Velocity changes at a steady rate whenever the acceleration is constant: a car pulling away from a stop sign, a train braking into a station, a cyclist coasting down an even slope. The equation v = u + at says the velocity after a time t is the starting velocity plus the acceleration times that time. This calculator solves it for any of the four quantities, which makes it handy for questions like “how fast after 6 seconds?” or “how long until it stops?”. It also charts how the velocity changes over the interval.
How to use the velocity calculator
- Under Solve for, choose Final velocity, Initial velocity, Acceleration or Time.
- Enter the other three: Final velocity (v), Initial velocity (u), Acceleration (a) and Time (t). Velocities accept m/s, km/h, mph, ft/s and knots. Acceleration accepts m/s², ft/s², g, km/h per second and mph per second.
- Choose an output unit under Show the result in.
- Read the result with its conversions, the velocity change Δv, the displacement covered during the interval (plus the distance traveled if the object turns around) and a line chart of velocity against time, plotted in your velocity unit.
The v = u + at formula
This is just the definition of acceleration, the change in velocity per unit time, rearranged. Velocities and acceleration are signed along the direction you choose as positive. Time is always positive. The displacement in the result comes from s = ½(u + v)t, which holds for the same constant-acceleration motion.
Worked example: pulling away from rest
Given: u = 0, a = 3 m/s², t = 6 s.
v = 0 + 3 × 6 = 18 m/s (64.8 km/h or 40.26 mph)
Displacement in that time: ½ × (0 + 18) × 6 = 54 m
The chart is a straight line climbing from 0 to 18 m/s. Its slope, 3 m/s per second, is the acceleration, and the triangle under it has an area of 54 m, the displacement. A steeper line means harder acceleration. A flat line means none.
Braking and time to stop
Stopping problems are where the solve-for-time option earns its keep. Take a car at 90 km/h (25 m/s) braking at 6 m/s². Call the direction of travel positive, so the acceleration is −6 m/s², and set the final velocity to 0.
t = (v − u) ÷ a = (0 − 25) ÷ (−6) = 4.17 s
Distance covered while braking: ½ × (25 + 0) × 4.167 = 52.08 m
Keep Final velocity as the unknown instead and you can watch the speed fall. After 2 s of braking the car is still doing 25 − 6 × 2 = 13 m/s (46.8 km/h) and has already used 38 m of road.
The signs matter. If you enter the braking acceleration as +6 m/s² while asking when v reaches 0, the calculator stops with an error: a positive acceleration would only make a forward-moving car go faster, so the velocity could never drop to zero in positive time. In general, solving for time only works when (v − u) and a have the same sign. Zero acceleration is rejected too, because the velocity would never change.
What “deceleration” really means
Everyday speech treats deceleration as negative acceleration, but physics cares about direction. An object slows down when its acceleration points opposite to its velocity. A puck sliding in the negative direction at −12 m/s slows down under a positive acceleration of +3 m/s². After 4 s its velocity is 0.
If the acceleration keeps acting after the object stops, the velocity changes sign and the object comes back. A ball rolled up a slope at u = 6 m/s with a = −2 m/s² has v = −4 m/s after t = 5 s. The calculator notes the reversal at t = 3 s, where the chart line crosses the time axis. The listed displacement, 5 m, is net: the ball went 9 m up and 4 m back down, 13 m of travel in all, which the calculator lists separately as the distance traveled.
To work with distances instead of times, use the final velocity calculator for v² = u² + 2as. To track position along the way, the s = ut + ½at² calculator plots the full position curve.
Frequently asked questions
How do I find the starting velocity?
Choose Initial velocity under Solve for and use u = v − at. A car doing 30 m/s after accelerating at 2 m/s² for 5 s started at 30 − 2 × 5 = 20 m/s (72 km/h), and it covered 125 m in those 5 seconds.
Is deceleration the same as negative acceleration?
Not always. Deceleration means the acceleration points opposite to the velocity, so the object slows down. If you move in the negative direction, slowing down requires a positive acceleration. The sign alone depends on which direction you chose as positive.
Why does the calculator refuse to solve for time?
Time must be positive, so the velocity change (v − u) and the acceleration must have the same sign. It also rejects zero acceleration, since the velocity then never changes, and equal initial and final velocities, which give zero elapsed time.
What is the difference between v = u + at and v² = u² + 2as?
Both describe constant acceleration. Use v = u + at when you know the time. Use v² = u² + 2as when you know the distance but not the time, as in stopping-distance problems.