Can Distance Traveled be Calculated Mathematically in Average Speed?

[PFuser1232]

It is my understanding that while instantaneous speed is the magnitude of instantaneous velocity, average speed is , in general, not the magnitude of average velocity, since average speed is total distance traveled divided by change in time. Why is the mathematical definition of average speed as such? Why can't average speed be defined so that it is equal in magnitude to average velocity?

 

[Maged Saeed]

what is the definition of average velocity?

 

[jtbell]

Suppose you drive your car around and then come back to your starting point. Wouldn't it seem a bit strange to say that the average speed of the car was zero?

The value you get using standard definition of "average speed" can be approximated by reading your car's speedometer once every minute, and then finding the average of all those readings at the end of the trip. If you read the speedometer more frequently, you get more measurements and a better approximation.

 

[Maged Saeed]

average velocity is defined as the following:
$$\frac{initial \space velocity\space +\space final\space velocity}{2}$$

But , initial velocity can be greater than final velocity(if the object is decreasing) so the velocity here would be negative while average speed cannot be negative.
other thing ' which is more important' if you drive a car and in a circular path , the average velocity here is zero {from initial point to the same point } but average speed isn't.

By the way :
Is instantaneous speed equal to instantaneous velocity or the absolute value of the instantaneous velocity? I'm not sure

 

[jtbell]

Is instantaneous speed equal to instantaneous velocity or the absolute value of the instantaneous velocity?

In English, we usually say that instantaneous speed is the magnitude of the instantaneous velocity. In an equation we often indicate the magnitude by using vertical bars (the same as for absolute value), e.g. $v = |\vec v|$, but we don't usually call it "absolute value". We usually say "absolute value" only when we are talking about scalar quantities, e.g. |-5| = 5, "the absolute value of -5 is 5".

 

[Khashishi]

If you drive a car around a circuit, the average velocity is zero. Do you really want to say the average speed is zero?

 

[nasu]

average velocity is defined as the following:
$$\frac{initial \space velocity\space +\space final\space velocity}{2}$$

But , initial velocity can be greater than final velocity(if the object is decreasing) so the velocity here would be negative while average speed cannot be negative.
other thing ' which is more important' if you drive a car and in a circular path , the average velocity here is zero {from initial point to the same point } but average speed isn't.

By the way :
Is instantaneous speed equal to instantaneous velocity or the absolute value of the instantaneous velocity? I'm not sure

No, this is not the usual definition of average velocity.
Average velocity is the displacement (a vector) divided by the time it takes for that displacement.
What you wrote is an average of two velocities (not the average velocity, in general).

A scalar (instantaneous speed) cannot be equal to a vector (instantaneous velocity).

 

[jtbell]

What you wrote is an average of two velocities (not the average velocity, in general).

It does work when the acceleration is constant. Many students learn that formula in the context of constant acceleration, then try to apply it (incorrectly) to situations where the acceleration is not constant.

 

[Redoctober]

The definition comes from the following mathematical model,

$V_{avg} = \frac{\int_{t1}^{t2} \vec{V} . dt }{t_2 - t_1}$...

1 dimensional so $\vec{v} = V \vec{i}$

It happens that the integral of the Velocity vector dot the time to be net distance between starting and ending point. therefore,

$V_{avg} = \frac{\Delta s}{ \Delta t}$

 

[jbriggs444]

A time weighted average is not the only possibility. A more general formula can include a weight, e.g.

$V_{avg} = \frac{\int_{t1}^{t2} \vec{V(t)} \, w(t) \, dt}{\int_{t1}^{t2} 1 \, w(t) \, dt}$

If $w(t) = 1$ then this reduces to the formula provided by RedOctober above. For a distance-weighted average, $w(t) = \frac{|ds|}{dt} = |\vec{V(t)|}$ and the result becomes

$V_{avg} = \frac{\int_{t1}^{t2} \vec{V(t)} \cdot \vec{V(t)} \, dt}{\int_{t1}^{t2} |\vec{V(t)}| \, dt}$

However, such averages do not have the very useful property that $V_{avg} = \frac{\Delta s}{ \Delta t}$, so they are not often encountered. The default meaning of "average velocity" is the time-weighted average as RedOctober showed above.

 

[Maged Saeed]

If you drive a car around a circuit, the average velocity is zero. Do you really want to say the average speed is zero?

No I didn't say that

 

[Maged Saeed]

I get it now ..
My formula is a special case where the acceleration is constant , Thanks for help

 

[nasu]

It does work when the acceleration is constant. Many students learn that formula in the context of constant acceleration, then try to apply it (incorrectly) to situations where the acceleration is not constant.

Yes, it does. I mentioned that this is not the average velocity, in general.
I should have mentioned that it works in special cases, right.

 

[robphy]

Here's an old post of mine defining average-velocity [agreeing with the previous responses involving the integral] and average-speed [answering the OP]
https://www.physicsforums.com/threads/average-speed.44394/#post-322150
You can generalize the piecewise-constant-velocity case to an integral.

and old comment on how strange it is for textbooks to
define the "average of a quantity" before defining the actual quantity.
https://www.physicsforums.com/threads/instantaneous-velocity.40569/#post-295735

Here's how I motivate "average velocity"...
With a generally varying velocity, go from point A to point B.
Now, do it with a constant velocity (i.e. speed and direction) so that you start at A when the original traveler starts and arrive at B when the original traveler arrives. That constant velocity is the [time-weighted]-average-velocity between those endpoints.

For "average speed", do the analogous thing:
Do it with a constant speed so that you start at A when the original traveler starts and arrive at B when the original traveler arrives. That constant speed is the [time-weighted]-average-speed between those endpoints.

To see a distinction between the two averages, make the original trip involve changes in direction, e.g. a two-leg piecewise-constant-velocity trip involving a change in direction.

 

[PFuser1232]

The problem is, "distance traveled" seems like a somewhat odd variable to me, mathematically. How can one represent distance traveled, using the language of calculus, as opposed to "change in distance from the origin" or the "magnitude of the displacement vector"? Does this involve inexact differentials?

 

[jbriggs444]

Total distance traveled is the integral of incremental distance travelled. It's exact, not inexact.

path length = $\int_{t0}^{t1}\sqrt{\frac{dx}{dt}^2+\frac{dy}{dt}^2+\frac{dz}{dt}^2} dt$ where the curve is parameterized by t and the normal Euclidean metric is used.

 

[sophiecentaur]

The problem is, "distance traveled" seems like a somewhat odd variable to me, mathematically. How can one represent distance traveled, using the language of calculus, as opposed to "change in distance from the origin" or the "magnitude of the displacement vector"? Does this involve inexact differentials?

If the distance traveled is the sum of the incremental steps along the path travelled, then what is the problem? It's a very useful practical quantity. The mpg in your car's fuel performance is a pretty useful thing for you to know and that involves total distance traveled - doesn't it?