In uniform acceleration, mean velocity ##\bar v = \frac{v_0+v}{2}##?

[brotherbobby]

TL;DR Summary

(I must begin by admitting that I don't know what what's TL and DR. Or DNR. I assume Relevant Equations)
1. Velocity $v = \frac{dx}{dt}$ and acceleration $a = \frac{dv}{dt}$.
2. Initial velocity $v(0) = v_0$, initial position $x(0) = x_0$ and acceleration $a(t)=a_0\;\; \forall \text{times}\;t$.

Question : For uniformly accelerated motion $a(t)=a_0\;\; \forall \text{times}\;t$, we can say that the average velocity for the entire motion $\bar v = \frac{v_0+v}{2}$, where $v(t)$ is the final velocity at some time $t$ and $v_0$ is the initial velocity. How do we show that?

Issue : We note that the final velocity $v(t) = v_0+a_0 t$. This would not be the case if the motion had varying acceleration, dependent on time, say some $a(t)$. Likewise, the average velocity $\bar v \ne \frac{v_0+v}{2}$ if the motion was not uniformly accelerated.

Note : We know that the net displacement $x-x_0 = v_0 t+\frac{1}{2}a_0 t^2$. If we can assume that the average velocity for the entire motion $\bar v = \frac{v_0+v}{2}\mathbf(?)$, we can use it to find the net displacement as above.
$$
\begin{eqnarray}
x-x_0 &=& \frac{v+v_0}{2}t\\
&=& \frac{2v_0+a_0t}{2}t\\
&=&v_0t+\frac{1}{2} at^2
\end{eqnarray}
$$

In the first line I have used the formula for average velocity and its definition $\bar v = \frac{\Delta x}{\Delta t}$. Question remains, how do we show that average velocity $\bar v = \frac{v_0+v}{2}$?

 

[PeroK]

In the first line I have used the formula for average velocity and its definition $\bar v = \frac{\Delta x}{\Delta t}$. Question remains, how do we show that average velocity $\bar v = \frac{v_0+v}{2}$?

The displacement is the signed area under a velocity vs time graph. For constant acceleration the graph is composed of a rectangle (of height $v_0$ and length $t$) and a triangle (of height $v - v_0$ and length $t$). Calculate the area and divide by the time $t$.

 

[brotherbobby]

Thank you @PeroK. Yes I know of that method; my issue was a bit different. Nevertheless, let me do as you said first.

Attempt : I use Autodesk Sketchbook$^\circledR$ for easy writing, hoping am not violating anything.

Issue : There are three ways to derive the well-known three formulas in galilean kinematics, viz. (1) $v=v_0+a_0t$, (2) $x=x_0+v_0t+1/2 a_0t^2$, and (3) $v^2 = v_0^2+2a_0(x-x_0)$. Those three ways are (a) Algebraic ?, (b) Graphs $\checkmark$, (c) Calculus $\checkmark$. I put checkmarks against the last two because I can derive formulas (1)-(3) in those ways. I am stuck at method (a), algebraic. Only because I cannot show, algebraically, how is it that the average velocity $\bar v = \frac{v_0+v}{2}$.

Note : Formula (1) follows simply enough from algebra, because the acceleration is uniform : $a_0 = \frac{v-v_0}{t}\Rightarrow \boxed{v=v_0+a_0t}$.

If I could show how $\bar v = \frac{v_0+v}{2}$ (?), I could find the formula for net displacement (2) (see my post #1 where I have done it) and with some algebra, the final velocity in terms of initial velocity and net displacement (3).

Request : Please show me how is $\bar v = \frac{v_0+v}{2}$, using methods of algebra. To be honest, I don't even know if it's possible. Or I can't see how.

 

[PeroK]

Request : Please show me how is $\bar v = \frac{v_0+v}{2}$, using methods of algebra. To be honest, I don't even know if it's possible. Or I can't see how.

$$\bar v = \frac d t = \frac{v_0t + \frac 1 2 at^2}{t} = v_0 + \frac 1 2 at= v_0 + \frac 1 2(v - v_0) = \frac {v_0 + v}{2}$$Give me a nomination for the Nobel Prize!

 

[brotherbobby]

$$\bar v = \frac d t = \frac{v_0t + \frac 1 2 at^2}{t} = v_0 + \frac 1 2 at= v_0 + \frac 1 2(v - v_0) = \frac {v_0 + v}{2}$$Give me a nomination for the Nobel Prize!

Smiles. I am afraid you are assuming that the displacement $d = v_0t+\frac{1}{2}at^2$. How do you show that, algebraically? If you read my post #3 above, I arrive at the formula for displacement using the value for the average velocity. Which is the opposite of what you have shown.

The average velocity $\bar v = \frac{v_0+v}{2}$ is more native but difficult to show (for uniform acceleration).

Is there a way to show that the average velocity $\bar v = \frac{v_0+v}{2}$ using small increments in time from start to finish, $\delta t$, over the whole interval? For example, the initial velocity is $v_0$ and the final $v$. The velocity a second after time $t=0$ is $v_0+a_0$. The velocity a second before the final velocity is $v-a_0$. If the average velocity is $\bar v = \frac{v_0+v}{2}$, then we can (easily) show it to be the case for both the first and last instants as well as any of the instances in between, with equal amounts added to and taken off the first and last instants.

Could this happen if the acceleration was non-uniform? Surely my method above, even if accepetable, cannot be considered a proof.

 

[gmax137]

@brotherbobby Can you say in words what you mean by "average velocity"?

 

[brotherbobby]

@brotherbobby Can you say in words what you mean by "average velocity"?

Average velocity is the total displacement divided by the total time taken.

I get your point. How do you find the total displacement (without finding the average velocity)?

 

[gmax137]

do you know the integral formula for the average of a continuous function (over the interval [a,b])?
$$f_{av}= \frac {1}{b-a} \int_a^b f(x)\,dx $$
in our case, the function f(x) is the velocity, and the integral of velocity is the position (replace f(x) with f(t) and dx with dt). Not sure if this helps answer your question or not.

 

[kuruman]

Average velocity is the total displacement divided by the total time taken.

I get your point. How do you find the total displacement (without finding the average velocity)?

You don't need the displacement to find the average velocity. Using the definition of average posted in #8 by @gmax137, $$v_{\text{avg}}=\frac{\int_{t_1}^{t_2}v(t)~dt}{\int_{t_1}^{t_2}dt}=\frac{\int_{t_1}^{t_2}at~dt}{t_2-t_1}=\frac{a}{2}\frac{t_2^2-t_1^2}{t_2-t_1}=\frac{at_2+at_1}{2}=\frac{v_2+v_1}{2}.$$I would consider this method algebraic because it uses algebra, not diagrams. You might object that I have used calculus as evidenced by the integrals. For this objection to be sustained, you have to come up with an non-calculus definition for finding an average that works and does not involve integration, i.e. the "area under the curve" idea. I don't think it can be done because finding the average of a function over an interval is, by definition, finding the height of a rectangle that has the same area as the area under the curve over the interval. Area under the curve means integration and integration means calculus.

 

[jack action]

[Others answered before I finished, but I began my answer before stopping for dinner time and I'm finishing it anyway! ]

Arithmetic mean:
$$\bar{v} = \frac{1}{n}\sum_{i=1}^n v_i$$
Or as a continuous function:
$$\bar{v} = \frac{1}{t_f - t_0}\int_{t_0}^{t_f} v(t)dt$$
where $v(t) = v_0 + at$.

Thus:
$$\bar{v} = \frac{v_0(t_f-t_0) + \frac{1}{2}a(t_f^2-t_0^2)}{t_f-t_0}$$
$$\bar{v} = v_0 + \frac{1}{2}a\frac{(t_f^2-t_0^2)}{t_f-t_0}$$
$$\bar{v} = v_0 + \frac{1}{2}a(t_f+t_0)$$
$$\bar{v} = v_0 + \frac{1}{2}(at_f+at_0)$$
Since $v_i = v_0 + at_i$:
$$\bar{v} = v_0 + \frac{1}{2}((v_f - v_0)+(v_0 - v_0))$$
Or:
$$\bar{v} = \frac{v_0 + v_f}{2}$$

 

[gmax137]

I began my answer before stopping for dinner time

It isn't dinner time here yet, but "honey - do's" were calling me and they win. So my #8 got shortchanged to "hope this helps" LOL.

 

[PeroK]

The average velocity $\bar v = \frac{v_0+v}{2}$ is more native but difficult to show (for uniform acceleration).

Constant acceleration means that velocity against time is a straight line. The average height of a straight line segment is the average of the heights at the two ends. That's fairly elementary.

You're claiming that's difficult to show?

 

[brotherbobby]

You don't need the displacement to find the average velocity. Using the definition of average posted in #8 by @gmax137, $$v_{\text{avg}}=\frac{\int_{t_1}^{t_2}v(t)~dt}{\int_{t_1}^{t_2}dt}=\frac{\int_{t_1}^{t_2}at~dt}{t_2-t_1}=\frac{a}{2}\frac{t_2^2-t_1^2}{t_2-t_1}=\frac{at_2+at_1}{2}=\frac{v_2+v_1}{2}.$$I would consider this method algebraic because it uses algebra, not diagrams. You might object that I have used calculus as evidenced by the integrals. For this objection to be sustained, you have to come up with an non-calculus definition for finding an average that works and does not involve integration, i.e. the "area under the curve" idea. I don't think it can be done because finding the average of a function over an interval is, by definition, finding the height of a rectangle that has the same area as the area under the curve over the interval. Area under the curve means integration and integration means calculus.

Thank you. Yes the matter's cleared. Though I would reply with a few of my ideas presently, you're correct that when you speak of continuously varying quantities like $v(t)$, the use of calculus is inevitable in calculating averages.

 

[PeroK]

Thank you. Yes the matter's cleared. Though I would reply with a few of my ideas presently, you're correct that when you speak of continuously varying quantities like $v(t)$, the use of calculus is inevitable in calculating averages.

Calculus is not needed when the motion is uniform acceleration.

 

[vanhees71]

But as you see in this thread, physics without calculus makes the subject much more difficult! It's a fairy tale to claim that physics without calculus is easier than with it!

 

[brotherbobby]

Constant acceleration means that velocity against time is a straight line. The average height of a straight line segment is the average of the heights at the two ends. That's fairly elementary.

You're claiming that's difficult to show?

I was claiming that it's difficult to show if you don't use graph-analogy and restrict yourself only to algebra. As @kuruman's post #9 above shows, you really cannot restrict yourself to algebra when you're speaking of continuously varying quantities. The use of calculus is inevitable.

But this is what I had in mind.

The velocities in uniform acceleration form an arithmetic series. So, say after every one second interval, we have

$$[v_0], [v_0+a], [v_0+2a,] \dots, [v_0+(n-1) a]\;\;\; [= v]$$

Here I have taken starting time $t_0=0$ and final time $t_f = t$.

We will note that the average of all the $n$ quantities is the same as the arithmetic mean of the first and the last quantitiy. Thus, we have $$\bar v = \sum_{i=1}^{i=n} v_i = \frac{v_0+v}{2}$$

I realise this is not a proof. I have taken discrete values after every 1s intervals. The proof of this lies in posts # 9 and 10 above.

Thank you all for the trouble in clearing this little matter up.

 

[PeroK]

But as you see in this thread, physics without calculus makes the subject much more difficult! It's a fairy tale to claim that physics without calculus is easier than with it!

No one is claiming that. But, delaying the study of motion with uniform acceleration until the student has studied the integral calculus is unnecessary, IMO.

 

[PeroK]

I was claiming that it's difficult to show if you don't use graph-analogy and restrict yourself only to algebra. As @kuruman's post #9 above shows, you really cannot restrict yourself to algebra when you're speaking of continuously varying quantities. The use of calculus is inevitable.

This is wrong. You don't need calculus for uniform acceleration.

Saying that velocity is a straight line is not a "graph analogy". It's a fundamental property of the motion. Ironically, you are talking about the physical motion of a particle, which is as geometric as it gets!

 

[brotherbobby]

No one is claiming that. But, delaying the study of motion with uniform acceleration until the student has studied the integral calculus is unnecessary, IMO.

Fair enough. But then you have to resort to the graph of an uniformly acceleration object, v versus t, which is a straight line.
Without the aid of graphs, how would you find out the average, purely using algebra?

 

[PeroK]

Without the aid of graphs, how would you find out the average, purely using algebra?

I don't draw a fundamental distinction between algebra and geometry when it comes to functions of a single variable.

How do you justify your algebraic model in the first place without this mapping between motion in space and functions? Displacement is a vector, which is a geometric concept to begin with. At least in this context.

If you disallow geometry, what is the motion of a particle? What is that physically?

 

[vanhees71]

Even the didactics experts now agree that mechanics should be taught from the very beginning in terms of vectors. It's of course a challenge to teach it without calculus, but it's doable to a certain extent, and it's a great motivation to (heuristically!) introduce calculus in the first place. Particularly motion at constant acceleration is well suited for this. The only problem is to avoid calculus at all costs. It's not by accident that Newton discovered calculus when dealing with the foundations of mechanics. In fact it was the big breakthrough, leading to the development of physics in the sense we know it today.

 

[vela]

Assume $a>0$ for simplicity. Then $v_{\rm min} = v_0$ and $v_{\rm max} = v_0 + at$, so $\Delta x_{\rm min} = v_0 t$ and $\Delta x_{\rm max} = v_0 t + a t^2$. Then we can say that $\Delta x_{\rm actual}$ is somewhere in between, so $\Delta x_{\rm actual} = v_0 t + k a t^2$ for some $0 \le k \le 1$. Then show that $$v(t) = \lim_{h \to 0} \frac{x(t+h)-x(t)}{h} = v_0 + 2kat$$ and compare this result to $v=v_0 + at$ to deduce that $k=1/2$. Then use what @PeroK posted in post #4.

 

[malawi_glenn]

I must begin by admitting that I don't know what what's TL and DR. Or DNR.

Too Long Didn't read

Anyway for any function y = f(x) = kx + m the mean y-value between x₁ and x₂ is (f(x₁) + f(x₂))/2
BUT we have to be careful, what do we MEAN by the mean of a function? Well, it is the integral formula that was given in #8 which is a generalization of what I wrote above. It means that the area of the parallellogram and the rectangle with the same base x₂ - x₁ should be the same. In the case of any continous function, you can figure out the correspondence there.

 

[PeroK]

One final thought. The development of the Riemann Integral that I have seen uses the known area of rectangles and triangles - in order to justify the convergence of the upper and lower sums of areas etc.

The position that the area of a rectangle or triangle is actually defined by an integral is, therefore, not totally satisfactory.