Many motion problems give you a distance but no time. How fast is a sprinter after a 50 m run-up? How much road does a car need to stop? What speed could leave skid marks this long? The equation v² = u² + 2as answers all of these directly, because it links velocity, acceleration and displacement without time. This calculator rearranges it for whichever of the four quantities is missing, then adds the time taken and, when an object is braking, its stopping distance.
How to use the final velocity calculator
- Pick the unknown under Solve for: Final velocity, Initial velocity, Acceleration or Displacement.
- Enter the remaining values in Final velocity (v), Initial velocity (u), Acceleration (a) and Displacement (s), each with its own unit menu.
- Decide how the answer is displayed with Show the result in.
- The result shows the answer with conversions, the time taken (assuming the object does not reverse, with a note when it turns around or passes the same point twice), the change in speed and, if the acceleration opposes the initial velocity and the object has not yet stopped, the full stopping distance at that deceleration. When the acceleration is about 1 g, as for a throw, that line is labeled as the peak height or turnaround distance.
The time-free equation
a = (v2 − u2) ÷ 2s · s = (v2 − u2) ÷ 2a
It follows from eliminating t between v = u + at and s = ½(u + v)t. Multiply both sides by half the mass and it becomes the work–energy theorem: the change in kinetic energy, ½mv² − ½mu², equals the work done, mas. Take the direction of the displacement as positive. Then a positive a means speeding up along the way and a negative a means slowing down.
Worked example
Given: starting from rest (u = 0), a = 4 m/s² over s = 50 m.
v = √(0² + 2 × 4 × 50) = √400 = 20 m/s (72 km/h or 44.74 mph)
Time taken: (20 − 0) ÷ 4 = 5 s
Stopping distance and a braking table
Set v = 0 and the equation gives the distance needed to stop from speed u:
Distance grows with the square of speed, so doubling your speed quadruples the braking distance. The table uses a deceleration of 7 m/s², about 0.71 g. That is a firm stop on dry pavement. Wet or icy roads, worn tires and heavy loads all lower it.
| Speed | Braking distance | Braking time |
|---|---|---|
| 20 mph | 5.71 m (18.7 ft) | 1.28 s |
| 30 mph | 12.85 m (42.1 ft) | 1.92 s |
| 40 mph | 22.84 m (74.9 ft) | 2.55 s |
| 50 mph | 35.69 m (117.1 ft) | 3.19 s |
| 60 mph | 51.39 m (168.6 ft) | 3.83 s |
| 70 mph | 69.95 m (229.5 ft) | 4.47 s |
| 80 mph | 91.36 m (299.7 ft) | 5.11 s |
These figures cover the braking phase only. Before the brakes engage, the car keeps rolling at full speed for the driver’s reaction time. At 60 mph, a 1.5 s reaction adds another 132 ft, which you can check with the s = vt calculator.
Running the formula backward gives a rough speed estimate from skid marks: 30 m of skid at 7 m/s² means u = √(2 × 7 × 30) ≈ 20.5 m/s, about 45.8 mph.
Inputs the calculator rejects
Square roots of negative numbers have no real value, so the calculator stops when u² + 2as (or v² − 2as, when solving for u) comes out negative. That is a physical message, not a glitch. A cyclist braking from 8 m/s at 2 m/s² stops after 16 m. Asking for the speed after 25 m gives 8² + 2 × (−2) × 25 = −36 m²/s², because the bike is already stopped before covering that distance. It also refuses s = 0 when solving for acceleration, since speed cannot change over no distance. When solving for displacement it refuses a = 0, because the speed then never changes.
A speed, not a velocity
Because only v² appears, the result is a magnitude. Drop a stone from a ledge with up as positive (u = 0, a = −9.80665 m/s², s = −20 m) and the calculator returns 19.81 m/s. You supply the direction: the stone is moving down, so its velocity is −19.81 m/s. The time taken, 2.02 s, comes from the same no-reversal assumption. A ball thrown up at 10 m/s (a = −9.80665 m/s²) is moving at 7.80 m/s when it is 2 m above your hand, on the way up and again on the way down. The time shown, 0.225 s, is for the first pass only; the result notes that the ball is back at that height at 1.81 s and lists its peak height, 5.10 m. To follow an object through its turnaround, use the velocity calculator or the SUVAT calculator.
Frequently asked questions
How do I calculate braking distance?
Set the final velocity to 0 and solve for displacement: s = u² ÷ (2|a|). From 60 mph (26.82 m/s) at a deceleration of 7 m/s² the car needs about 51.4 m, or 169 ft, of braking. Add the distance covered during the driver's reaction time to get the full stopping distance.
Why is the answer always positive?
The equation contains only v², and every square root the calculator takes is positive, so it returns the speed. The direction has to come from the situation. For motion that never turns around, the velocity has the same sign as the displacement.
Why do I get an error saying no real velocity fits?
If u² + 2as is negative, the object stops, and under a constant acceleration turns back, before it can cover the displacement you entered. For example, from 8 m/s with a = −2 m/s² it stops after 16 m, so its speed after 25 m has no answer.
How fast was a car going if it left 30 m of skid marks?
Solve for initial velocity with v = 0, s = 30 m and a = −7 m/s²: u = √(2 × 7 × 30) ≈ 20.5 m/s, or about 45.8 mph. Real crash reconstruction uses a measured drag factor for the road surface instead of an assumed deceleration.