Velocity Calculator — final velocity under constant acceleration

How it works

This is the first kinematic equation — it gives the velocity of an object moving under constant acceleration after a given time:

v = u + a·t

where u is the initial velocity, a is the (constant) acceleration and t is the elapsed time. Starting from speed u, each second the acceleration adds a to the velocity, so after t seconds the velocity has changed by a·t.

The acceleration can be negative, in which case the velocity decreases over time (and may even reverse). In SI units — m/s for velocity, m/s² for acceleration and seconds for time — the answer is in metres per second. This calculator is unit-agnostic, so any consistent set of units works.

Note the difference from the speed calculator: speed is distance ÷ time, an average rate over a journey. This finds the instantaneous velocity reached after accelerating for a set time, and it only holds while the acceleration stays constant.

Worked example

OpenStax's example starts at u = 70 m/s and decelerates at a = −1.5 m/s² for t = 40 s: v = 70 + (−1.5)(40) = 70 − 60 = 10 m/s. An object dropped from rest (u = 0) under gravity for 3 s reaches 0 + 9.8 × 3 = 29.4 m/s, and one starting at 5 m/s that speeds up at 2 m/s² for 10 s reaches 5 + 2 × 10 = 25 m/s.

Frequently asked questions

What is the formula for final velocity?

Under constant acceleration, v = u + a·t, where u is the starting velocity, a is the acceleration and t is the time. It is the first of the standard kinematic ("SUVAT") equations of motion.

What is the difference between this and the speed calculator?

Speed = distance ÷ time gives an average rate over a whole trip. This equation gives the velocity at the end of a period of constant acceleration, built from the starting velocity and how long the acceleration acts — not from a distance.

Does this work if the acceleration changes?

No. v = u + a·t assumes the acceleration is **constant** over the whole time. If the acceleration varies, split the motion into intervals where it is roughly constant, or use calculus. For free fall near Earth's surface, a ≈ 9.8 m/s² is a good constant.

Can the final velocity be negative?

Yes. If the acceleration opposes the motion long enough, the object stops and then moves backwards, giving a negative velocity. For example u = 10, a = −4, t = 3 gives v = 10 − 12 = −2 m/s.

What does a negative acceleration mean here?

It means the velocity is decreasing — the object is slowing down (or speeding up in the negative direction). Free fall taken as positive-downward uses a = +9.8 m/s²; an object thrown upward uses a = −9.8 m/s² until it falls back.

How is this related to acceleration and force?

Acceleration itself is (v − u) ÷ t, which this equation rearranges to solve for v. The acceleration usually comes from a force via Newton's second law (a = F ÷ m), so force → acceleration → velocity is the chain from cause to motion.