Kinematic equations ##\textbf{purely}## from graphs

[brotherbobby]

TL;DR

Derive the three well known equations of kinematics for uniform acceleration $a_0$ using graphs only.

1. The first equation between velocity $v$ and time $t$ can be derived using the graph I have drawn for the purpose as shown on the right. Since acceleration $a_0$ is a constant, the graph of $v-t$ is a straight line. The slope of the line is $\dfrac{v-v_0}{t} = a_0\Rightarrow \boxed{v = v_0+a_0t}$ $\checkmark$.

2. The second equation between displacement $(x-x_0)$ and time $t$ can be derived by finding the area under the same graph. The area is shown shaded in $\mathbf{\color{orange}{orange}}$. The area is in the shape of a trapezium whose area is given by the formula $\left(\dfrac{1}{2}\text{Sum of parallel sides}\times\text{height}\right)$. The area represents the displacement $\small{x-x_0 =\dfrac{1}{2}(v_0+v)t = \dfrac{1}{2}(v_0+v_0+a_0t)t = v_0t+\dfrac{1}{2}at^2}$ which becomes the second kinematical equation $\boxed{x=x_0+v_0t+\dfrac{1}{2}a_0t^2}\quad\checkmark$.

3. Now for the last equation, which represents the change in velocity $\Delta v$ in terms of the displacement $\Delta x$. This is where am stuck. How do we draw a graph between $v-x$ without knowing the equation in advance? Can this third equation, well known to be $\boxed{v^2=v_0^2+2a_0(x-x_0)}$ be even derived purely using graphs?A hint or help deriving the last of the three equations using graphs would be very welcome.

 

[Orodruin]

Just combine results 1 and 2?

 

[Hill]

Take the area like in #2, but instead of substituting for $v$ substitute for $t$.

 

[PeroK]

3. Now for the last equation, which represents the change in velocity $\Delta v$ in terms of the displacement $\Delta x$. This is where am stuck. How do we draw a graph between $v-x$ without knowing the equation in advance? Can this third equation, well known to be $\boxed{v^2=v_0^2+2a_0(x-x_0)}$ be even derived purely using graphs?

Probably not purely using graphs. Note that velocity and acceleration are the first and second time derivatives of position, which is a function of time. And that time ($t$) is a parameter. This ensures simple position against time and velocity against time graphs.

There is no simple way, in general, to define velocity as a function of displacement without doing the necessary algebra. And, in any case, in general you may have a multi-valued function: two different velocities at the same displacement (at different times). Unless you restrict yourself to the case where displacement is only increasing, say.

In fact, it's simpler to express displacement as a function of velocity. But, in that case, you have derived the required algebraic relationship in order to draw the graph in the first place.

 

[brotherbobby]

Take the area like in #2, but instead of substituting for $v$ substitute for $t$.

I did not understand - could you elucidate?

 

[brotherbobby]

Just combine results 1 and 2?

I suppose that method should do, because both (1) and (2) were derived from graphs. However, it resorts to the use of algebra than a purely graphical solution.

 

[Hill]

I did not understand - could you elucidate?

In #2, you've calculated the area, $x-x_0 =\frac{1}{2}(v_0+v)t$. Then, you've used $\dfrac{v-v_0}{t} = a_0$ from #1 to replace $v$ in this expression.
If instead of replacing $v$, you use the same $\dfrac{v-v_0}{t} = a_0$ from #1 to replace $t$ in the same area expression of #2, you get the expression you're looking for in #3.

 

[EcSolticia]

Thinking in reverse, deriving the last equation, $v^2 - v_0^2 = 2a_0(x - x_0)$, purely graphically would require finding a way to represent the squares of initial and final velocity values and doing the same for the product $2a_0(x - x_0)$, to find a relation between them. The product $a_0(x - x_0)$ in particular being there entails that the graph would have to represent displacement and acceleration in a way both can be multiplied together. An additional requirement is that there are two time derivatives of position to consider (velocity and acceleration) as well as position itself in the same graph.

Assuming that the graph is two-dimensional, there then are ${3 \choose 2} = 3$ possible graphs (which feature two of the derivatives) with respect to the pair of variables each axis represents:

1. $v$ and $a$
2. $v$ and $x$
3. $x$ and $a$

In representation 1, areas are $v*a$, which does not appear particularly useful in this context. In representation 2, areas are $v*x$, which shares a similar fate as 1. With representation 3, it appears that areas are $x*a$, which allows us to represent the product $a_0(x - x_0)$ as an area in the graph. However, acceleration is constant, so if acceleration is the dependent variable, then the graph is a boring horizontal line. It contains little meaningful information about the relationship of $x$ and $a$ to $v$.
Additionally, even if we can represent the product of acceleration and displacement change in a way that connects v and a*x, how do we relate that to this product of velocities? $v^2 - v_0^2 = (v - v_0)(v + v_0)$? $a_0(x - x_0)$ is an area, so, to show equivalence, we would want $v^2 - v_0^2$ to also be an area in the same graph as the product of displacement and acceleration. However, the velocity-including product is an area only in a graph where the independent and dependent variable both are velocities, essentially yielding an useless line with a slope of 1.

My analysis is not comprehensive in the sense that I did not explore how the problem could be solved by getting more creative. For example, what if we use 3D graphs? However, for two dimensions, it does not seem like a feasible goal.

 

[brotherbobby]

In #2, you've calculated the area, $x-x_0 =\frac{1}{2}(v_0+v)t$. Then, you've used $\dfrac{v-v_0}{t} = a_0$ from #1 to replace $v$ in this expression.
If instead of replacing $v$, you use the same $\dfrac{v-v_0}{t} = a_0$ from #1 to replace $t$ in the same area expression of #2, you get the expression you're looking for in #3.

Sure, but that would be algebraic. You are suggesting the same as did @Orodruin in post #2.
I am looking to derive the expression involving $\Delta x$ and $\Delta v$ purely from graph, without resorting to expressions obtained earlier from graphs. It seems that it is not possible to do.

 

[Ibix]

Sure, but that would be algebraic.

Well yeah, but as @Hill points out, it's algebra you already used yourself.

 

[brotherbobby]

Well yeah, but as @Hill points out, it's algebra you already used yourself.

Yes, which makes my finding the second equation $\Delta x = v_0t+\frac{1}{2}at^2$ also suspect. I did use algebra from (1) in (2), so (2) isn't purely graphical either.
Thanks.

 

[Orodruin]

No algebraic expression is purely graphical …

 

[Hill]

Yes, which makes my finding the second equation $\Delta x = v_0t+\frac{1}{2}at^2$ also suspect. I did use algebra from (1) in (2), so (2) isn't purely graphical either.
Thanks.

No algebraic expression is purely graphical …

... (1) isn't purely graphical either: to go from $\dfrac{v-v_0}{t} = a_0$ to $v = v_0+a_0t$ you did use algebra.

 

[brotherbobby]

... (1) isn't purely graphical either: to go from $\dfrac{v-v_0}{t} = a_0$ to $v = v_0+a_0t$ you did use algebra.

I admit. I suppose it's impossible to "see" the equations coming out naturally from graphs. However while (1) and (2) can be argued for, as slopes and areas of the $v-t$ curve, (3) has no graphical equivalent at all.

 

[Hill]

I admit. I suppose it's impossible to "see" the equations coming out naturally from graphs. However while (1) and (2) can be argued for, as slopes and areas of the $v-t$ curve, (3) has no graphical equivalent at all.

I struggle to see the difference. Just repeating the same argument as in (2):

The area represents the displacement $\small{x-x_0 =\dfrac{1}{2}(v_0+v)t = \dfrac{1}{2}(v_0+v)\dfrac {v-v_0} {a_0} = \dfrac{v^2-v_0^2}{2a_0}}$ which becomes the third kinematical equation
$\boxed{v^2=v_0^2+2a_0(x-x_0)}$

 

[EcSolticia]

I struggle to see the difference. Just repeating the same argument as in (2):

The area represents the displacement $\small{x-x_0 =\dfrac{1}{2}(v_0+v)t = \dfrac{1}{2}(v_0+v)\dfrac {v-v_0} {a_0} = \dfrac{v^2-v_0^2}{2a_0}}$ which becomes the third kinematical equation
$\boxed{v^2=v_0^2+2a_0(x-x_0)}$

In the derivation of (1) as shown by the OP as well as in (2), there appears to be a visible direct correspondence between the components of the equations and the graphs. For example, $\frac{v - v_0}{t}$ is the magnitude of the slope. Additionally, whilst the formula for the area of a trapezium was used in (2), it still exhibits a correspondence to the graph presented. However, whilst the proposed derivation of (3) can be done following the prior two derivations just algebraically, it does not appear possible to represent the equation completely as a graph in the sense of each component corresponding to some geometric figure and they having certain fixed relations to each other, in this case an equality. The latter appears to be what OP is asking for (correct me if wrong), the question might be interpreted as: "Is there a graph such that equation (3) can be independently derived from it by representing relationships between present geometric figures as mathematical expressions and equations? How can I find it?"
The focus seems to be on representations and utilization of equivalences between relations between geometric figures and mathematical expressions directly, and in the same graph.
Then the proposed solution fails to satisfy this criterion, since it uses prior results and does not use a graph that represents the equations geometrically. In this case, the equation $\frac{v - v_0}{a_0} = t$ was used, however, no correspondence between itself and any graph was established. It seems that $\frac{v - v_0}{a_0} = t \iff \frac{v - v_0}{t} = a_0$ (assuming $v - v_0 \neq 0$), so a solution would be possible (and #8 would have to contain at least one false premise or logical error) if the equation of form $A/B = C \iff A/C = B$ can be shown to be true (where all of them are nonzero) "purely graphically" for A, B and C being reals. However, in this case $A/B$ represents the slope magnitude, with $A$ and $B$ being certain lengths in the graph, and $C$ the same value as the slope but not exhibiting any corresponding geometric figure in itself in the graph. Switching from $A/B = C$ to $A/C = B$ like this would entail switching from treating a slope magnitude value in the same graph as a line, which goes beyond the constraints of the "purely graphical" approach the OP appears to be seeking.

It does not appear that there is such a solution for a 2 dimensional graph to begin with, as shown in #8, assuming the premises I used model OP's own requirements of a "purely graphical" solution accurately.

 

[brotherbobby]

However, whilst the proposed derivation of (3) can be done following the prior two derivations just algebraically, it does not appear possible to represent the equation completely as a graph in the sense of each component corresponding to some geometric figure and they having certain fixed relations to each other, in this case an equality. The latter appears to be what OP is asking for (correct me if wrong), the question might be interpreted as: "Is there a graph such that equation (3) can be independently derived from it by representing relationships between present geometric figures as mathematical expressions and equations? How can I find it?"

Yes, that is correct. I suppose we should know how to derive the three equations using (a) algebra, (2) calculus and (3) graphs. Of course, as we have seen, each method will involve a little use of the others, so in that sense nothing is pure. However, the first method [algebra] is pure in the sense it's the most elementary.

As an exercise, try deriving equation $\color{blue}{\mathrm{III}\rightarrow \boxed{v^2=v^2_0+2a_0(x-x_0)}}$ using purely algebra.

It's straightforward if you use equation $\color{green}{\mathrm{II}\rightarrow {\boxed{x=x_0+v_0t+\dfrac{1}{2}a_0t^2}}}$ and equation $\color{green}{\mathrm{I}\rightarrow \boxed{v=v_0+a_0t}}$, but far from trivial if you start from first principles using only concepts of displacement, velocity and acceleration.

 

[sophiecentaur]

In the derivation of (1) as shown by the OP as well as in (2), there appears to be a visible direct correspondence between the components of the equations and the graphs

It doesn't surprise me that there is a "correspondence" between the results of the two perfectly valid branches of Maths. We all happily use numerical methods in preference to analytical methods because computers allow us to get the answer quicker than slaving over a page of symbols. And, of course, many times we just cannot get our answers by analytical methods. This isn't a matter of Physics.

 

[nasu]

Yes, that is correct. I suppose we should know how to derive the three equations using (a) algebra, (2) calculus and (3) graphs. Of course, as we have seen, each method will involve a little use of the others, so in that sense nothing is pure. However, the first method [algebra] is pure in the sense it's the most elementary.

You cannot derive the equations of motion by algebra. At least, not if you use instantaneous velocity. Even by using graphical methods, you are using calculus.

 

[Hill]

You cannot derive the equations of motion by algebra. At least, not if you use instantaneous velocity. Even by using graphical methods, you are using calculus.

Unless you are Galileo :

Galileo ... wrote that the distance of the fall was proportional to the square of the fall duration, which, if you think of it, is a mind-blowing discovery.

...

This famous equation was found by Galileo Galilei in the first half of the seventeeth century, before the invention of calculus. In fact, it is one of the equations that triggered the invention of calculus by Newton and Leibniz.

(Susskind, Leonard; Cabannes, André. General Relativity: The Theoretical Minimum.)

 

[nasu]

Using graphical methods means using calculus even if you don't formally call it so. Archimedes also used "calculus" methods to derive some of his results. Calculus is not just a bunch a formulas but a way of thinking. About limits, mainly.

 

[brotherbobby]

You cannot derive the equations of motion by algebra. At least, not if you use instantaneous velocity. Even by using graphical methods, you are using calculus.

Both in response to your (@nasu ) quote above and to @sophiecentaur, let me derive the three well-known equations of kinematics using only algebra. But first let's have our problem situation.

A body moves with a uniform acceleration $a_0$ along the $x-$axis, having a velocity $v_0$ at a position $x_0$ when we start the clock. It accelerates to a velocity $v$ at a position $x$ at a later time $t$.

$\mathrm{I.}\quad$ Average acceleration for the body is defined $\bar{a}= \frac{\Delta v}{\Delta t}$. Since the acceleration is a constant, $\bar{a}=a_0$. Between the given points of motion, we can write $a_0 = \dfrac{v-v_0}{t}\Rightarrow \boxed{v = v_0+a_0t}\quad\color{blue}{(1)}$.

$\mathrm{II.}\quad$ In order to find the displacement of the body, let us divide its velocity over intervals as small as we please, say $\Delta t$. The velocities over this interval would look like something below :
$$v_0,v_0+a_0\Delta t, v_0+2a_0\Delta t, \dots, v_0+(n-1)a_0\Delta t$$
Since the acceleration $a_0$ is a constant, the terms of the above series form an A.P. We know that the average of all terms of an A.P. is equal to the that of the first and the last terms; hence $\bar{v}=\dfrac{v_0+v}{2}$.
The net displacement $\Delta x = \bar{v}t=\dfrac{v_0+v}{2}t=\dfrac{2v_0+a_0t}{2}t=v_0t+\dfrac{1}{2}a_0t^2\Rightarrow\boxed{x=x_0+v_0t+\dfrac{1}{2}a_0t^2}\quad\color{blue}{(2)}$

$\mathrm{III.}\quad$ Squaring $\mathrm{I}$, we have $v^2=v_0^2+2a_0v_0t+a_0^2t^2 = v_0^2+2a_0\left(v_0t+\dfrac{1}{2}a_0t^2\right)=v_0^2+2a_0(x-x_0)$ upon using (2). We thus have our third equation : $\boxed{v^2=v_0^2+2a_0(x-x_0)}\quad\color{blue}{(3)}$

I fail to see where have I used the ideas of modern calculus. I am willing to be corrected on that score, however.

 

[nasu]

You call it "just algebra" but the name does not change anything. You are using implicit assumptions which are not algebra. How do you define the quantites $v_0$ and $v$ for example? "Without calculus" you can only define an average velocity but not an instantaneous velocity. Assuming instanatneous velocities you assume an operation of taking the limit and infinitesimal quantities.
You assume that for constant acceleration the average velocity is equal with the arithmetic mean of the velocities but this is a result of calculus. The fact that the area under the graph of velocity is equal to the displacement is a result of calulus too. So don't say that it can be obtained "by graphs" and not by calculus.

Same things applies to instantaneous acceleration. These methods (as in your posts) are used in algebra based courses to justify or provide an intuition for the results that can be rigourously proven by the methods (or ways of thinking) included in calculus.

 

[Hill]

the average of all terms of an A.P. is equal to the that of the first and the last terms

The last term of this A.P. is not equal to $v$, but rather to $v-a_0 \Delta t$. So, the average is not equal to $\frac {v_0+v} 2$, but rather to $\frac {v_0+v-a_0 \Delta t} 2$. Isn't it? Looks like you have to take limit then. Which is not algebra.

 

[sophiecentaur]

I fail to see where have I used the ideas of modern calculus.

Here:

Average acceleration for the body is defined ä=Δv/Δt.

Strongly implied here but glossed over. But who really cares? When we did SUVAT at school (not by that actual modern name in 1962) the need for constant acceleration was not stressed enough and I, being a practical thinker, couldn't reconcile it with motor cars.

 

[sophiecentaur]

This all brought o mind the school investigation with a ticker timer and tape. The IOP have a page which makes my point. The (rather poor) result of one run of the timer are presented as a graph

of distance against time. It includes the shole journey from start to stop and the area is the total length of tape (total distance). It's totally graphical.
The best of my student groups did a lot better - but maybe they were older.

 

[brotherbobby]

You call it "just algebra" but the name does not change anything. You are using implicit assumptions which are not algebra. How do you define the quantites v0 and v for example? "Without calculus" you can only define an average velocity but not an instantaneous velocity. Assuming instanatneous velocities you assume an operation of taking the limit and infinitesimal quantities.

Yes I agree with you entirely.

Same things applies to instantaneous acceleration. These methods (as in your posts) are used in algebra based courses to justify or provide an intuition for the results that can be rigourously proven by the methods (or ways of thinking) included in calculus.

True; however, what you prefer depends strongly on where you're coming from. I work in physics teaching where intuition is the starting point, unless for those (famous) cases where intuition fails. The whole idea is to simplify a subject for the time being in order to give the student an introduction to it. That happens in secondary (junior high) school. Later on, in sernior high school, the student is taught the methods of calculus for a thorough understanding of the subject.

The last term of this A.P. is not equal to $v$, but rather to $v-a_0 \Delta t$. So, the average is not equal to $\frac {v_0+v} 2$, but rather to $\frac {v_0+v-a_0 \Delta t} 2$. Isn't it? Looks like you have to take limit then. Which is not algebra.

A question for @Hill : could you explain the above please? The body moves from a velocity $v_0$ to a velocity $v$ in a time $t$ with uniform acceleration $a_0$. I plot the velocities of the body as under:
$$v_0,v_0+a_0\Delta t, v_0+2a_0\Delta t, \dots, v_0+(n-1)a_0\Delta t$$
The last term is the final speed $v$, isn't it? Which would imply that there are a total of $n$ terms in the sequence.
Where am I going wrong? Many thanks.

 

[Hill]

The last term is the final speed v, isn't it?

I don't think so. Take an example, say, 3 seconds. At $t=0$ the speed is $v_0$, and at $t=3$, the speed is $v$. So, $v=v_0+3a_0$.
Now, divide it into 3 intervals of 1 second: $\Delta t=1$. The A.P. is : $$v_0, v_0+a_0, v_0+2 a_0$$ The last term is not the final speed $v$.

 

[Lnewqban]

3. Now for the last equation, which represents the change in velocity $\Delta v$ in terms of the displacement $\Delta x$. This is where am stuck. How do we draw a graph between $v-x$ without knowing the equation in advance? Can this third equation, well known to be $\boxed{v^2=v_0^2+2a_0(x-x_0)}$ be even derived purely using graphs?A hint or help deriving the last of the three equations using graphs would be very welcome.

Could this work?

 

[sophiecentaur]

I work in physics teaching where intuition is the starting point,

Wow! And whose intuition would that be? Take a class of twenty students and you could find twenty different intuitions at the start of the process. Or would it be a voting procedure?