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🔭 Kinematics

Acceleration Calculator

From the initial and final velocity and the time, find the acceleration with a = (v − u)/t, the distance covered, and the equivalent g-force — using the SUVAT equations of motion.

Acceleration
Distance covered
g-force
SUVAT
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Acceleration — Quick answer

Acceleration is how fast velocity changes: the change in speed divided by the time. Distance follows from the average speed.

a = (v − u) / t · v = u + a·t
s = (u + v)/2 · t = u·t + ½·a·t² · g-force = a / 9.81

Worked example: 0 → 30 m/s in 10 s. a = (30−0)/10 = 3 m/s², distance 150 m, 0.31 g.

0 → 30 m/s in different times

TimeAccelerationDistance
5 s6.0 m/s²75 m
10 s3.0 m/s²150 m
15 s2.0 m/s²225 m

Used for: motion problems, vehicle 0–60, braking, g-force, SUVAT.

🔭 Acceleration Calculator

Enter the initial velocity, final velocity and time. Distance and g-force are worked out for you.

Acceleration
Distance covered
g-force
Change in velocity

⚠️ Assumes constant (uniform) acceleration — the standard SUVAT case. A negative result means deceleration (slowing down). For changing acceleration, use calculus or split the motion into uniform segments.

Acceleration measures how quickly velocity changes — the difference between the starting and finishing speed, spread over the time it took. That is the whole of a = (v − u)/t. Once you have it, the rest of the motion falls out of the SUVAT equations: the distance covered is just the average speed times the time, and dividing the acceleration by 9.81 gives the g-force, an intuitive way to feel how hard the change really is. A positive answer means speeding up; a negative one is braking.

Reviewed: June 19, 2026 · Author: Naveen P N, Founder — AI Calculator · Verified against: constant-acceleration kinematics (SUVAT).

The acceleration equations

Acceleration & final velocity
a = (v − u) / t · v = u + a·t
Distance covered
s = (u + v)/2 × t = u·t + ½·a·t²
g-force
g-force = a / 9.81

These hold whenever the acceleration is constant. The distance can be found two ways that always agree: the average-velocity form (u + v)/2 × t needs only the speeds and time, while s = ut + ½at² uses the acceleration directly. The g-force conversion turns a bare m/s² figure into something tangible — 3 m/s² is a brisk but comfortable 0.31 g, whereas an emergency stop can briefly hit 1 g or more.

Worked example — a car pulling away

Scenario: A car accelerates from rest to 30 m/s (about 108 km/h) in 10 seconds.

Acceleration
a = (30 − 0) / 10 = 3 m/s² (≈ 0.31 g)
Distance
s = (0 + 30)/2 × 10 = 150 m

The car accelerates at a steady 3 m/s² and covers 150 m reaching 30 m/s. Do it in half the time — 5 s — and the acceleration doubles to 6 m/s² (0.61 g) but the distance halves to 75 m, because the car spends less time at lower speeds. Slamming the brakes to stop from 30 m/s in 3 s would be a deceleration of −10 m/s², just over 1 g — firmly into "hold on" territory.

Frequently Asked Questions

How do you calculate acceleration?

a = (v − u)/t — change in velocity over time. 0→30 m/s in 10 s = 3 m/s². Negative means slowing.

What are the units of acceleration?

m/s² — velocity changes that many m/s each second. Also g (÷9.81) for big accelerations.

Speed vs velocity vs acceleration?

Speed = how fast; velocity = speed + direction; acceleration = rate velocity changes. Turning is acceleration too.

How to find distance from acceleration?

s = ut + ½at², or s = (u+v)/2 × t. 0→30 m/s in 10 s at 3 m/s² → 150 m.

What is g-force?

Acceleration as multiples of 9.81 m/s². 3 m/s² ≈ 0.31 g. Pilots pull up to ~9 g.

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