nasu said:
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.