The kinematic equations
What you'll learn
Derive the constant-acceleration equations from one picture: the area under a straight v–t line.
Most textbooks hand you the kinematic equations as a list to memorize. You're going to read them off a picture instead — because when acceleration is constant, the v–t graph is a straight line, and everything about straight lines is easy.
The setup
Constant acceleration a, starting velocity v₀. The velocity line is:
That's not a new fact — it's just "slope × run": start at v₀ and add a for every second. First equation done.
Displacement is the area
How far did you go by time t? The area under the velocity line — and that area splits into two friendly shapes:
The rectangle is the distance you'd cover just coasting at v₀. The triangle — base t, height at — is the extra distance the acceleration adds:
The famous ½ isn't mysterious. It's the area of a triangle.
The shortcut without time
Sometimes you know speeds and distance but not the time ("how long a runway does a plane need to reach takeoff speed?"). Solve v = v₀ + at for t, plug it into the area formula, and t cancels:
Worth checking the algebra once by hand — after that, it's yours.
The toolkit
| Equation | Use it when |
|---|---|
| v = v₀ + at | you don't care about distance |
| x = v₀t + ½at² | you don't know the final velocity |
| v² = v₀² + 2aΔx | you don't know the time |
Three equations, one straight line, zero memorization: the first is the line's slope, the second is the area under it, the third is the two combined with t eliminated.
Why this matters
Constant acceleration isn't a toy case — it's gravity. Everything in the next module (free fall, projectiles) is these three equations with a = g pointed down.
Check your understanding
Question 1 of 2
In x = v₀t + ½at², where does the ½ come from?