Deriving the "third" kinematic equation using graphs

[brotherbobby]

TL;DR

The three equations describing one-dimensional kinematics are well known from high school. The third connects velocity to displacement : ${\boldsymbol{(3)}}\;\boxed{v^2(x)=v^2_0+2a_0(x-x_0)}$, where $v_0,\, x_0$ are the velocity and position of the body with respect to the origin at time $t=0$ and $a_0$ is the (constant) acceleration. (3) can be derived using both algebra and calculus. Question is - how to derive it using graph? (!)

Problem Statement : For a particle moving in one dimension along the $x$ axis with constant acceleration $a_0$ (see figure alongside), the three well known equations of high school kinematics are $(1)\; v_(t)=v_0+a_0t\quad (2)\; x(t)=x_0+v_0t+\dfrac{1}{2}a_0t^2\quad \text{and}\quad\boxed{(3)\; v^2(x) = v^2_0+2a_0(x-x_0)}$. These equations can be derived using the methods of (1) algebra and (2) calculus - and I have been able to do it [I would be happy to provide the details if asked]. Additionally, there's a third method, the method of graphs. I have been able to derive (1) and (2) using the graph depicting $v(t)-t$, its slope and the area under the straight line lead to (1) and (2). [I would be happy to provide the details if asked].

I am stuck with (3). Knowing it already (or beforehand), I can see that the graph of $v(x)\;\text{versus}\; x$ will be a parabola that "curves" with a decreasing slope $-$ an attempt of which I have made in the graph alongside. But that doesn't help. It is "how" to derive the equation using graphs where I am at a loss. More specifically, which graph to use? It cannot be the one I have shown, for it was sketched using the equation that I seek to derive. I am left with $v\;\text{versus}\; t$ graph which is a straight line, or the $x\;\text{versus}\; t$ graph which is parabolic. For reference, I put both graphs to the left.

Can either of the two graphs above be used to derive (3), namely : $v^2(x)=v^2_0+2a_0(x-x_0)$?

Many thanks.

 

[Ibix]

I suspect all the same questions from your previous similar thread will arise here too. Basically, what do you regard as a "using a graph"? Because some algebra is inevitable.

 

[brotherbobby]

Yes, am afraid I entirely forgot about that thread.
True, you'd need some algebra but that can be readily done with equations (1) and (2). With (3), none seems to be possible.

 

[kuruman]

As @Ibix remarked, some algebra is inevitable when you translate a graphic form into an algebraic form. Here is the simplest way I can think of to do this using a velocity vs. time plot (see figure on the right).

Total distance $x$ traveled in time $t$ is "the area under the curve", in this case the area $A_1$ of the red triangle plus the area $A_2$ of the yellow rectangle.
Note that the acceleration $a$ is the slope, $$a=\dfrac{\text{rise}}{\text{run}}=\dfrac{(v-v_0)}{t} \tag{1}.$$ The areas are $A_1=\frac{1}{2}\left({v-v_0}\right)t~$ and $~A_2=v_0t~$ so that $$x=A_1+A_2=\frac{1}{2}(v+v_0)t. \tag{2}$$Multiply equations (1) and (2) to get $$ax=\frac{(v-v_0)}{t}\times\frac{1}{2}(v+v_0)t \implies 2ax=v^2-v_0^2 $$ QED.

 

[Dale]

which graph to use?

Obviously, you should use a graph of $v^2$ vs $x$. That will give you the point-slope form of a straight line $(y-y_0)=m(x-x_0)$ where $y=v^2$, $m=2a$, and $x=x$. You can just directly read it off the graph.

 

[Herman Trivilino]

Basically, what do you regard as a "using a graph"? Because some algebra is inevitable.

I can't speak for the OP and I don't know if you're referring to this equation in particular or in general. But this is what I would mean for the equation $v=v_o +at$.

I would start with students performing guided activities in front of an ultrasonic motion detector or the like, connected to a computer that displays graphs of $v$ versus $t$.

Then I would draw a $v$ versus $t$ graph on the board for the case of constant acceleration. From there the intercept is $v_o$ and the slope is $a$ so the equation of the graph must be
$v=v_o +at$.

Can such a thing be done for a $v$ versus $x$ graph?

 

[kuruman]

Obviously, you should use a graph of $v^2$ vs $x$. That will give you the point-slope form of a straight line $(y-y_0)=m(x-x_0)$ where $y=v^2$, $m=2a$, and $x=x$. You can just directly read it off the graph.

I think that this approach requires collection of experimental values for $v$ and $x$, however . . .

Question is - how to derive it using graph? (!)

Can a fit to the data be considered a "derivation"?

 

[Dale]

I think that this approach requires collection of experimental values for v and x, however

I had assumed that all the graphs were obtained from experimental values.

 

[A.T.]

Yes, am afraid I entirely forgot about that thread.
True, you'd need some algebra but that can be readily done with equations (1) and (2). With (3), none seems to be possible.

It's not clear what amount of algebra is acceptable and what graphs can be used.

 

[kuruman]

I had assumed that all the graphs were obtained from experimental values.

The opening statement of the problem declares "The three equations describing one-dimensional kinematics are well known from high school." Based on my high school experience, these equations are usually given as "that's the way it is" instead of solving the ODE $~\ddot x = \text{const.}~$ which constitutes a proper derivation.

A justification (derivation?) of $v=v_0+at$ without using calculus can be obtained as follows.
Consider an object starting from rest under constant acceleration $a$. By the definition of the acceleration, its velocity $v$ at any time $t$ would be the area under the $a$ vs. $t$ curve, namely $~v=at.$ Now if this object is on a train moving with constant velocity $v_0$ relative to an observer O at rest by the tracks, the object's velocity at any time $t$ relative to O would be the sum $$v=v_0+ at. \tag{1}$$For the second equation, one would argue that the distance traveled, relative to an observer on the train, is the area of the triangle, $~A_1=\frac{1}{2}at^2.$ Relative to the ground observer O, one would have to add to that the distance traveled by the train at constant velocity $A_2=v_0t$ in which case $$x=v_0t+\frac{1}{2}at^2. \tag{2}$$ Thus, the first two equations are "derived" using graphical considerations, namely the idea of "area under the curve" without recourse to experimental values.

OP's question is, I think, whether the third equation can be obtained by similar considerations. Algebraically, we all know that one needs to solve for time in equation (1) and substitute in equation (2). This is essentially what I did combining "areas under the curve" in post #4 in what is probably a faux derivation. A real derivation would have to use simple graphic ideas to arrive at an expression for $v^2(x)$ as you suggested in post #5. It would require establishing that $~adx=vdv~$ using simple arguments and without calculus. I think that is the gist of OP's query. It might be done, but I don't see how right now.

 

[Dale]

I don't see what the difficulty is. If you can know and plot $v$ then you can know and plot $v^2$. It isn't even algebra, it is just arithmetic.

I assumed they were all experimental, but it doesn't matter. Whatever means gives you $v$ for one plot necessarily also gives you $v^2$ for this plot.

 

[kuruman]

I don't see what the difficulty is. If you can know and plot v then you can know and plot $v^2$. It isn't even algebra, it is just arithmetic.

That's my point which, I assume, is also OP's query. Is there a way to know directly and plot $v^2(x)$ independently of knowing $v(t)$ and/or doing arithmetic that relies on the previous two equations?

 

[Dale]

Is there a way to know directly and plot $v^2(x)$ independently of knowing $v(t)$

If you know $v$ then you know $v^2$, but the reverse is not always true. It is mathematically possible to know $v^2$ without knowing $v$ since $v^2$ lacks the information about direction.

 

[A.T.]

That's my point which, I assume, is also OP's query. Is there a way to know directly and plot $v^2(x)$ independently of knowing $v(t)$ and/or doing arithmetic that relies on the previous two equations?

Yes, by multipling $KE = KE_0 + Fs$ with $2/m$.

But seriously, since eq (3) can be easily derived from eq (1) and eq (2), what is the point of deriving it while avoiding them? I don't even remember learing eq (3) in school as "one of the three equations describing one-dimensional kinematics", because it is so easily defived from from eq (1) and eq (2).

 

[kuruman]

Yes, by multipling $KE=KE_0+Fs$ with $2/m$.

Isn't this a circular argument? How does one get this expression to begin with? The work-energy theorem is a rewritten "third equation" once Newton's second law is established and comes after kinematics in the standard curriculum.

But seriously, since eq (3) can be easily derived from eq (1) and eq (2), what is the point of deriving it while avoiding them? I don't even remember learing eq (3) in school as "one of the three equations describing one-dimensional kinematics", because it is so easily defived from from eq (1) and eq (2).

The point is one of idle curiosity. The question is not to do it, but whether it can be done by simple geometric considerations like the other two equations.

 

[Dale]

The question is not to do it, but whether it can be done by simple geometric considerations like the other two equations.

And the answer is very clearly yes.

Like the other two equations means you can pick which quantities to plot. So plot $v^2$ and $x$. The OP clearly considers $v$ to be known, so then $v^2$ is necessarily also known.

 

[kuruman]

And the answer is very clearly yes.

Like the other two equations means you can pick which quantities to plot. So plot $v^2$ and $x$. The OP clearly considers $v$ to be known, so then $v^2$ is necessarily also known.

So when OP writes

It is "how" to derive the equation using graphs where I am at a loss. More specifically, which graph to use? It cannot be the one I have shown, for it was sketched using the equation that I seek to derive.

what does that mean? To me it says that OP rejects plotting $v^2$ vs. $x$ and is seeking an alternate way.

 

[Dale]

So when OP writes

what does that mean? To me it says that OP rejects plotting $v^2$ vs. $x$ and is seeking an alternate way.

No, he says that it cannot be the one he showed ($v$ vs $t$). He didn't show $v^2$. So all he needs is that pointer: use $v^2$.

 

[pbuk]

So all he needs is that pointer: use $v^2$.

Indeed.

To me it says that OP rejects plotting $v^2$ vs. $x$ and is seeking an alternate way.

If the OP is satisfied that they can "derive" $v = v_0 + at$ by plotting $v \ vs \ t$ and observing the slope then how can they reject "deriving" $v^2 = v_0^2 + 2 a x$ by plotting $v^2 \ vs \ x$ and observing the slope?

 

[brotherbobby]

I am the OP of this thread, sorry coming in almost two days late. I have read some of the posts of the thread and the best way to clear the doubts is to show you what I have been able to do so far. But first I must start with the problem statement.

Statement of the problem : The third of the well known equations of constant acceleration kinematics relates velocity to displacement thus : $\boxed{(3).\; v^2(x) = v^2_0+2a_0(x-x_0)}$. Please see the figure for the symbol details.

I have been able to derive this relation (3) using both algebra and calculus (separately). The latter involved noting that the (constant) acceleration $a_0 = v\dfrac{dv}{dx}$, a point you'd sometimes miss in the older textbooks. So far so good.

Additionally, I have been able to derive the two earlier equations using graphs, namely $(1)\; v(t)=v_0+a_0t,\quad (2)\; x(t)=x_0+v_0t+\dfrac{1}{2}a_0t^2$. Let me give you the details in brief.

(1) Let's imagine the graph alongside was drawn from data. We are required to find the equation from it. Since the slope of a $v-t$ graph is the acceleration and that is a constant here $(=a_0)$, we can take a point on the graph $(t,v(t))$. From the equation of a line, $y=mx+c\Rightarrow \boxed{v(t)=v_0+a_0t}\quad(1)$, where the $y$ intercept of the line $c = v_0$ here.

(2) For (2), the graph is the same $v-t$, with the points laballed. The net displacement $\Delta x = x-x_0$ is the area under the $v-t$ curve. Which is the area of the trapezium OCADB. The area of a trapezium is $\small{\dfrac{1}{2}\times(\text{sum of parallel sides})\times(\text{height}) = \dfrac{1}{2}(v_0+v(t))\times t= \dfrac{1}{2}(2v_0+a_0t)t=v_0t+\dfrac{1}{2}a_0t^2\Rightarrow \boxed{x=x_0+v_0t+\dfrac{1}{2}a_0t^2}\quad (2)}$

(3) I follow @Dale 's suggestion (post #5, above), with a slight change - plot $v^2 - (x-x_0)$. I use the values, $v_0=2, a_0=3, x_0=4$, and plot $v^2(x)=4+6(\Delta x)$ in the graph alongside. Of course $x-4>0$ and $v^2 = y>0$. Clearly, $v^2$ varies linearly with $x$ and $v_0^2=4$. Needless to say, the formula to be derived is being used to plot this graph, imagining that this is not the case and raw data was instead available. That aside, how does one show that the slope of this line is equal to $2a_0$? The slope $m = \dfrac{10-4}{1-0} = 6$. Assuming from the raw data the value of the acceleration can be obtained $(a_0 = 3)$, we can say that the slope turns out to be $2a_0$. In that case, $\dfrac{v^2-v_0^2}{\Delta x}=2a_0\Rightarrow\boxed{v^2=v_0^2+2a_0\Delta x}\quad(3)$.

I don't know if there's a better way. It just feels contrived and less convincing than deriving the earlier equations.

 

[Dale]

It just feels contrived and less convincing than deriving the earlier equations.

This one is no less convincing than the others. Indeed as @pbuk said, this is procedurally exactly the same as what you did for the first equation $v=v_0+a t$.

 

[A.T.]

I don't know if there's a better way.

Maybe not better in the sense of easier to unterstand, but fun:

You can get $v(x)$ by chaining the functions:
- $t(x)$ (eq. 2 solved for $t$)
- $v(t)$ (eq. 1)
yielding:
$v(x)=v(t(x))$
which you can visualize in a 3D plot with the axes: $x, t, v$ :
- plot $t(x)$ in the $xt$-plane and extrude the curve along the $v$ axis
- plot $v(t)$ in the $tv$-plane and extrude the curve along the $x$ axis
- $v(x)$ is then the 3D intersection of the two extruded graphs, projected onto the $xv$-plane