$$s_ extrmaverage = frac extrmtotal distance traveled extrmtotal time needed$$
which usually speaking is no equal to the magnitude of the equivalent average velocity.
What historical, technical or didactic factors are there to define average speed this means instead of as the magnitude of mean velocity?
kinematics velocity vectors speed
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edited Sep 2 "18 in ~ 20:47
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People currently answered your inquiry from a usefulness standpoint, but I just want to include that your thinking isn"t correct:
Speed is usually identified as the size of (instantaneous) velocity. For this reason one can assume that mean speed would certainly be characterized as the size of average velocity.
You are watching: Is average speed the magnitude of average velocity
That"s not how it works. If us have
then logic dictates the we must have
and, indeed, this is what we have.
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reply Sep 3 "18 in ~ 6:44
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Given a velocity as a role of time, the rate as a role of time is the magnitude of the velocity in ~ each point in time. The mean speed is then the median of this magnitude, together it would be for any function of time - together as density or temperature. The inquiry of even if it is the typical magnitude that the velocity is equal to the size of the typical of the velocity then becomes a conjecture to check. Since an item can move roughly at high rate while returning to the exact same place, and so have actually an zero mean velocity with a high typical speed, this shows by counter example that the average speed, defined just like any type of other average, is actually not equal come the size of the median velocity.
It is very common to find that the typical of some duty is not the duty of the average. For example, the typical $x^2$ is not frequently the square the the mean of $x$.
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answered Sep 2 "18 in ~ 20:39
Ponder StibbonsPonder Stibbons
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You deserve to compute the median value of the velocity vector.
However, it turns out to be useless sometimes. A trivial example is a one motion.
The average velocity the a complete loop in a circular movement is $vec0$, together velocity is pointing in one direction in ~ first, and also $pi$ radians later it"s pointing in opposing one, so your contributions cancel out. So the median "velocity" is $vec0$.
Nevertheless, this is not giving us lot information. In contrast, the proportion of "circumference" to "time elapsed" provides us the yes, really "mean speed".
Sometimes, anyways, it deserve to be helpful to provide them both. The more information, the better.
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edited Sep 3 "18 in ~ 4:02
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reply Sep 2 "18 at 20:35
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The an easy reason is that velocity have the right to go negative, and also this will influence the average. The clearest example of the distinction would be a pendulum (or any other resonating system).
A pendulum swings backwards and forwards. It complies with its track in one direction, accelerating, climate decelerating to a momentary stop; and then reverses direction come repeat the precise same trajectory in reverse.
The pendulum adheres to the same course each time, in the contrary directions. Because velocity is signed, the typical velocity because that going one way is specifically equal in magnitude to and the opposite sign to the median velocity walk the various other way. The mean velocity is therefore zero.
The mean speed of course will certainly not be zero. It will certainly be same to the size of the median velocity because that one half of the pendulum"s trajectory, since both halves have the same magnitude that velocity.
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reply Sep 2 "18 in ~ 23:31
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There error is the magnitude of velocity is not in reality a definition, however is better appreciated as a consequence that the yes, really definition, and any products that say it is space problematic research materials.
The actual definition of rate in both cases yes, really is street traveled end the time taken to travel it. It"s simply that in the situation of instantaneous speed, the absolute value of velocity wake up to be equal to the speed since the size of the differential that arc size ($ds$) - i.e. The small increment the distance travel - is the very same as the magnitude of the differential that displacement ($dmathbfr$), i.e. The vector from beginning to existing position. Mathematically, the correct definition of instantaneous rate is
$$mathrmspeed = left| fracdsdt ight|$$
$$mathrmvelocity = fracdmathbfrdt$$
However currently (for 2 dimensions in ~ least) we have that $ds = sqrtdx^2 + dy^2$, but also $dmathbfr = dx mathbfi + dy mathbfj$. What is $|dmathbfr|$?