Acceleration Calculator (SUVAT equations)
Enter any three of initial speed, final speed, acceleration, time, and distance, and get the other two from the SUVAT equations.
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Braking distance by speed
Distance to a full stop at a steady 0.7 g (6.86 m/s²), which is roughly firm braking on dry tarmac. This is the pure physics only — it excludes reaction time, which adds a further second or so of travel before the brakes are even applied.
| Speed | mph | Braking distance | Feet | Time to stop |
|---|---|---|---|---|
| 30 km/h | 18.6 | 5.058 m | 16.59 | 1.21 s |
| 50 km/h | 31.1 | 14.05 m | 46.1 | 2.02 s |
| 70 km/h | 43.5 | 27.54 m | 90.35 | 2.83 s |
| 90 km/h | 55.9 | 45.52 m | 149.4 | 3.64 s |
| 110 km/h | 68.4 | 68 m | 223.1 | 4.45 s |
| 130 km/h | 80.8 | 94.98 m | 311.6 | 5.26 s |
Doubling the speed quadruples the distance — 100 km/h needs four times the room of 50 km/h, not twice.
Acceleration and the SUVAT equations
Acceleration is the rate at which velocity changes. When it is constant, five quantities describe the motion completely — initial velocity u, final velocity v, acceleration a, time t, and distance s — and any three of them determine the other two. The four standard relationships between them are known as the SUVAT equations.
v = u + a·t · s = u·t + ½a·t² · v² = u² + 2a·s · s = (u + v)·t / 2
Rather than making you pick the right equation, this calculator takes whichever three values you know and selects the pair that fits. Leave the two you want blank.
Worked example
A car accelerating from rest at 2 m/s² for 5 seconds: v = 0 + 2 × 5 = 10 m/s, and s = 0 × 5 + ½ × 2 × 25 = 25 m. Enter u = 0, a = 2, t = 5 above and those are the two values that come back. Now suppose you instead know it covered 25 m and finished at 10 m/s — enter those two with u = 0 and you get the same acceleration and time back out.
Negative acceleration is not the same as reversing
A negative value of a simply means the acceleration points backwards along whichever direction you called positive. Braking from 30 m/s to a stop over 60 m gives a = −7.5 m/s², and the object is still travelling forwards the whole time. It only genuinely reverses if the final velocity itself comes out negative — which is exactly what happens to something thrown straight up once it passes the top of its arc.
Why braking distance grows so fast
The v² = u² + 2as relationship is the one worth internalising, because it says stopping distance scales with the square of speed. Going from 50 to 100 km/h does not double the distance needed, it quadruples it. The table above is that fact made concrete, and it is the reason speed limits have such a disproportionate effect on collision severity — see alsokinetic energy, which scales the same way.
When constant acceleration is the wrong model
These equations assume a is genuinely constant. Real vehicles accelerate hardest at low speed and tail off; real braking varies with tyre and road conditions; and anything falling far enough is limited byair resistance rather than by gravity. For a constant push over a modest interval they are exact, and for most rough estimates they are close enough.