How are the SUVAT equations derived? I hope everyone is doing well and staying safe. So instead of simply memorizing the SUVAT equations, I wanted to find out how the equations are derived to broaden my knowledge. I'm currently a high school student who is completing university-level mathematics and somehow ended up doing questions on SUVAT and constant accelerations (haha). Are there any mnemonics or easy way to memorize/understand the equations. Regardless, I am very keen to know how the SUVAT equations are derived.

 A: If the acceleration $a$ is constant, then from the calculus definition of acceleration as the time derivative of velocity
$$\frac{dv}{dt} = a$$
we see that integrating the left-hand side with respect to $t$ gives
$$\int_0^t \frac{dv}{dt} dt = \int_{v(0)}^{v(t)} dv = v(t) - v(0) = v - u,$$
where we set the initial time as $t = 0$ and initial velocity as $v(0) = u$, while on the right-hand side we have
$$\int_0^t a dt = a \int_0^t dt = at$$
so
$$v = u + at.$$
This is the first equation. Using the definition of velocity as the time derivative of position
$$
\frac{dx}{dt} = v$$
we see that integrating the left-hand side gives
$$ \int_0^t \frac{dx}{dt} dt = \int_{x(0)}^{x(t)} dx = x(t) - x(0) = s(t)$$
where we set $s(t) = x(t) - x(0)$ as the displacement, and integrating the right-hand side gives
$$
\int_0^t v(t) dt = \int_0^t [u + at] dt = u t + \frac{1}{2} at^2$$
so that
$$s(t) = u t + \frac{1}{2} at^2.$$
This is the third equation. This is expressed in terms of $u$, we can express it in terms of the final velocity $v$ by taking $v = u + at$ and setting $u = v - at$ to find
$$s(t) = ut + \frac{1}{2} a t^2 = (v - at)t + \frac{1}{2} a t^2 = vt - at^2 + \frac{1}{2} a t^2 = vt - \frac{1}{2} a t^2 .$$
This is the fourth equation. We can eliminate $t$ from the third equation $s = ut + \frac{1}{2} at^2$ by using the first equation $v = u + at$ to solve for $t = \frac{v-u}{a}$ and then use this in the third equation
$$s = ut + \frac{1}{2} at^2 = u \frac{v-u}{a} + \frac{1}{2} a(\frac{v-u}{a})^2 = u \frac{v-u}{a} + \frac{(v-u)^2}{2a} $$
reduces to
$$2a s = 2u(v-u)+ (v-u)^2 = 2uv-2u^2+ v^2 - 2vu+u^2 = -u^2+ v^2  $$
or
$$v^2 = u^2 + 2a s.$$
This is the second equation. We can eliminate $a$ from the third equation $s = ut + \frac{1}{2} at^2$ by using the first equation $v = u + at$ to solve for $a = \frac{v-u}{t}$ so that
$$s = ut + \frac{1}{2} at^2  = ut + \frac{1}{2} \frac{v-u}{t} t^2 = ut + \frac{1}{2} (v-u) t =  \frac{1}{2} (u+v) t .$$
This is the fifth equation. Summarizing:

*

*Calculus lets you derive the first and third;

*The fourth is just a trivial re-write of the third using the first;

*The second is a rewrite of the third with $t$ eliminated (using the first);

*The fifth is a rewrite of the third with $a$ eliminated (using the first).

Thus, using calculus to get the first and the third from the assumption of constant acceleration is the key.
Things naturally change for non-constant acceleration, we can no longer do the step $\int_0^t a dt = a \int_0^t dt$ we have $\int_0^t a(t) dt \neq a(t) \int_0^t dt$ i.e. $\int_0^t a(t) dt$ might not even be something we can compute or we might not even know $a(t)$ as a function of $t$ (for example in the Harmonic oscillator problem $F = - kx$, what is $a(t)$ based on this information? We are forced from $F = ma$ i.e. $-kx = m \ddot{x}$ with $a = \ddot{x}$ to solve the 'second order linear differential equation' $\ddot{x} + k x = 0$ rather than working out a simpler integral, thus you see why constant acceleration problems are important).
You can think of SUVAT as saying

*

*No S: $v = u + at$

*No U: $s = vt - \frac{1}{2} at^2$

*No V: $s = ut + \frac{1}{2} at^2$

*No A: $s = \frac{u+v}{2}t$

*No T: $v^2 = u^2 + 2as$.

