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.

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.

brotherbobby said:

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.

Dale said:

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 . . .

brotherbobby said:

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

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

Dale said:

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

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 said:

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?

A.T. said:

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.

A.T. said:

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 said:

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

Indeed.

kuruman said:

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 said:

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##.

brotherbobby said:

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