Equations of motion

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2023-01-01

A quick derivation of kinematic equations

Written by: Ned

When I took introductory physics during my math degree, I’d already taken differential equations and vector calculus. I took the time to learn the units for everything, and the math tended to follow pretty easily from that by applying the transformations needed to reach the correct units at the end. It’s been a while since I’ve looked at any physics, but integrating acceleration with respect to time yields veloctity, and integrating velocity yields the position. Today is just a simple illustration of that to derive

$v(t) = \int a(t) \mathrm{d}t$

$v(t) = at + c$, where c is the initial velocity. By convention c is usually written as $v_0$

Thus, $v(t) = at + v_0$

$x(t) = \int v(t) \mathrm{d}t = \int at + v_0 \mathrm{d}t$

$x(t) = \frac{1}{2}at^2 + v_0t + c$, where c here is the initial position. By convention that’s typically written as $x_0$, its value is often 0 so it’s generally omitted.

Thus, $x(t) = x_0 + v_0t + \frac{1}{2}at^2$

Other kinematic equations

The first I’ll discuss occurs when acceleration is constant, so the velocity graph will be a line with the value of its slope equal to the acceleration $a$. That means the definite integral, which is how we calculate displacement, evaluated between $t_0$ and $t_f$ will be the average velocity multiplied by the time. That yields:

$x(t_f) = \frac{v_0 + v_f}{2}\Delta t = \frac{v_0 + v_f}{2}(t_f - t_0)$

For the last of the usual kinematic equations, we take the equations we derived from $a(t)$ for $x(t)$ and $v(t)$ and use $x_0 = 0$, let $v = v_f$ then combine to eliminate t and solve for $v_f^2$

$x = 0 + v_0t + \frac 1 2 at^2$

$v_f = v = v_0 + at$

Then: $t = \frac{v_f - v_0}{a}$

Substitute into the first equation to get:

$x = \frac{v_0(v_f - v_0)}{a} + \frac{a}{2} (\frac{v_f - v_0}{a})^2$

Simplify: $x = \frac{v_fv_0 - v_0^2}{a} + \frac{a}{2} \frac{v_f^2 -2v_fv_0 + v_0^2}{a^2}$

$2x = \frac{2v_fv_0 - 2v_0^2}{a} + \frac{v_f^2 -2v_fv_0 + v_0^2}{a}$

$2xa = 2v_fv_0 - 2v_0^2 + v_f^2 -2v_fv_0 + v_0^2$

$2xa = v_f^2 - v_0^2$

Finally,

$v_f^2 = v_0^2 + 2xa$

Techniques and Observations

Nothing fancy here, just some polynomial integration and evaluations at particular values to determine other equations. Physics problems are a great reminder to add the constant term though since it often matters. Otherwise, there’s a little bit of averaging and some brief discussion of how slopes, derivatives, and areas relate. The last equation uses some substition and initial values to apply specific conditions onto a system of equations to derive a particular formula for those conditions.