How this instrument works
Displacement is the net change in position: the straight-line vector from where a body started to where it finished, not the length of whatever path it wandered along the way. Under constant acceleration it splits into two honest pieces. The term u·t is the ground that would have been covered had the opening speed simply been held; the term ½·a·t² is everything the acceleration added on top. Drawn on a velocity-versus-time graph those pieces are a rectangle and the triangle perched on it, and the displacement is the area underneath.
The relation was settled long before anyone could measure it. Scholars at Merton College, Oxford, in the 1330s stated the mean speed theorem — a uniformly accelerating body travels as far as one moving steadily at its average speed — and around 1350 Nicole Oresme, in Paris, proved it by drawing precisely that triangle. Galileo supplied the measurement three centuries later, rolling bronze balls down a grooved inclined plane and finding that the ground covered in successive equal intervals ran 1, 3, 5, 7: the fingerprint of a length growing with the square of the time. He published it in 1638.
Two assumptions carry the formula, and both can fail quietly. Acceleration must hold constant across the whole interval, which rules out air-resisted falls, rockets lightening as they burn, and anything that stops accelerating partway. And the output is a displacement, not an odometer reading: hand the sheet a positive Initial velocity against a negative Acceleration and it will march straight past the turning point at t = −u ⁄ a and begin subtracting. A car at 30 m/s braking at 6 m/s² halts after 5 s and 75 m; ask for 10 s and the answer is 0 m, the formula having assumed the braking continued and hauled the car back to its starting line.
- Enter Initial velocity — leave it at 0 for a body starting from rest. The unit menu takes km/h, mph, or ft/s if that is how your figure arrived.
- Enter Acceleration in m/s², or switch that field to g0 to work in multiples of standard gravity, 9.80665 m/s².
- Set Time elapsed. Milliseconds, minutes, and hours are on the menu; the arithmetic runs in seconds regardless of what you type.
- Read Displacement, given to four decimals, and restate it in feet, kilometres, or miles without re-entering anything.
- Watch the signs: a positive velocity with a negative acceleration means the body is slowing, and the answer stays physical only up to the instant it stops.
Worked example — five seconds into the takeoff roll
Brakes off at the runway threshold. A loaded narrow-body airliner pushes down the roll at roughly 2 m/s², so set Initial velocity to 0, Acceleration to 2, and Time elapsed to 5. The first term vanishes — a body at rest contributes no u·t — leaving s = ½ × 2 × 5² = ½ × 2 × 25 = 25 metres exactly. Five full seconds of thrust, and the aircraft has shifted less than its own fuselage length.
The following five seconds are the ones that matter. At t = 10 s the same formula gives 100 m, so the second interval swallowed 75 m — three times the first, precisely the 1 : 3 ratio Galileo counted off on his inclined plane. That quadratic bite is why runway length is dictated by the closing seconds of a roll rather than its opening ones, and why an engine failure late in the sequence is the expensive kind.
Questions
Is displacement the same as distance travelled?
No. Displacement is the net change in position and carries a sign; distance is total path length and never decreases. The two agree only while motion keeps a single direction. Throw a ball straight up and catch it again and the displacement is zero while the distance is twice the peak height. This sheet reports displacement, so a body that reverses inside the interval will always show less than its odometer would.
What if the acceleration is not constant?
Then the formula does not apply, and no rearrangement rescues it — the derivation fixes a across the entire interval. The general statement is that displacement is the integral of velocity over time. For motion that changes its rate, such as a parachute opening, a rocket shedding propellant mass, or a car working up through its gears, break the run into stretches where a is near enough steady and sum the pieces, or integrate properly.
Why is my answer coming out negative?
Because the body finished behind where it began. That happens when Acceleration opposes Initial velocity for long enough to stop the motion and drive it back; the turning point sits at t = −u ⁄ a. Enter u = 30 m/s, a = −6 m/s², t = 10 s and the readout is 0 m — the car braked to rest at 5 s and 75 m, then, in the formula's tidy world, reversed over the identical stretch. Trim Time elapsed to the stopping time for the answer you meant.
What acceleration should I use for a dropped object?
Standard gravity, 9.80665 m/s², or pick g0 on the Acceleration menu and enter 1. With Initial velocity at 0, one second of fall covers 4.9033 m, two seconds 19.613 m, three seconds 44.130 m. Those figures assume vacuum; real drag starts trimming them visibly once a compact object passes about 20 m/s, which arrives after roughly two seconds.
How does this relate to the mean speed form?
They are one equation wearing two hats. Substituting v = u + a·t into s = ½(u + v)·t reproduces s = u·t + ½·a·t² term for term, so reach for whichever pair of quantities you actually measured. Knowing the opening and closing velocities but not the rate, the mean speed form is quicker; knowing the rate but never having recorded the final velocity, this instrument's form is the one that answers.
Does this work for motion in two dimensions?
Yes, one axis at a time. Displacement is a vector, and the relation holds independently for each component. Run the sheet once with the horizontal Initial velocity and Acceleration, once with the vertical pair, then combine the two answers with Pythagoras. For a thrown object the horizontal acceleration is 0 and the vertical is −9.80665 m/s², which is the whole of projectile motion in two passes.