The Constant-Acceleration Equations gave you the constant-acceleration equations as tools. Velocity and Position from Acceleration lets you derive them in a few lines, and the derivation shows exactly when they are allowed.
Why it works
Set $a$ equal to a constant and integrate. $$v(t) = v_0 + \int_0^t a\,dt' = v_0 + at.$$ Integrate again: $$x(t) = x_0 + \int_0^t (v_0 + at')\,dt' = x_0 + v_0t + \tfrac{1}{2}at^2.$$ For the third equation, use the chain rule: $a = \dfrac{dv}{dt} = \dfrac{dv}{dx}\dfrac{dx}{dt} = v\dfrac{dv}{dx}$. Then $\int_{v_0}^{v} v\,dv = \int_{x_0}^{x} a\,dx$ gives $\tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 = a(x - x_0)$, or $$v^2 = v_0^2 + 2a\,\Delta x.$$
Key idea
$v = v_0 + at$, $x = x_0 + v_0t + \tfrac{1}{2}at^2$ and $v^2 = v_0^2 + 2a\Delta x$ hold only when the acceleration is constant, because pulling $a$ out of each integral assumed it never changes. When $a$ depends on time, position or velocity, integrate the real $a$ instead.
Worked example: Using the equation without time
A cart moving at $6\ \text{m/s}$ speeds up at a constant $2\ \text{m/s}^2$ over $16\ \text{m}$. How fast is it moving then?
- The acceleration is constant, so $v^2 = v_0^2 + 2a\Delta x$ applies.
- $v^2 = 36 + 2(2)(16) = 100\ \text{m}^2/\text{s}^2$.
- $v = 10\ \text{m/s}$.
Answer: $10\ \text{m/s}$.
Worked example: When the shortcut fails
A sled starts from rest with $a(t) = 3t$ (m/s², s). How far does it go in 2 s? Compare with $\tfrac{1}{2}at^2$ using $a(2)$.
- The acceleration grows with time, so integrate: $v(t) = \tfrac{3}{2}t^2$ and $x(t) = \tfrac{1}{2}t^3$.
- $x(2) = \tfrac{1}{2}(8) = 4\ \text{m}$.
- The shortcut with $a(2) = 6\ \text{m/s}^2$ gives $\tfrac{1}{2}(6)(2)^2 = 12\ \text{m}$, three times too far: it pretends the sled had its final acceleration all along.
Answer: $4\ \text{m}$ (the constant-acceleration formula would wrongly give 12 m).
Common mistake
Using the constant-acceleration equations for a spring, air drag, a rocket burn or any force that changes. They will give a number, but a wrong one. Check before you start: is the net force, and so the acceleration, the same at every moment of the motion? If not, write $a(t)$, $a(x)$ or $a(v)$ and integrate.
One more equation falls out of the area picture: under a straight $v$–$t$ line, the area is a trapezoid, so $\Delta x = \tfrac{1}{2}(v_0 + v)t$. The average velocity is the plain average of the starting and final velocities, but only because the graph is a straight line. For a curved $v$–$t$ graph, the average velocity is $\dfrac{1}{t}\int_0^t v\,dt'$ instead.
- Free fall without air resistance: $a = -g$, constant. The equations apply.
- A block sliding down a straight frictionless ramp: $a = g\sin\theta$, constant. They apply.
- A mass on a spring: $a = -(k/m)x$ changes with position. They don't.
- A falling raindrop with drag: $a$ shrinks as it speeds up. They don't.