal to the magnitude of the average velocity, Instantaneous speed at an instant is equal to the magnitude of the instantaneous velocity at that instant. Why so?

 

[Ibix]

What is the definition of speed and velocity?

 

[Huzaifa]

What is the definition of speed and velocity?

The speed (or the average speed) is defined as the total distance traveled divided by the total time interval during which the motion has taken place.

The velocity (or the average velocity) is defined as the change in position (or displacement) divided by the time intervals in which the displacement occurs.

Thread 'Average speed and average velocity' https://www.physicsforums.com/threads/average-speed-and-average-velocity.986630/

Now consider the motion of an object along a straight line, the magnitude of the displacement is equal to the total distance. In this case, the magnitude of average velocity is equal to the average speed. But if the motion involves change in direction then the distance is greater than the magnitude of displacement. Thus, in this case the average speed is not equal to the magnitude of the average velocity.

 

[PeroK]

Now consider the motion of an object along a straight line, the magnitude of the displacement is equal to the total distance. In this case, the magnitude of average velocity is equal to the average speed. But if the motion involves change in direction then the distance is greater than the magnitude of displacement. Thus, in this case the average speed is not equal to the magnitude of the average velocity.

Any good maths student could construct an example where the limits do not agree. Generally, this will involve so-called pathological functions, which are not possible as a physical trajectory. E.g. some sort of infinite oscillation.

 

[Ibix]

Thus, in this case the average speed is not equal to the magnitude of the average velocity.

Right. And which is larger, the displacement or the distance?

 

[Huzaifa]

Right. And which is larger, the displacement or the distance?

Distance is greater than or equal to to the magnitude of the displacement... I have understood the distinction between the magnitude of average velocity over an interval of time and the average speed over the same interval. The average speed it is defined as the total distance divided by time and not as magnitude of average velocity.

But then why is the instantaneous speed always equal to the magnitude of instantaneous velocity?

 

[Huzaifa]

Any good maths student could construct an example where the limits do not agree. Generally, this will involve so-called pathological functions, which are not possible as a physical trajectory. E.g. some sort of infinite oscillation.

Can you please explain a little bit more about the pathological functions that you mentioned? seems interesting, and I would like to learn more about it.

 

[Ibix]

But then why is the instantaneous speed always equal to the magnitude of instantaneous velocity?

You can approximate a curve as a series of straight lines. More straight lines is a better approximation. Infinitely many infinitely short straight lines is a perfect approximation - so at an instant there is no difference between following a curve and following an infinitely short straight line (the tangent to the line, in fact).

Can you please explain a little bit more about the pathological functions that you mentioned?

If I teleport from one point to another then I didn't follow a path between them so neither displacement nor distance are well defined for the jump. Teleportation is science fiction, but discontinuous curves can certainly be defined in maths. For example $y=1/x$ is discontinuous at the origin. You can't approximate that jump by a short straight line so the whole argument in my previous paragraph falls apart.

 

[PeroK]

Can you please explain a little bit more about the pathological functions that you mentioned? seems interesting, and I would like to learn more about it.

Imagine a particle spiralling into the origin in smaller and smaller circles in finite time. The average velocity depends on the decreading radius of the circle. But the average speed depends in how many times it goes round the circle.

You should be able to construct it so the the final velocity is zero but the final speed is infinite.

Something like that.

Mathematically, there will need to be some constraints on the trajectory so that the limits exist and are equal.

 

[Herman Trivilino]

... I have understood the distinction between the magnitude of average velocity over an interval of time and the average speed over the same interval. The average speed it is defined as the total distance divided by time and not as magnitude of average velocity.

Imagine what happens as the duration of those intervals of time approach zero.

 

[Delta2]

Though average speed over a finite interval of time is greater or equal to the magnitude of the average velocity, Instantaneous speed at an instant is equal to the magnitude of the instantaneous velocity at that instant. Why so?

As far as I know instantaneous speed is equal to the magnitude of instantaneous velocity by definition. That's how instantaneous speed is defined. Period. OR if you want instantaneous speed to be the quotient of the infinitesimal distance to the infinitesimal time its almost the same thing cause infinitesimal distance $$ds=|d\vec{r}|=|\frac{d\vec{r}}{dt}|dt=|\vec{v}|dt$$ and hence $$instantaneousspeed=\frac{ds}{dt}=|\vec{v}|$$

The other inequality is a consequence of the general inequality $$|\int \vec{v} dt|\leq \int|\vec{v}|dt$$ (which holds for any vector $\vec{v}$ and is a consequence of the triangular inequality for the norm || (modulo) of vectors and of the definition of the integral as a sum of infinitesimal products). In the case the vector $\vec{v}$ is the velocity vector the above inequality divided by the total duration T tell us that the magnitude of the average velocity is less or equal to the average speed.

(In the case vector $\vec{v}$ is the acceleration it tell us that the magnitude of the average acceleration is less or equal to the average of the magnitude of acceleration.)