The constant-acceleration formulae you've used since AS level aren't a separate set of rules — they fall straight out of integrating a constant acceleration, and calculus proves them in a couple of short steps.
12 min read3 worked examples
Deriving v = u + at
When acceleration is constant, integrating $a=\dfrac{dv}{dt}$ recovers the first SUVAT formula directly.
ProofA particle has constant acceleration a and initial velocity u
1
$v = \displaystyle\int a\,dt = at+c$, since $a$ is constant.
Integrating a constant with respect to t.
2
At $t=0$, the velocity is $u$ (its initial value): $u = a(0)+c$, so $c=u$.
Use the initial condition to fix the constant.
3
Hence $v = u+at$.
Substitute c = u back in.
▪ QED
Deriving s = ut + ½at²
Integrating the formula just proved, $v=u+at$, recovers the second SUVAT formula in exactly the same way.
At $t=0$, the position is $s_0$: $s_0 = 0+0+k$, so $k=s_0$.
Use the initial position to fix k.
3
Hence $s = s_0+ut+\tfrac12at^2$ — or simply $s=ut+\tfrac12at^2$ when the particle starts at the origin, $s_0=0$.
The familiar SUVAT formula is the special case s₀ = 0.
▪ QED
Worked example 1
Deriving v² = u² + 2asChallenge
Using $a=\dfrac{dv}{dt}$ and the chain rule, show that for constant acceleration $a$, $v^2 = u^2+2as$, where $s$ is measured from the point where the velocity is $u$.
MethodRewrite dv/dt in terms of ds using the chain rule, then integrate with respect to s
1
By the chain rule, $a = \dfrac{dv}{dt} = \dfrac{dv}{ds}\cdot\dfrac{ds}{dt} = v\dfrac{dv}{ds}$.
A particle starts at position $s_0=4\ \text{m}$ with velocity $u=3\ \text{m s}^{-1}$ and constant acceleration $a=2\ \text{m s}^{-2}$. Find its position at $t=5$ seconds, using the general form derived above.
MethodSubstitute directly into s = s₀ + ut + ½at²
1
$s(5) = 4 + 3(5) + \tfrac12(2)(25) = 4+15+25$.
Substitute the given values.
2
$s(5) = 44\ \text{m}$.
Add the three terms.
▪ solved
Worked example 3
A second route to v² = u² + 2as
By substituting $t=\dfrac{v-u}{a}$ (rearranged from $v=u+at$) into $s=ut+\tfrac12at^2$, show algebraically that $v^2=u^2+2as$ — giving a second proof, independent of the calculus argument in example 1.
Example 1 derived $v^2=u^2+2as$ from the calculus definition of acceleration directly. This example reaches the identical formula by pure algebra, starting only from the two formulae already proved. Agreement between two genuinely different methods is strong evidence the result is correct.
Common mistake
Assuming the constant of integration is automatically zero. It represents whatever the initial condition of the problem actually is — the constant is $u$ when integrating $a$, and $s_0$ when integrating $v$, and is only zero in the special case that the particle starts at rest, or starts at the origin.
Check yourself
A particle has constant acceleration $a=-4\ \text{m s}^{-2}$ and velocity $u=10\ \text{m s}^{-1}$ at $t=0$. Using $v=u+at$, find the time at which the particle is instantaneously at rest, then use $s=ut+\tfrac12at^2$ to find its displacement at that time. (Hint: this is just the two proved formulae applied numerically.)