Displacement (Uniform Acceleration)

Also known as second kinematic equation · d = v₀t + ½at² · SUVAT s = ut + ½at²

d=v0t+12at2d = v_0 t + \tfrac{1}{2} a t^2

Worked example: Plane from rest, 2.5 m/s² for 30 s → 1125 m — press Try an example to run it live, then adjust anything.

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Displacement (Uniform Acceleration) explained

v0atd

When acceleration is constant, displacement has two parts: the distance you would cover at your initial velocity alone, plus the extra distance contributed by speeding up — and that extra grows with the square of time. A jet starting its takeoff roll from rest and holding 2 m/s² covers d = 0 + ½(2)(30²) = 900 m in 30 seconds, which is why runways are measured in kilometres.

The ½ appears because the acceleration term is built from the average of a speed that grows linearly from zero. Galileo uncovered the underlying pattern — distances in successive equal time intervals follow the odd numbers 1, 3, 5, 7 — by rolling bronze balls down inclined planes. Note that solving for t would mean solving a quadratic with potentially two positive roots, so this calculator rearranges only for d, v₀, and a, where the answer is always single-valued.

Displacement (Uniform Acceleration) formula

d=v0t+12at2d = v_0 t + \tfrac{1}{2} a t^2
Where
  • dd= Displacement (m)
  • v0v_0= Initial velocity (m/s)
  • aa= Acceleration (m/s²)
  • tt= Time (s)