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.

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  • $\begingroup$ They are derived using simple algebra. That's all. $\endgroup$ Sep 16, 2020 at 14:35
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    $\begingroup$ Does this answer your question? How to obtain the distance traversed by a free falling body equation? $\endgroup$ Sep 16, 2020 at 14:43
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    $\begingroup$ The question asked for mnemonics and ways to understand each of the SUVAT relations, the linked question derives two of them and just mentions loads of identities related to the core two at the end but the question here is about a way to think about the ones just referenced in the other question and how they are derived also, this is hardly a duplicate. $\endgroup$
    – bolbteppa
    Sep 16, 2020 at 15:37

1 Answer 1


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:

  1. Calculus lets you derive the first and third;
  2. The fourth is just a trivial re-write of the third using the first;
  3. The second is a rewrite of the third with $t$ eliminated (using the first);
  4. 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

  1. No S: $v = u + at$
  2. No U: $s = vt - \frac{1}{2} at^2$
  3. No V: $s = ut + \frac{1}{2} at^2$
  4. No A: $s = \frac{u+v}{2}t$
  5. No T: $v^2 = u^2 + 2as$.
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    $\begingroup$ Wow this is really in-depth. Thank you heaps for the explanation! Really appreciate it! You explanation has really broadened my knowledge by a lot. Again, thank you so much for the clear and concise clarification. I will accept it as an answer once i get 15 reps on this stack. Stay safe. :) $\endgroup$
    – Munchies
    Sep 16, 2020 at 15:19

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